Class 11 Physics

Newtonian Gravitation And Planetary Motion Short Answer Type Questions

WBBSE Class 11 Short Answer Questions on Gravitation

Question 1. At what height above the surface of the earth is the gravitational potential energy of a body equal to that on the surface of the moon? Assume that the mass of the earth is 80 times that of the moon and the radius of the earth is 4 times that of the moon. Given the radius of the earth = 6400 km.
Answer:

Given the radius of the earth = 6400 km.

Mass of the earth =M, mass of the moon = \(\frac{M}{80}\)

Radius of the earth, R = 6400 km and radius of the moon = \(\frac{R}{4}\)

The gravitational potential energy of a body of mass m at a distance r from the centre of the earth = –\(\frac{G M m}{r}\)

Again, gravitational potential energy on the surface of the moon

= \(-\frac{G \cdot \frac{M}{80} \cdot m}{\frac{R}{4}}=-\frac{G M m}{20 R}\)

Now, \(-\frac{G M m}{r}=-\frac{G M m}{20 R} \quad \text { or, } r=20 R\)

So, height above the surface of the earth = r – R = 20R – R = 19R = 19 x 6400 = 121600 km

Question 2. Does the motion of a satellite obey Kepler’s laws?
Answer:

Yes, it obeys. As the eccentricity of the elliptical orbit is very small, it can be treated as circular.

Question 3. The orbital velocity of a satellite in a circular orbit close to the earth is v₀. If the escape velocity of an object projected from the earth be v_e, then the ratio of these two velocities is

1. 1:1
2. √2:1
3. √3:1
4. 2:1

Answer:

Given

The orbital velocity of a satellite in a circular orbit close to the earth is v₀. If the escape velocity of an object projected from the earth be v_e,

⇒ \(\frac{v_e}{v_0}=\sqrt{\frac{2 g R}{\sqrt{g R}}}=\frac{\sqrt{2}}{1}\)

The option 2 is correct.

Question 4. Find expressions for the potential energy (V) and the kinetic energy (K) of the moon in the gravitational field of the earth. Hence find the total energy of the moon and state the significance of its negative sign.
Answer:

Let mass of the earth = M, mass of the moon = m, distance between centres of earth and moon = r, orbital speed of moon, v = \(\sqrt{\frac{G M}{r}}\)

So, kinetic energy of moon = \(K=\frac{1}{2} m v^2=\frac{G M m}{2 r}\)

Again, gravitational potential energy of moon = V = –\(\frac{G M m}{r}\)

So, total energy, E = K + V = \(\frac{G M m}{2r}\) – \(\frac{G M m}{r}\) = –\(\frac{G M m}{2r}\)

The significance of the negative sign is that, due to gravitational attraction between the moon and the earth, a closed system has formed. If \(\frac{G M m}{2r}\) amount of energy can be given to the moon from outside, the total energy of the moon will be zero, i.e., it will be free from earth’s attraction force.

Question 5. An artificial satellite of mass m is moving around the earth in an orbit of radius 2R. How much work is to be done to transfer the satellite to an orbit of radius 4i?? How will the potential energy of the satellite change? (R = Radius of the earth)
Answer:

Given

An artificial satellite of mass m is moving around the earth in an orbit of radius 2R.

If mass of the earth=M, potential energy of artificial satellite at distance 2R = \(\frac{G M m}{2R}\)

Total energy = –\(\frac{G M m}{2. 2R}\) = –\(\frac{G M m}{4R}\)

Potential energy at a distance 4R = –\(\frac{G M m}{4R}\)

Total energy = –\(\frac{G M m}{2.4 R}\) = –\(\frac{G M m}{8R}\)

So, work done = increase in energy = –\(\frac{G M m}{8R}\)– (-\(\frac{G M m}{4R}\))

= \(\frac{G M m}{4R}\)(1- \(\frac{1}{2}\)) = \(\frac{G M m}{8R}\)

Change in potential energy = –\(\frac{G M m}{4R}\)-(-\(\frac{G M m}{2R}\)) = \(\frac{G M m}{2R}\)(1-\(\frac{1}{2}\)) = \(\frac{G M m}{4R}\)

Key Concepts of Planetary Motion: Short Answers

Question 6. Suppose the gravitational force varies inversely as the nth power of the distance. Thus the time period of a planet moving in a circular orbit of radius r around the sun will be proportional to

1. \(r^{\frac{n+1}{2}}\)
2. \(r^n\)
3. \(r^{\frac{n-1}{2}}\)
4. \(r^{-n}\)

Answer:

Gravitational force = \(\frac{k}{r^n}\) [k is constant]

This gravitational force gives rise to centripetal force.

If the mass of the planet is m and its velocity is v, centripetal force = \(\frac{m v^2}{r}\)

So, \(\frac{m v^2}{r}=\frac{k}{r^n} \quad or, v=\sqrt{\frac{k}{m r^{n-1}}}\)

Time period, \(T=\frac{2 \pi r}{\nu}=2 \pi \sqrt{\frac{m}{k}} \sqrt{r^2 \cdot r^{n-1}}=\text { constant } \times r^{\frac{n+1}{2}}\)

∴ \(T \propto r^{\frac{n+1}{2}}\)

The option 1 is correct

Question 7. What are the characteristics of geostationary satellites and polar satellites? Why polar satellites are called “weather satellites”-Explain.
Answer:

Geostationary satellites are placed on the equatorial plane at a height of nearly 36000 km from the earth’s surface. Since the motion of such a satellite is the same as that of the diurnal motion of the earth, the satellite seems to be stationary with respect to the earth’s surface.

On the other hand, polar satellites are placed at a height of 800 km from the surface of the earth, aligned with the polar plane. To revolve around the earth once, a polar satellite takes about 2 hours.

Polar satellites are placed comparatively nearer to the earth’s surface and travel over a vast area while revolving. Thus they are suitable for observing the weather and hence, are called ‘weather satellites.’

Question 8. A planet is rotating around a heavy star along a circular path of radius R with time T. If the gravitational force of attraction between planet and star is proportional to \(R^{\frac{5}{2}}\), then T² ∝ R^x. What is the value of x?
Answer:

Given

A planet is rotating around a heavy star along a circular path of radius R with time T. If the gravitational force of attraction between planet and star is proportional to \(R^{\frac{5}{2}}\), then T² ∝ R^x.

Gravitational force =  \(k R^{\frac{5}{2}}\) [K = constant]

If the mass of the planet is m and its velocity is v, then centripetal force = \(\frac{m v^2}{R}\).

So, \(\frac{m v^2}{R}=k r^{-5 / 2} \quad or, v=\sqrt{\frac{k}{m} R^{-3 / 2}}\)

Time period, T = \(\frac{2 \pi R}{v}=2 \pi \sqrt{\frac{m}{k}} \sqrt{R^2 R^{3 / 2}}=\text { constant } \times \sqrt{R^{7 / 2}}\)

∴ \(T^2 \propto R^{7 / 2}\)

So, the value of \(x=\frac{7}{2}\)

Question 9. A man on earth can jump to a maximum height of 2m. To what maximum height man will be able to jump to another planet whose density is \(\frac{1}{3}\) of the density of earth and radius is \(\frac{1}{4}\) of the radius of earth?
Answer:

Given

A man on earth can jump to a maximum height of 2m.

In the case of Earth, \(g=\frac{4}{3} \pi {GR} \rho\)

In the case of the other planet, \(g^{\prime}=\frac{4}{3} \pi \mathrm{GR}^{\prime} \rho^{\prime}\)

∴ \(\frac{g}{g^{\prime}}=\frac{R}{R^{\prime}} \cdot \frac{\rho}{\rho^{\prime}}=\frac{R}{R / 4} \cdot \frac{\rho}{\rho / 3}=12\)

If the man jumps with the highest initial velocity u in both cases, u² = 2gh = 2g’h’

i.e., \(h^{\prime}=\frac{g}{g^{\prime}} h\) = 12 x 2 = 24m

Question 10. If the radius of the earth becomes half of the present radius, mass being constant, the weight of a body will be

1. Halved
2. One-fourth
3. Four times
4. Doubled

Answer: 4. Doubled

The option 4 is correct.

Newton’s Law of Gravitation Short Answer Questions

Question 11. An object is projected vertically upwards with a velocity u from the surface of the earth. Show that the maximum height reached by the object is h = \(\frac{u^2 R}{2 g R-u^2}\), where R is the radius of the earth. Calculate escape velocity from it.
Answer:

Given

An object is projected vertically upwards with a velocity u from the surface of the earth. Show that the maximum height reached by the object is h = \(\frac{u^2 R}{2 g R-u^2}\), where R is the radius of the earth.

Let m be the mass of the projected object.

∴ Initial kinetic energy = \(\frac{1}{2}m u^2\)

and potential energy at the surface, U = –\(\frac{G M m}{R}\)

At the maximum height h its kinetic energy, K’ = 0 and potential energy,

U’ = \(=-\frac{G M m}{(R+h)}\)

According to the law of conservation of energy, total energy at the earth’s surface = total energy at height h

∴ \(\frac{1}{2} m u^2+\left(-\frac{G M m}{R}\right)=-\frac{G M m}{R+h}+0\)

or, \(\frac{1}{2} m u^2=G M m\left[\frac{1}{R}-\frac{1}{R+h}\right]\)

or, \(u^2=\frac{2 G M h}{R(R+h)}\)

or, \(u^2=\frac{2 g R^2 h}{R(R+h)} \quad\left[G M=g R^2\right]\)

or, \(u^2(R+h)=2 g R h\) or, \(h\left(2 g R-u^2\right)=u^2 R\)

or, \(h=\frac{u^2 R}{2 g R-u^2}\) [Proved]

From equation (1), \(2 g R-u^2=\frac{u^2 R}{h}\)

If the object starts with escape velocity then \(h \rightarrow \infty\)

or, \(\frac{v_e^2 R}{h} \rightarrow 0\)

∴ \(2 g R-v_e^2=0\)

∴ \(v_e=\sqrt{2 g R}\)

Question 12. Keeping the mass fixed, if the radius of the earth is halved, the acceleration due to gravity at any place will be

1. Half of the original
2. One-fourth of the original
3. Double of the original
4. Four times of the original

Answer:

Acceleration due to gravity, g = \(\frac{G M}{R^2}\)

Keeping the mass fixed, if the radius of the earth is halved, the acceleration due to gravity becomes,

g’ = \(\frac{G M}{\left(\frac{R}{2}\right)^2}=\frac{4 G M}{R^2}=4 g \text { or, } \frac{g^{\prime}}{g}=4\)

The option 4 is correct.

Question 13. An artificial satellite moves in a circular orbit around the earth. The total energy of the satellite is given by E. The potential energy of the satellite is

1. -2E
2. 2E
3. \(\frac{2E}{3}\)
4. \(\frac{-2E}{3}\)

Answer:

Potential energy = 2 x total energy.

The option 2 is correct.

Question 14. Two particles of mass m₁ and m₂, approach each other due to their mutual gravitational attraction only. Then

1. Accelerations of both particles are equal
2. Acceleration of the particle of mass m₁ is proportional to m₁
3. Acceleration of the particle of mass m₁ is proportional to m₂
4. Acceleration of the particle of mass m₁ is inversely proportional to m₁.

Answer:

Force, \(F=\frac{G m_1 m_2}{r^2}\)

So, acceleration of the particle with mass \(m_1\), \(a_1=\frac{F}{m_1}=\frac{G m_2}{r^2}\)

i.e., \(a_1 \propto m_2\)

The option 3 is correct.

Applications of Gravitation: Short Answer Format

Question 15. A satellite has kinetic energy K, potential energy V and total energy E. Which of the following statements is true?

1. K = -V/2
2. K = V/2
3. E = K/2
4. E = -K/2

Answer: 1. K = -V/2

The option 1 is correct

Question 16. The ratio of accelerations due to gravity g₁: g₂ on the surfaces of two planets is 5: 2 and the ratio of their respective average densities ρ₁:ρ₂ is 2: 1. What is the ratio of respective escape velocities v₁: v₂ from the surface of the planets?

1. 5: 2
2. √5:√2
3. 5:2√2
4. 25:4

Answer:

Given

The ratio of accelerations due to gravity g₁: g₂ on the surfaces of two planets is 5: 2 and the ratio of their respective average densities ρ₁:ρ₂ is 2: 1.

⇒ \(g_1: g_2=5: 2\)

and \(\rho_1: \rho_2=2: 1\)

Escape velocity, \(v=\sqrt{2 g R}\)

From (1), \(\frac{\frac{G M_1}{R_1^2}}{\frac{G M_2}{R_2^2}}=\frac{5}{2} \quad \text { or, } \frac{M_1 R_2^2}{M_2 R_1^2}=\frac{5}{2} \quad \text { or, } \frac{M_1}{M_2}=\frac{5 R_1^2}{2 R_2^2}\)

From (2), \(\frac{\frac{M_1}{\frac{4}{3} \pi R_1^3}}{\frac{M_2}{\frac{4}{3} \pi R_2^3}}=\frac{2}{1} \text { or, } \frac{M_1 \times R_2^3}{M_2 \times R_1^3}=2\)

or, \(\frac{R_2^3}{R_1^3}=2 \times \frac{M_2}{M_1} \text { or, } \frac{R_2^3}{R_1^3}=2 \times \frac{2 R_2^2}{5 R_1^2} \text { or, } \frac{R_2}{R_1}=\frac{4}{5}\)

∴ \(\frac{v_1}{v_2}=\sqrt{\frac{R_1}{R_2} \times \frac{g_1}{g_2}}=\sqrt{\frac{5}{4} \times \frac{5}{2}}=\frac{5}{2 \sqrt{2}}\)

The option 3 is correct.

Question 17. Four particles, each of mass M and equidistant from each other move along a circle of radius R under the action of their mutual gravitational attraction, the speed of each particle is

1. \(\sqrt{\frac{G M}{R}}\)
2. \(\sqrt{2 \sqrt{2} \frac{G M}{R}}\)
3. \(\sqrt{\frac{G M}{R}(1+2 \sqrt{2})}\)
4. \(\frac{1}{2} \sqrt{\frac{G M}{R}(1+2 \sqrt{2})}\)

Answer:

The centripetal force on the particle at A, i.e., the force along AO due to the particles at B, C and D, is

F= \(\frac{G M \cdot M}{(\sqrt{2} R)^2} \cos 45^{\circ}+\frac{G M \cdot M}{(2 R)^2}+\frac{G M \cdot M}{(\sqrt{2} R)^2} \cos 45^{\circ}\)

= \(\frac{G M^2}{4 R^2}+2 \cdot \frac{G M^2}{2 R^2} \cdot \frac{1}{\sqrt{2}}\)

= \(\frac{G M^2}{4 R^2}\left(1+\frac{4}{\sqrt{2}}\right)=\frac{G M^2}{4 R^2}(1+2 \sqrt{2})\)

Again, \(F=\frac{M v^2}{R}\)

∴ \(\frac{M v^2}{R}=\frac{G M^2}{4 R^2}(1+2 \sqrt{2})\)

or, \(v=\frac{1}{2} \sqrt{\frac{G M}{R}(1+2 \sqrt{2})}\)

The option 4 is correct.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Short Answer Questions on Gravitational Potential Energy

Question 18. From a solid sphere of mass M and radius R, a spherical portion of radius \(\frac{R}{2}\) is removed, as shown. Taking gravitational potential V = 0 at r = ∞, the potential at the centre of the cavity thus formed is (G = gravitational constant)

1. \(\frac{-G M}{2 R}\)
2. \(\frac{-G M}{R}\)
3. \(\frac{-2 G M}{3 R}\)
4. \(\frac{-2 G M}{R}\)

Answer:

The potential of a solid sphere of mass m and radius r at a distance x(x < r) = \(\frac{G m}{2 r^3}\) (3r² – x²)

Here, the combination of a solid sphere of mass M and radius R with another solid sphere of mass \(\frac{M}{8}\) and radius \(\frac{R}{2}\) can be considered as the final system. The centre of the hollow sphere is situated at a distance of x = \(\frac{R}{2}\) from the centre of the solid sphere. Now potential at that point due to the solid sphere,

⇒ \(V_1=-\frac{G M}{2 R^3}\left[3 R^2-\left(\frac{R}{2}\right)^2\right]=-\frac{G M}{2 R^3} \cdot \frac{11}{4} R^2=-\frac{11}{8} \frac{G M}{R}\)

At the centre of the hollow sphere for the sphere, x = 0

So, the potential at that point, \(V_2=-\frac{G \times(-M / 8)}{2\left(\frac{R}{2}\right)^3}\left[3\left(\frac{R}{2}\right)^2-0\right]\)

= \(\frac{G M}{2 R^3} \cdot \frac{3}{4} R^2=\frac{3}{8} \frac{G M}{R}\)

∴ V = \(V_1+V_2=-\frac{11}{8} \frac{G M}{R}+\frac{3}{8} \frac{G M}{R}=-\frac{G M}{R}\)

Question 19. A satellite is revolving in a circular orbit at a height h from the earth’s surface (radius of earth R; h<<R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth’s gravitational field, is close to: (Neglect the effect of the atmosphere)

1. \(\sqrt{2 g R}\)
2. \(\sqrt{g R}\)
3. \(\sqrt{\frac{g R}{2}}\)
4. \(\sqrt{g R}(\sqrt{2}-1)\)

Answer:

Given

A satellite is revolving in a circular orbit at a height h from the earth’s surface (radius of earth R; h<<R). The minimum increase in its orbital velocity required,

Orbital velocity of the satellite, \(\nu=\sqrt{\frac{G M}{R+h}}=\sqrt{\frac{G M}{R}}[because R \gg h]\)

[Here, M= mass of the earth]

If the escape velocity of the satellite is \(v^{\prime}\), \(\frac{1}{2} m v^{\prime 2}=\frac{G M m}{R+h}[m=\text { mass of the satellite }]\)

∴ \(v^{\prime}=\sqrt{\frac{2 G M}{R+h}}=\sqrt{\frac{2 G M}{R}}\) [because h <<R]

∴ Increment of orbital velocity \(=v^{\prime}-v=\sqrt{\frac{2 G M}{R}}-\sqrt{\frac{G M}{R}}\)

= \(\sqrt{2 g R}-\sqrt{g R}=\sqrt{g R}(\sqrt{2}-1)\)

The option 4 is correct.

Question 20. The variation of acceleration due to gravity g with distance d from the centre of the earth is best represented by (R = earth’s radius):

Answer:

For d < R, g = \(\frac{G M d}{R^3}\) (g d graphic is linear)

For d = R, g = \(\frac{G M}{R^2}\) (highest value of g)

And for d > R, g = \(\frac{G M}{d^2}\) (g d graph is a rectangular parabola)

The option 4 is correct.

Short Answer Questions on Escape Velocity

Question 21. A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would earth (mass = 5.98 x 10²⁴ kg) have to be compressed to be a black hole?

1. 10^-9 m
2. 10^-6 m
3. 10^-2 m
4. 100m

Answer:

Given

A black hole is an object whose gravitational field is so strong that even light cannot escape from it.

⇒ \(v_e=\sqrt{\frac{2 G M}{R}}=C or, R=\frac{2 G M}{C^2}\) ; where, C= velocity of light

The option 3 is correct.

Question 22. The dependence of intensity of the gravitational field (E) of the earth with distance (r) from the centre of the earth is correctly represented by

Answer: Option 1 is correct

Question 23. Kepler’s third law states that the square of the period of revolution (T) of a planet around the sun, is proportional to the third power of the average distance between the sun and the planet i.e. T² = Kr³ here K is constant. If the masses of the sun and planet are M and m respectively, then as per Newton’s law of gravitation, the force of attraction between them is F = \(\frac{G M m}{r^2}\) here Q is the gravitational constant. The relation between G and K is described as

1. GK = 4π²
2. GMK = 4π²
3. K = G
4. K = \(\frac{1}{G}\)

Answer:

Given

Kepler’s third law states that the square of the period of revolution (T) of a planet around the sun, is proportional to the third power of the average distance between the sun and the planet i.e. T² = Kr³ here K is constant. If the masses of the sun and planet are M and m respectively, then as per Newton’s law of gravitation, the force of attraction between them is F = \(\frac{G M m}{r^2}\) here Q is the gravitational constant.

The required centripetal force for the rotation of a planet is provided by the gravitational force.

∴ \(\frac{m v^2}{r}=\frac{G M m}{r^2} \text { or, } v^2=\frac{G M}{r}\)

So, time period, \(T=\frac{2 \pi r}{v} \text { or, } T^2=\frac{4 \pi^2 r^2}{G M / r}=\frac{4 \pi^2}{G M} r^3=K r^3\)

Hence, \(\frac{4 \pi^2}{G M}=K or, G M K=4 \pi^2\)

The option 2 is correct

Question 24. The ratio of escape velocity at Earth (v_e) to the escape velocity at a planet (v_p) whose radius and mean density are twice as that of Earth is

1. 1:2:√2
2. 1:4
3. 1:√2
4. 1:2

Answer:

Escape velocity = \(\sqrt{\frac{2 G M}{R}}=\sqrt{\frac{8}{3} \pi G R^2 \rho}\)

∴ \(\frac{v_e}{v_p}=\sqrt{\left(\frac{R_e}{R_p}\right)^2\left(\frac{\rho_e}{\rho_p}\right)}=\sqrt{\left(\frac{R_e}{2 R_e}\right)^2\left(\frac{\rho_e}{2 \rho_e}\right)}=\sqrt{\frac{1}{4} \cdot \frac{1}{2}}=\frac{1}{2 \sqrt{2}}\)

The option 1 is correct

Question 25. At what height from the surface of the earth the gravitation potential and the value of g are -5.4 x 10⁷ J · kg^-1 and 6 m · s^-2 respectively? Take the radius of the earth as 6400 km

1. 1600 km
2. 1400 km
3. 2000 km
4. 2600 km

Answer:

Let, at height h from the surface of the earth the gravitation potential and the value of g are respectively equal to the given values.

∴ \(V_h=-\frac{G M}{R+h}=-5.4 \times 10^7 \mathrm{~J} \cdot \mathrm{kg}^{-1}\)

⇒ \(g_h=\frac{G M}{(R+h)^2}=6 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

∴ \(\frac{V_h}{g_h}=\frac{(R+h)^2}{R+h}=R+h=\frac{5.4 \times 10^7}{6}=0.9 \times 10^7\)

∴ h =\(\left(0.9 \times 10^7-R\right)=\left(0.9 \times 10^7-6400 \times 10^3\right)\)

= \(2.6 \times 10^6 \mathrm{~m}=2600 \mathrm{~km}\)

Question 26. Imagine earth to be a solid sphere of mass M and radius R. If the value of acceleration due to gravity at a depth d below earth’s surface is the same as its value at a height h above its surface and equal to \(\frac{g}{4}\)(where g is the value of acceleration due to gravity on the surface of earth), the ratio of \(\frac{h}{d}\) will be

1. 1
2. \(\frac{4}{3}\)
3. \(\frac{3}{2}\)
4. \(\frac{2}{3}\)

Answer:

At depth d below the earth’s surface \(\frac{g}{4}\) = g(1-\(\frac{d}{R}\))

or, \(\frac{d}{R}=1-\frac{1}{4}\) or, \(d=\frac{3}{4} R\)

At height h above the earth’s surface, \(\frac{g}{4}=g\left(\frac{R}{R+h}\right)^2 \text { or, } \frac{R}{R+h}=\frac{1}{2}\)

or, \(R+h=2 R or, h=R\)

∴ \(\frac{h}{d}=\frac{4}{3}\)

The option 2 is correct.

Question 27. A satellite of mass m is in a circular orbit of radius 3 R_E about earth (mass of earth M_E, radius of earth R_E). How much additional energy is required to transfer the satellite to an orbit of radius 9 R_E?

1. \(\frac{G M_E m}{3 R_E}\)
2. \(\frac{G M_E m}{18 R_E}\)
3. \(\frac{3 G M_E m}{2 R_E}\)
4. \(\frac{G M_E m}{9 R_E}\)

Answer:

The total energy of the satellite in orbit of radius r is given by,

E= \(-\frac{G M m}{2 r}\)

Therefore, \(E_1=-\frac{G M_E m}{2 \times 3 R_E}\) and \(E_2=-\frac{G M_E m}{2 \times 9 R_E}\)

∴ additional energy required, \(E_2-E_1=-\frac{G M_E m}{6 R_E}\left(\frac{1}{3}-1\right)=\frac{G M_E m}{6 R_E} \times \frac{2}{3}\)

= \(\frac{G M_E m}{9 R_E}\)

The option 4 is correct.

Question 28. The kinetic energies of a planet in an elliptical orbit about the sun, at positions A, B and C are K_A, K_B and K_C, respectively. AC is the major axis and SB is perpendicular to AC at the position of the sun S as shown. Then

1. K_B<K_A<K_C
2. K_A>K_B>K_C
3. K_A<K_B<K_C
4. K_B>K_A>K_C

Answer: 

Given

The kinetic energies of a planet in an elliptical orbit about the sun, at positions A, B and C are K_A, K_B and K_C, respectively.

The speed of the planet will be maximum at point A and will keep decreasing while moving towards point C.

If the speed at points A, B and C are respectively V_A, V_B and V_C,

V_A>V_B>V_C

∴ K_A>K_B> K_C

The option 2 is correct.

Real-Life Examples of Gravitation: Short Answer Questions

Question 29. If the mass of the sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct?

1. Time period of a simple pendulum on the earth would decrease
2. Walking on the ground would become more difficult
3. Raindrops will fall faster
4. g on the earth will not change

Answer:

g_earth= \(\frac{G M}{R^2}\) (M = mass of the earth, R = radius of the earth)

If G’ = 10G, \(g_{\text {earth }}^{\prime}=\frac{G^{\prime} M}{R^2}=\frac{10 G M}{R^2} \text { or, } g_{\text {earth }}^{\prime}=10 g_{\text {earth }}\)

So, the value of g will change.

The option 4 is correct.

Question 30. What is the reason for the absence of an atmosphere in some planets?
Answer:

The reason for the absence of atmosphere in some planets is a very low escape velocity of 2.38 km · s^-1.

Question 31. Explain the way the three laws can be proved.
Answer:

If r₁ and r₂ are the shortest and the longest distances of the planet from the sun, then the semimajor axis is given by \(\left(\frac{r_1+r_2}{2}\right)\)

1st law is proved from angular momentum conservation in a circular path.

Since, L = 2m x areal velocity, for equal time, the area swept will be the same. So, 2nd law is proved.

From \(F=\frac{G M_0 m}{r^2}, \frac{m \nu^2}{r}=F \text { and } T=\frac{2 \pi r}{\nu}\) one can prove T² ∝ r³. So, 3rd law is also proved.

Question 31. A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?
Answer:

Let R = radius of the earth

So, distance of the body from the centre of the earth in the first case, r₁ = R and that in the second case

r₂ = R + \(\frac{R}{2}\) = \(\frac{3}{2} R\)

Now, gravitational force, F = \(\frac{G M m}{R^2}\)

So, \(\frac{F_1}{F_2}=\left(\frac{r_2}{r_1}\right)^2\)

∴ \(F_2=F_1 \cdot\left(\frac{r_1}{r_2}\right)^2=63 \times\left(\frac{R}{3 / 2 R}\right)^2=63 \times \frac{4}{9}=28 \mathrm{~N}\)

Question 32. Deduce Kepler’s second law of planetary motion for a planet.
Answer:

Kepler’s second law of planetary motion for a planet

S is the fixed position of the sun. In a very small interval of time dt, a planet makes an angular displacement dθ, moving from A to B in its orbit. If dθ is sufficiently small, SAB is effectively a triangle with SA ≈ SB = r. Also, AB = rdθ = altitude of the triangle.

So the area described by the planet in time dt is, dA = area of the triangle SAB

= \(\frac{1}{2}\) x base SA x altitude AB

= \(\frac{1}{2}\)r·rdθ = \(\frac{1}{2}\)r² dθ

So the areal velocity of the planet, \(\)

(\(\frac{d A}{d t}=\frac{1}{2} r^2 \frac{d \theta}{d t}=\frac{1}{2} r^2 \omega\) = angular velocity) (ω= dθ/dt = angular velocity)

Now, if m = mass of the planet, then its angular momentum,

L = Iω – mr²ω

The angular momentum is conserved as there is no external torque on the system.

So, mr²ω = constant

or, r²ω= constant

or, \(\frac{d A}{d t}=\frac{1}{2} r^2 \omega\) = constant

Question 33. A Saturn year is 29.5 times the Earth year. How far is Saturn from the sun if the Earth is 1.5 x 10⁸ km away
Answer:

Hence, a line joining the sun with any planet in its orbit describes equal areas in equal intervals of time, i.e., the areal velocity is a constant. This is Kepler’s second law.

Question 34. Why does a satellite not need any fuel to circle around the Earth?
Answer:

The gravitational force between the Earth and the satellite provides the necessary centripetal force to move around the Earth. Also, this centripetal force acts normally with the direction of motion of the satellite. So, the work done is zero. So a satellite does not need any fuel to move around the earth.

Question 35. What would be the weight of a person, if he goes to a height equal to the radius of the earth from its surface?
Answer:

Weight at the surface of the earth W = mg = \(\frac{G M m}{R^2}\)

Weight at height h, \(W^{\prime}=m g^{\prime}=\frac{G M m}{(R+h)^2}=\frac{G M m}{(R+R)^2}\)

[∵ h=R]

So, \(\frac{W^{\prime}}{W}=\frac{R^2}{2 R^2}=\frac{1}{4} or, W^{\prime}=\frac{W}{4}\)

Question 36. A remote-sensing satellite of Earth revolves in a circular orbit at a height of 0.25 x 10⁶ m above the surface of Earth. Find the orbital speed and the period of revolution of the satellite. Given: Earth’s radius R_e = 6.38 x 10⁶ m and g = 9.8 m/s²
Answer:

Given

A remote-sensing satellite of Earth revolves in a circular orbit at a height of 0.25 x 10⁶ m above the surface of Earth.

Earth’s radius R_e = 6.38 x 10⁶ m and g = 9.8 m/s²

Orbital speed, \(v=R \sqrt{\frac{g}{R+h}}=6.38 \times 10^6 \sqrt{\frac{9.8}{(6.38+0.25) \times 10^6}}\)

= \(\frac{6.38 \times 10^6 \times 3.13}{2.57 \times 10^3}=7.8 \times 10^3 \mathrm{~m} / \mathrm{s}\)

Period of revolution, T = \(\frac{2 \pi(R+h)^{3 / 2}}{R \sqrt{g}} \)

=\(\frac{2 \times 3.142}{6.38 \times 10^6} \sqrt{\frac{6.63 \times 6.63 \times 6.63 \times 10^{18}}{9.8}}\)

= \(\frac{0.98 \times 17.07 \times 10^9}{3.13 \times 10^6}=5.34 \times 10^3 \mathrm{~s}\)

Question 37. According to Newton’s law of gravitation, everybody in this universe attracts every other body with a force, which is directly proportional to the product of their masses and is inversely proportional to the square of the distance between their centres, i.e., \(F \propto \frac{m_1 m_2}{r^2}\) or \(F=G \frac{m_1 m_2}{r^2}\) where G is universal gravitational constant = 6.67 x 10^-11 N · m² · kg^-2.

Read the above passage and answer the following questions:

1. What is the value of G on the surface of the moon?
2. How is the gravitational force between two bodies affected when the distance between them is halved?
3. What values of life do you learn from it?

Answer:

1. The value of G at the moon will be the same as G = 6.67 x 10^-11 N m^{2 ·}  kg^-2

2. Given: \(F=G \frac{m_1 m_2}{r^2}\)

Now, \(r^{\prime}=\frac{r}{2}\)

Then, new force of attraction, \(F^{\prime}=G \frac{m_1 m_2}{r^{\prime 2}}=G \frac{m_1 m_2}{\left(\frac{r}{2}\right)^2}=4\left[\frac{G m_1 m_2}{r^2}\right]\)

or, \(F^{\prime}=4 F\) [using equation (1)]

3. Everybody in the universe attracts every other body with a certain amount of force

4. Everybody in the universe attacks every other body with a certain amount of force.

Question 38. A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?
Answer:

Given

A body weighs 63 N on the surface of the earth.

Let g’ be the acceleration due to gravity at a height h above the earth’s surface. \(g^{\prime}=\frac{g}{\left(1+\frac{h}{R_e}\right)^2}\)

Let m be the mass of body. Then its weight at a height h above the earth’s surface, \(m g^{\prime}=\frac{m g}{\left(1+\frac{h}{R_e}\right)^2}\)

But h = \(\frac{R_e}{2}\)

∴ \(m g^{\prime}=\frac{m g}{\left[1+\frac{R_e{ }^{\prime 2}}{R_e}\right]^2}=\frac{63}{\left(1+\frac{1}{2}\right)^2}=63 \times \frac{4}{9}=28 \mathrm{~N}\)

Newtonian Gravitation And Planetary Motion Multiple Choice Questions And Answers

WBBSE Class 11 Gravitation MCQs with Answers

Question 1. The density of the earth is about

1. \(\frac{1}{5.5}\) times the density of water
2. 2 times the density of water
3. 5.5 times the density of water
4. 10 times the density of water

Answer: 3. 5.5 times the density of water

Question 2. A body of mass m is divided into two parts. The mass of one part is xm and that of the other part is (1- x) m. For a definite distance of separation between them, if the gravitational force of attraction has to be the maximum, the value of x should be

1. \(\frac{1}{2}\)
2. \(\frac{1}{4}\)
3. \(\frac{1}{\sqrt{2}}\)
4. \(\frac{\sqrt{3}}{2}\)

Answer: 1. \(\frac{1}{2}\)

Read And Learn More WBCHSE Class 11 Physics MCQs

Question 3. Two bodies of masses m₁ and m₂ are initially at rest and an infinite distance apart. Due to mutual attraction, they approach each other. When they are r distance apart, their relative velocity of approach is

1. \(\left[\frac{2 G\left(m_1+m_2\right)}{r}\right]^{1 / 2}\)
2. \(\left[\sqrt{\frac{2 G}{r}}\left(\frac{m_1+m_2}{2}\right)\right]^{1 / 2}\)
3. \(\left[\frac{r}{2 G\left(m_1 \cdot m_2\right)}\right]^{1 / 2}\)
4. \(\left[\frac{2 G}{r} \cdot\left(m_1 m_2\right)\right]^{1 / 2}\)

Answer: 1. \(\left[\frac{2 G\left(m_1+m_2\right)}{r}\right]^{1 / 2}\)

Question 4. Gravitational force is

1. Repulsive
2. Conservative
3. Electrical
4. Non-conservative

Answer: 3. Electrical

Question 5. Two small but heavy spheres of mass M each are kept at a distance r on a horizontal plane. The magnitude of the gravitational potential at the mid-point of the line joining the centres of the two spheres will be

1. Zero
2. –\(\frac{G M}{r}\)
3. –\(\frac{2G M}{r}\)
4. –\(\frac{4G M}{r}\)

Answer: 4. –\(\frac{4G M}{r}\)

Key Concepts of Newtonian Gravitation: MCQs

Question 6. An infinite number of masses, each of mass M, are placed along a straight line at distances of R, 2R,4R, 8 R, etc. from a reference point O. The magnitude of the gravitational potential at point O will be

1. \(\frac{G M}{2 R}\)
2. \(\frac{G M}{R}\)
3. \(\frac{2 G M}{R}\)
4. \(\frac{3 G M}{4 R}\)

Answer: 3. \(\frac{2 G M}{R}\)

Question 7. In the case of a freely falling spherical body, its acceleration due to gravity depends on

1. Mass of the body
2. Radius of the body
3. Density of the material of the body
4. None of the above

Answer: 4. None of the above

Question 8. If the earth is assumed to be a sphere of radius R, the height above the surface of the earth where the value of the acceleration due to gravity will be half its value on the earth’s surface is

1. h = \(\frac{R}{2}\)
2. h = \(\frac{R}{\sqrt{2}}\)
3. h = (√2 + 1)R
4. h = (√2 – 1)R

Answer: 4. h = (√2 – 1)R

Question 9. The mass of a planet is 4 times the mass of the earth and its radius is 2 times the radius of the earth. The acceleration due to gravity on that planet is

1. 9.8 m · s^-2
2. 19.6 m · s^-2
3. 4.9 m · s^-2
4. 39.2 m · s^-2

Answer: 1. 9.8 m · s^-2

Question 10. The mass of the moon is 1/10 th of that of the earth and the radius of the moon is 1/4 th of that of the earth. The ratio of the acceleration due to gravity on the moon and that on the earth is

1. 2:1
2. 1:2
3. 1:5
4. 5:1

Answer: 3. 1:5

Question 11. If the diurnal motion of the earth ceases all of a sudden, then the value of the acceleration due to the gravity of a body at the equator will

1. Remain the same
2. Be zero
3. Increase
4. Decrease

Answer: 3. Increase

Practice Questions on Kepler’s Laws of Planetary Motion

Question 12. Keeping the mass of the earth constant, if the radius of the earth is made 80% of its present value, the acceleration due to gravity on the earth’s surface would

1. Remain unchanged
2. Decrease by 36% (approx.).
3. Increase by 36% (approx.)
4. Increase by 56% (approx.)

Answer: 4. Increase by 56% (approx.)

Question 13. If the radius of the earth is R, the height above the face of the earth where the acceleration due to gravity will be 1 % of its value on the earth’s surface is

1. 8 R
2. 9R
3. 10 R
4. 20 R

Answer: 2. 9R

Question 14. If the change in the acceleration due to gravity (g) at an altitude h above the earth’s surface is equal to the change in g at a depth x below the earth’s surface [assume that both x and h are significantly smaller than the radius of the earth, then

1. x = h
2. x = 2h
3. x = \(\frac{h}{2}\)
4. x = h²

Answer: 2. x = 2h

Question 15. The acceleration due to gravity on the surface of the earth is 9.8 m · s^-2. The size of a planet is the same as that of the Earth, but its density is twice the density of the Earth. The value of the acceleration due to gravity on that planet is

1. 19.6 m · s^-2
2. 9.8 m · s^-2
3. 4.9 m · s^-2
4. 2.45 m · s^-2

Answer: 1. 19.6 m · s^-2

Question 16. A rocket is projected vertically upwards from the surface of the earth (radius = R) with a velocity v. To what height will the rocket rise? (Neglect air friction)

1. \(h=\frac{R}{\left(\frac{2 g R}{v^2}-1\right)}\)
2. \(h=\frac{R}{\left(\frac{2 g R}{\nu^2}+1\right)}\)
3. \(h=R\left(\frac{2 g R}{v^2}-1\right)\)
4. \(h=R\left(\frac{2 g R}{v^2}+1\right)\)

Answer: 1. \(h=\frac{R}{\left(\frac{2 g R}{v^2}-1\right)}\)

Question 17. If one moves from the equator to the pole, the value of g

1. Remains unchanged
2. Decreases
3. Increases
4. Increases first and then decreases

Answer: 3. Increases

Newton’s Law of Gravitation MCQs for Class 11

Question 18. If the radius of the earth were to shrink by 1%, its mass remaining the same, the acceleration due to gravity on the surface of the earth would

1. Increase
2. Decrease
3. Remain unchanged
4. Be zero

Answer: 1. Increase

Question 19. If G is the universal gravitational constant and g is the acceleration due to gravity then the unit of the quantity \(\frac{G}{g}\) is

1. kg · m²
2. kg/m
3. kg/m²
4. m²/kg

Answer: 4. m²/kg

Question 20. 20. At what altitude (h) above the earth’s surface would the acceleration due to gravity be one-fourth of its value at the earth’s surface? Where R is the radius of the earth.

1. h =R
2. h =4R
3. h =2R
4. h = 16R

Answer: 1. h = R

Question 21. A planet has the same density and acceleration due to gravity as of earth and the universal gravitational constant G is twice of Earth. The ratio of thin radii is

1. 1:4
2. 1:5
3. 1:2
4. 3:2

Answer: 3. 1:2

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 22. The escape velocity from the earth is v_e. If both the mass and the radius of a planet are twice that of the Earth, then the escape velocity from that planet will be

1. v_e
2. 2v_e
3. 4v_e
4. 16v_e

Answer: 1. v_e

Question 23. The escape velocity of a particle of mass m is

1. Directly Proportional To m²
2. Directly Proportional To m
3. Direcdy Proportional To m⁰
4. Directly Proportional To m^-1

Answer: 3. Direcdy Proportional To m⁰

Question 24. The value of the escape velocity of a body thrown vertically upwards from the surface of the earth is v. If the body is thrown making an angle θ with the vertical, then the value of the escape velocity will be

1. v
2. vcosθ
3. vsinθ
4. vtanθ

Answer: 1. v

Question 25. The value of the escape velocity from the earth is v_e. If the radius of a planet is 4 times that of the Earth and its density is 9 times the density of the Earth, then the value of the escape velocity from that planet will be

1. 6v
2. 12v
3. 20v
4. 36v

Answer: 2. 12v

Question 26. The escape velocity of a planet is v_e. From this planet a particle is projected upwards with a velocity v. The particle will revolve like a satellite if

1. \(\frac{v_e}{\sqrt{2}}<v<2 v_e\)
2. \(\frac{v_e}{v_2}<v<v_e\)
3. \(v_e<v<\sqrt{2} v_e\)
4. \(\frac{v_e}{v_2}<v<\frac{v_e}{2}\)

Answer: 2. \(\frac{v_e}{v_2}<v<v_e\)

Question 27. Escape speed on the surface of a planet varies with the mass m of a body as

1. m⁰
2. m
3. m^-1
4. m²

Answer: 1. m⁰

Question 28. The escape velocity of a body on the surface of the earth is 11.2 km/s. If the earth’s mass increases to twice its present value and the radius of the earth becomes half, the escape velocity would become

1. 5.6 km · s^-1
2. 11.2km · s^-1
3. 44.8 km · s^-1
4. 22.4 km · s^-1

Answer: 4. 22.4 km · s^-1

Question 29. Keeping the mass constant, if the radius of the earth is halved, then the span of a day

1. Will decrease
2. Will increase
3. Will remain unchanged
4. No conclusion can be arrived at

Answer: 1. Will decrease

Question 30. The weight of a body on the surface of the earth is W. The weight of that body at an altitude equal to half the radius of the earth will be

1. \(\frac{W}{2}\)
2. \(\frac{2W}{3}\)
3. \(\frac{4W}{9}\)
4. \(\frac{W}{4}\)

Answer: 3. \(\frac{4W}{9}\)

Question 31. A planet in elliptical orbit around a star moves form the point in its orbit furthest from the star to the closest point. The work done by the force of gravity during this movement is

1. Zero
2. Positive
3. Negative
4. Infinite

Answer: 3. Negative

Question 32. The height of a geostationary satellite from the surface of the earth is

1. 100 km
2. 5 km
3. 36000 km
4. 2 x 10⁵ km

Answer: 3. 36000 km

Sample Questions on Gravitational Potential Energy

Question 33. Two satellites of masses m₁ and m₂ (m₁ > m₂) are revolving around the earth in orbits of radii r₁ and r₂ (r₁ > r₂) velocities v₁ and v₂ respectively. In this case

1. v₁ = v₂
2. v₁ < v₂
3. v₁ > v₂
4. \(\frac{v_1}{r_1} = {v_2}{r_2}\)

Answer: 2. v₁ < v₂

Question 34. Two satellites A and B are revolving along circular paths of the same radius. The mass of A is 16 times the mass of B. The ratio of the period of revolution of B to that of A is

1. 1:16
2. 1:4
3. 1:2
4. 1:1

Answer: 4. 1:1

Question 35. Two planets are revolving around the sun. Their time periods of revolution and the average radii of the orbits are respectively (T₁, T₂) and (r₁, r₂). The ratio T₁/T₂ is

1. \(\left(\frac{r_1}{r_2}\right)^{1 / 2}\)
2. \(\frac{r_1}{r_2}\)
3. \(\left(\frac{r_1}{r_2}\right)^2\)
4. \(\left(\frac{r_1}{r_2}\right)^{3 / 2}\)

Answer: 4. \(\left(\frac{r_1}{r_2}\right)^{3 / 2}\)

Question 36. The centripetal force necessary for an artificial satellite revolving along its orbit around the earth is delivered by

1. The combustion of the engine fuel
2. The ejection of exhausted hot gas
3. The gravitational attraction of the sun
4. The gravitational attraction of the earth

Answer: 4. The gravitational attraction of the earth

Question 37. Keeping the sun at the focus, a planet revolves around the sun in an elliptical orbit. The shaded areas shown are equal, but the path DC is less than the path AB. If the time taken by the planet to go from A to B is t₁, and that from C to D is t₂, then

1. t₁ < t₂
2. t₁ > t₂
3. t₁ = t₂
4. t₁ = 3t₂

Answer: 3. t₁ = t₂

WBBSE Class 11 Practice Tests on Gravitation Concepts

Question 38. The radii of the orbits of two satellites A and B revolving around a planet are 4R and R respectively. If the velocity of A is 3 v, the velocity of B will be

1. \(\frac{4}{3}\) v
2. \(\frac{3}{2}\) v
3. 6v
4. 12v

Answer: 3. 6v

Question 39. In the case of the motion of a planet

1. The orbital velocity in its orbit remains constant
2. The orbital angular velocity remains constant
3. The total angular momentum remains constant
4. The orbital radius remains constant

Answer: 3. The total angular momentum remains constant

Question 40. Two small artificial satellites are revolving around the earth in two circular orbits of radii r and (r+Δr). If their time periods of revolution are T and T+ ΔT, then(Δr<<1, ΔT<<T)

1. \(\Delta T=\frac{3}{2} T \frac{\Delta r}{r}\)
2. \(\Delta T=-\frac{3}{2} T \frac{\Delta r}{r}\)
3. \(\Delta T=\frac{2}{3} T \frac{\Delta r}{r}\)
4. \(\Delta T=T \cdot \frac{\Delta r}{r}\)

Answer: 1. \(\Delta T=\frac{3}{2} T \frac{\Delta r}{r}\)

Question 41. If the gravitational force is inversely proportional to the   n-th power of the distance, then the time period of the revolution of a planet around the sun in a circular orbit of radius R will be

1. Directly proportional to \(R^{\frac{1}{2}(n+1)}\)
2. Directly proportional to \(R^{\frac{1}{2^{(n-1)}}}\)
3. Directly proportional to \(R^n\)
4. Directly proportional to \(R^{\frac{1}{2}(n-2)}\)

Answer: 1. Directly proportional to \(R^{\frac{1}{2}(n+1)}\)

Question 42. A satellite of mass m is revolving around the earth at a height x above the surface of the earth. The radius of the earth is R. If the acceleration due to gravity is g, the orbital speed of the satellite will be

1. gx
2. \(\frac{g R}{R-x}\)
3. \(\frac{g R^2}{R+x}\)
4. \(\left(\frac{g R^2}{R+x}\right)^{1 / 2}\)

Answer: 3. \(\frac{g R^2}{R+x}\)

Question 43. The ratio of the magnitudes of the kinetic energy and the potential energy of an artificial satellite revolving around the Earth is

1. 1:2
2. 1:√2
3. 2:1
4. √2:1

Answer: 1. 1:2

Question 44. Kepler’s second law is the consequence of the law of conservation of

1. Linear momentum
2. Energy
3. Angular momentum
4. Mass

Answer: 3. Angular momentum

Interactive MCQs on Orbital Motion for Students

Question 45. A satellite is moving in an orbit around a planet with kinetic energy k and potential energy v. The satellite will escape from the gravitational pull of the planet if its kinetic energy becomes

1. Half
2. Double
3. Three times
4. Four times

Answer: 2. Double

Question 46. The angle between the equatorial plane and the orbital plane of a geostationary satellite is:

1. 0°
2. 60°
3. 90°
4. 120°

Answer: 1. 0°

Question 47. The angle between the equatorial plane and the orbital plane of a polar satellite is

1. 0°
2. 90°
3. 120°
4. 180°

Answer: 2. 90°

Question 48. A geostationary satellite orbits around the earth’s surface in a circular orbit of radius 36,000 km. Then, the time period of a spy satellite orbiting seventeen hundred kilometres above the earth’s surface (R_e = 6400 km ) will approximately be

1. 1/2 h
2. 1h
3. 2h
4. 4h

Answer: 3. 2h

Question 49. An artificial satellite moving in a circular orbit around the earth has a total energy -E₀. Its potential energy is

1. -E₀
2. 1.5 E₀
3. -2 E₀
4. E₀

Answer: 3. -2 E₀

Question 50. The reason of weightlessness in a satellite is

1. Zero gravity
2. No atmosphere
3. Zero reaction force by satellite surface
4. None of the above

Answer: 3. Zero reaction force by satellite surface

Question 51. A body projected vertically from the Earth reaches a height equal to the Earth’s radius before returning to the earth. The power exerted by the gravitational force is greatest

1. At the instant just before the body hits the earth
2. It remains constant all through
3. At the instant just after the body is projected
4. At the highest position of the body

Answer: 1. At the instant just before the body hits the earth

Question 52. An orbiting satellite has

1. Only Kinetic Energy
2. Only Potential Energy
3. Kinetic And Potential Energy
4. Zero Energy

Answer: 3. Kinetic And Potential Energy

Question 53. If the daily rotation of the earth ceases suddenly, then the weight of a body situated at the north pole will

1. Zero
2. Remain the same
3. Increase
4. Decrease

Answer: 2. Remain the same

Question 54. In the case of a freely falling body, which of the graphs will indicate accurately the distance vs time variation? (air resistance is neglected)

Answer: 1.

Question 55. The radius of a uniform sphere is R and its mass is M. The values of the gravitational intensity at distances r₁ and r₂ from the centre of the sphere are F₁ and F₂ respectively. Then,

1. \(\frac{F_1}{F_2}=\frac{r_1}{r_2}, if r_1<R\) and \(r_2<R\)
2. \(\frac{F_1}{F_2}=\frac{r_1^2}{r_2^2}, if r_1>R\) and \(r_2>R\)
3. \(\frac{F_1}{F_2}=\frac{r_1}{r_2}, if r_1>R\) and \(r_2>R\)
4. \(\frac{F_1}{F_2}=\frac{r_1^2}{r_2^2}, if r_1<R\) and \(r_2<R\)

Answer: 1. \(\frac{F_1}{F_2}=\frac{r_1}{r_2}, if r_1<R\) and \(r_2<R\)

Question 56. The dimensional formula of gravitational field intensity is

1. MLT^-1
2. MLT^-2
3. M⁰LT^-2
4. M⁰L²T^-1

Answer: 3. M⁰LT^-2

In this type of question, more than one option is correct.

Question 57. An orbiting satellite will escape if

1. Its speed is increased by 41%
2. Its speed in the orbit is made \(\sqrt{(1.5)}\) times of its initial value
3. Its kinetic energy is doubled
4. It stops moving in the orbit

Answer:

1. Its speed is increased by 41%

3. Its kinetic energy is doubled

Question 58. A comet revolves around the sun in a highly elliptical orbit. Which of the following will remain constant throughout its orbit?

1. Kinetic energy
2. Potential energy
3. Total energy
4. Angular momentum

Answer:

3. Total energy

4. Angular momentum

Question 59. Which of the following is correct?

1. Out of electrostatic, electromagnetic, nuclear and gravitational interactions, the gravitational interaction is the weakest
2. If Earth were to rotate faster than its present speed, the weight of an object would decrease at the equator but remain unchanged at the poles
3. The mass of the earth in terms of g, R and g is (gR²/G)
4. If Earth stops rotating in its orbit around the sun there will be no variation in the weight of a body on the surface of Earth

Answer:

Question 60. A small mass m is moved slowly from the surface of the earth to a height h above the surface. The work done (by an external agent) in doing this is

1. mgh, for all values of h
2. mgh, for h << R
3. 1/2 mgR, for h = R
4. -1/2 mgR, for h = R

Answer:

2. mgh, for h << R

3. 1/2 mg R, for h = R

Artificial Satellites

WBBSE Class 11 Artificial Satellites Notes

The moon is the only satellite of the earth. It has already been discussed that the motions of different planets and satellites can be determined by Newton’s laws of motion and of gravitation.

From these considerations, it was thought that, if a projectile from the surface of the earth was raised to an appropriate height and given an appropriate velocity, it would also revolve around the earth like the moon.

• After a considerable amount of research in this field, the first artificial satellite (Sputnik-I) of the Earth was launched on 4 October 1957. Presently technological expertise has advanced to such an extent that it has been possible to set up artificial satellites not only around the Earth but also around other planets and satellites.
• Any artificial satellite is projected vertically upwards or towards the east from the surface of the earth using rockets. The direction of motion of the satellite is changed using rockets arranged at the rear of the satellite in such a way that the satellite attains the desired horizontal velocity on reaching the predetermined height.  In that case, it would set it in the desired orbit and start revolving around the Earth.

Obviously, once in an orbit, any satellite follows Kepler’s laws.

Read and Learn More: Class 11 Physics Notes

Hence, the orbit of an artificial satellite can also be elliptical or circular like the orbits of planets and satellites. However, these orbits are assumed to be circular as the eccentricities of such orbits are usually very low; the errors that may occur during calculations are negligible.

Orbital speed And Period Of Revolution Of an Artificial Satellite: The speed with which the satellite revolves around the earth is called its orbital speed. Let the mass of the earth = M, radius of the earth (OP) = R, mass of the satellite = m, orbital speed of the satellite = v and the height of the orbit from the surface of the earth (PA) = h. Hence, the distance of the orbit from the centre of the earth, i.e., the radius of the orbit, r = R+h

Types of Artificial Satellites Explained

Considering the orbit to be circular, the centripetal force = \(\frac{m v^2}{r}=\frac{m v^2}{(R+h)}\)

The gravitational force between the earth and the satellite \(\frac{G M m}{r^2}\) supplies this centripetal force.

∴ \(\frac{m v^2}{r}=\frac{G M m}{r^2} \text { or, } v^2=\frac{G M}{r}=\frac{G M}{R+h}\)

or, \(v=\sqrt{\frac{G M}{R+h}}\)…(1)

If the value of the acceleration due to gravity on the earth’s surface is g, then

g = \(\frac{G M}{R^2} \text { or, } G M=g R^2\)

Inserting the value in equation (1)

v = \(\sqrt{\frac{g R^2}{r}}=R \sqrt{g}\)…(2)

If the time period of revolution of the satellite is T, the distance covered in time T by the satellite = the circumference of the orbit = \(2 \pi r\).

∴ T = \(\frac{2 \pi r}{v}=2 \pi r \cdot \frac{1}{R} \sqrt{\frac{r}{g}}=\frac{2 \pi}{R} \sqrt{\frac{r^3}{g}}\)…(3)

Equations (2) and (3) desired above can be written in terms of the distance of the satellite from the earth’s surface (h) as \(v=\sqrt{\frac{g R^2}{R+h}}=R \sqrt{\frac{g}{R+h}}\)…(4)

and T = \(\frac{2 \pi}{R} \sqrt{\frac{(R+h)^3}{g}}=\frac{2 \pi}{R} \sqrt{\frac{R^3\left(1+\frac{h}{R}\right)^3}{g}}=\frac{2 \pi}{R} \cdot R \sqrt{\frac{R\left(1+\frac{h}{R}\right)^3}{g}}\)

= \(2 \pi \sqrt{\frac{R}{g}\left(1+\frac{h}{R}\right)^{3 / 2}}\)….(5)

Equations (1) to (5) show clearly that the orbital speed or the time period of a satellite does not depend on the mass of the satellite at all. It should be mentioned that if the orbit of a satellite is an ellipse of eccentricity e, the highest and the lowest orbital speed will be

⇒ \(v=\sqrt{\frac{g R^2}{R+h}}=R \sqrt{\frac{g}{R+h}}\)

Where a is the semi-major axis of the ellipse.

Artificial Satellite Very Close To The Earth:

From equations (4) and (5), it is evident that the lower the height (h) of the orbit from the earth’s surface, the higher the value of the orbital speed (v) and the lower the time period (T). Hence, to set a satellite in an orbit very close to the earth’s surface, it is necessary to impart a very high velocity to the satellite.

In the case of such orbits, the value of h is negligible in comparison to the radius R of the earth. That means R+h ≈ R and h = 0 can be substituted in equations (4) and (5) without introducing much error.

Hence, the orbital speed v = \(\sqrt{g R}\)….(6)

and the time period, T = \(2 \pi \sqrt{\frac{R}{g}}\)….(7)

The time period of revolution for such types of satellites depends only on the average density (ρ) of the earth.

As, \(\rho=\frac{3 g}{4 \pi G R} \quad therefore \frac{R}{g}=\frac{3}{4 \pi G \rho}\)

Putting the value of \(\frac{R}{g}\) in equation (7), we get \(T=2 \pi \sqrt{\frac{3}{4 \pi G \rho}} \text { or } T \propto \frac{1}{\sqrt{\rho}}\)

The radius of the earth R = 6400 km = 6.4 x 10⁶ m, the acceleration due to gravity on the earth’s surface g = 9.8 m · s^-2. Substituting these values in equations (6) and (7),

v = \(\sqrt{9.8 \times 6.4 \times 10^6}=7.9 \times 10^3 \mathrm{~m} \cdot \mathrm{s}^{-1}=7.9 \mathrm{~km} \cdot \mathrm{s}^{-1}\)

and  T = \(2 \times \pi \times \sqrt{\frac{6.4 \times 10^6}{9.8}}=5078 \mathrm{~s} \text { (approx.) }\)

= \(1 \mathrm{~h} 24 \mathrm{~min} 38 \mathrm{~s}\)

Applications of Artificial Satellites in Daily Life

The escape velocity from the earth, \(v_e=\sqrt{2 g R}=11.2 \mathrm{~km} \cdot \mathrm{s}^{-1}\). It can be said that the orbital speed of an artificial satellite close to the earth’s surface \(v=\sqrt{g R}=7.9 \mathrm{~km} \cdot \mathrm{s}^{-1}\) is comparatively less than the escape velocity v_e from the earth’s surface.

The ratio of these two velocities, \(\frac{\nu}{v_e}=\frac{\sqrt{g R}}{\sqrt{2 g R}}=\frac{1}{\sqrt{2}}=0.707\)

As the value of h increases, the value of orbital speed v decreases. Hence, the orbital speed decreases further compared to the escape velocity. Hence, to set up an artificial satellite to revolve around the Earth, it is not necessary to impart an orbital speed equal to or greater than the escape velocity.

Possible Trajectories Of Satellite: For different velocities, the trajectory of the satellite would be different. Let us discuss these cases.

If v is the velocity given to a satellite, v₀ represents the orbital speed of the satellite and v’_e be the escape velocity at a distance h from the earth’s surface, then

⇒ \(v_o=\sqrt{\frac{G M}{R+h}}, v_e^{\prime}=\sqrt{\frac{2 G M}{R+h}}\) where M and R are the mass and radius of the earth respectively.

Possible Trajectories Of Satellite Notes: When, v < v₀, the satellite follows an elliptical path with the centre of the earth as the further foci. In this case, if the satellite is projected from near the surface of the earth, it will fall on the earth without completing the orbit.

1. If v = v₀, the satellite follows a circular orbit with the centre of the earth as the centre of the orbit.
2. If v₀ < v < v’_e, the satellite follows an elliptical path with the centre of the earth as the nearer foci.
3. If v = v’_e, then the satellite escapes the gravitational field of Earth along a parabolic trajectory.
4. If v > v’_e, the satellite escapes the gravitational field of Earth along a hyperbolic trajectory.

Uses Of An Artificial Satellite: A few uses of artificial satellites are mentioned below:

1. Determination of the air pressure, height and composition of the atmosphere at a higher altitude.
2. Observation and forecast of weather.
3. Defence surveillance.
4. Study of the shape and size of the earth.
5. Telecommunication. [Signals can be exchanged using artificial satellites to broadcast television shows, games, etc., and also to provide communication by telephone.]
6. Collection of data about the ionosphere, cosmic rays, Van Allen radiation belts, effects of solar flares etc.

Geostationary Or Parking Orbit Geostationary Satellite

The earth rotates about its own axis (line joining the north and south poles) once in every 24 hours. This is the diurnal motion of Earth. We may consider an artificial satellite, set in an orbit in such a way that,

1. The orbit of the satellite is circular,
2. The plane of the orbit coincides with the equatorial plane,
3. The satellite revolves in the direction of the diurnal motion of the earth (i.e., from west to east) and completes a revolution in 24 h.

Then, the satellite will seem to be stationary at a place in the sky over the equator, when observed from the earth’s surface. Such a satellite is called a geostationary satellite and its orbit is called a geostationary or parking orbit Clearly, the centres of such orbits coincide with the centre of the earth

Geostationary Satellite Definition: With reference to the diurnal motion of the earth, if the relative angular velocity of an artificial satellite is zero and the satellite is always on the equatorial plane so that it appears stationary at one place in the sky, as seen from the earth’s surface, the satellite is called a geostationary artificial satellite.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Height And Orbital Speed Of A Geostationary Satellite: Let the mass of the earth = M, the radius of the earth = R, the mass of the geostationary satellite = m, the distance of the satellite from the centre of the earth = r, the orbital speed of the satellite = v, time period of revolution = T

As the force of gravitation provides the necessary centripetal force for the revolution in a circular orbit, \(\frac{m v^2}{r}=\frac{G M m}{r^2} \text { or, } v^2=\frac{G M}{r}\)

The acceleration due to gravity on the earth’s surface is g.

So, g = \(\frac{G M}{R^2}\) or, \(G M=g R^2\)

∴ \(v^2=\frac{g R^2}{r} \text { or, } v=\sqrt{\frac{g R^2}{r}}\)….(1)

Also, during the time period of the revolution,

T = \(\frac{2 \pi r}{v}=2 \pi r \sqrt{\frac{r}{g R^2}}=2 \pi \sqrt{\frac{r^3}{g R^2}}\)

or, \(T^2=\frac{4 \pi^2 \cdot r^3}{g R^2} or, r^3=\frac{g R^2 T^2}{4 \pi^2}\)

or, \(r=\left(\frac{g R^2 T^2}{4 \pi^2}\right)^{1 / 3}\)….(2)

For a geostationary satellite, T = 24 h = 24 x 60 x 60 s , g = 9.8 m · s^-2, R = 6400 km = 6.4 x 10⁶m.

Substituting these values, we obtain, r = 4.234 x 10⁷m = 4.234 x 10⁴ km = 42340 km (approx.).

Hence, the height of the geostationary satellite at the equator from the earth’s surface,

h = r – R = 42340- 6400 = 35940 ≈ 36000 km

It is to be noted that the height of a geostationary orbit does not depend on the mass of the satellite.

The orbital speed of a geostationary satellite, \(\nu=\frac{2 \pi r}{T}=\frac{2 \pi \times 4.234 \times 10^7}{24 \times 60 \times 60}=3079 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

= 3. 079 km · s^-1

Clearly, this orbital speed is much less than the escape velocity from the earth and the orbital speed for an artificial satellite close to the earth’s surface.

Height And Orbital Speed Of A Geostationary Satellite Discussions:

Height And Orbital Speed Of A Geostationary Satellite Uses: Nowadays, geostationary satellites are in extensive use, especially for weather observation, TV and radio broadcasting, telecommunication, etc. Games and entertainment programmes performed anywhere in the world can be transmitted and telecast live at other places using parking relay satellites on a geostationary orbit.

Height And Orbital Speed Of A Geostationary Satellite Difficulties In Use:

1. As geostationary satellites are placed at a height of about 36000 km above the earth’s surface, a signal has to travel a minimum distance of 72000 km before it is received at the other end.
   • This causes a delay of about \(\frac{1}{4}\)s from the time of occurrence, which is negligible in the case of radio, TV broadcasting, but causes inconvenience in telephonic conversations between two countries. Due to the high orbit, the spatial resolution of data is not as great as for the polar satellites.
2. The plane of orbits of geostationary satellites coincides with the earth’s equatorial plane. Hence, they cannot function properly for the zones closer to the poles.

Importance of Artificial Satellites in Modern Science

Polar Satellite: Satellites placed 700-800 km above the surface of the earth and aligned with the polar plane are called polar satellites. Using these satellites, a signal can reach from one point on the earth to another point in approximately \(\frac{1}{100}\)s.

• Because of such a small time difference, there is practically no time lagfelt during telephonic communication. These satellites also do away with the problems of reception and transmission of signals in the polar regions.
• Obviously, these are not geostationary satellites and hence, are not apparently stationary above the earth’s surface. Polar satellites have a time period of 2 h, i.e.„ to revolve around the earth once, a polar satellite takes about 2 h.

Geosynchronous Satellite: If the orbit of a satellite is inclined with the equatorial plane and the satellite revolves around the earth once in 24 h in the direction of the diurnal motion of the earth, the angular velocity of the satellite equals that of the earth. If observed from the surface of the earth, it appears that the artificial satellite is oscillating from north to south along a longitude and completes one oscillation in 24 h. Such satellites are called geosynchronous satellites.

Unit 6 Gravitation Chapter 1 Newtonian Gravitation And Planetary Motion

Artificial Satellites Numerical Examples

Short Answer Questions on Artificial Satellites

Example 1. An artificial satellite revolves around the earth in a circular orbit and is at a height of 300km above the surface of the earth. Find the orbital speed and the time period of the satellite. The radius of the earth = 6400km, g = 980cm · s^-2.
Solution:

Given

An artificial satellite revolves around the earth in a circular orbit and is at a height of 300km above the surface of the earth.

Radius of the earth, R = 6400 km = 64 x 10⁷ cm

Distance of the artificial satellite from the centre of the earth (r) = 6400 + 300 = 6700 km = 67 x 10⁷ cm.

The orbital speed of the artificial satellite, \(\nu=R \sqrt{\frac{g}{r}}=64 \times 10^7 \sqrt{\frac{980}{67 \times 10^7}}\)

= \(7.74 \times 10^5 \mathrm{~cm} \cdot \mathrm{s}^{-1}=7.74 \mathrm{~km} \cdot \mathrm{s}^{-1}\)

Time period of revolution, T = \(\frac{2 \pi r}{\nu}=\frac{2 \times \pi \times 67 \times 10^7}{7.74 \times 10^5}=5439 \mathrm{~s}\)

= \(1 \mathrm{~h} 30 \mathrm{~min} 39 \mathrm{~s}\)

Example 2.  A man is positioned in a circular orbit at a height of 1.6 x 10⁵ m above the surface of the earth. The radius of the earth is 6.37 x 10⁶ m and the mass is 5.98 x 10²⁴ kg. What would be the orbital speed of the man? (G = 6.67 x 10^-11 N · m² · kg^-2)
Solution:

Given

A man is positioned in a circular orbit at a height of 1.6 x 10⁵ m above the surface of the earth. The radius of the earth is 6.37 x 10⁶ m and the mass is 5.98 x 10²⁴ kg.

Radius of the earth, R = 6.37 x 10⁶ m = 63.7 x 10⁵ m ; height, h = 1.6 x 10⁵ m

Distance of the man from the centre of the earth (r) = R+h = 63.7 x 10⁵ + 1.6 x 10⁵ = 65.3 x 10⁵ m

∴ Orbital speed \(v=\sqrt{\frac{G M}{R+h}}=\sqrt{\frac{G M}{r}}=\sqrt{\frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{65.3 \times 10^5}}\)

= 0.78 x 10⁴ m · s^-1 = 7.8 km · s^-1.

Example 3. Prove that, if the orbital speed of the moon increases by 42%, it will stop orbiting around the earth.
Solution:

Let the mass of the earth = M and the radius of the moon’s orbit = r.

Orbital speed of the moon (v) = \(\sqrt{\frac{G M}{r}}\)

The acceleration due to gravity, \(g^{\prime}=\frac{G M}{r^2} \text { or, } G M=g^{\prime} r^2\)

∴ \(\nu=\sqrt{\frac{g^{\prime} r^2}{r}}=\sqrt{g^{\prime} r}\)

Also, the escape velocity of the moon with respect to the earth’s surface, \(v_e=\sqrt{2 g^{\prime} r}\)

∴ \(\frac{v_e}{v}=\sqrt{\frac{2 g^{\prime} r}{g^{\prime} r}}=\sqrt{2}=1.414\)

This implies, v_e = 1.414 v = 141.4% of the orbital speed (v).

Hence, if the orbital speed of the moon increases by 42%, which means that it changes to 142% of its present value, it crosses the value of the escape velocity from the earth. As a result, the moon will move out of the Earth’s gravitational field and will not move around the Earth anymore.

Example 4. An artificial satellite is orbiting around the earth at a height of 3400 km above the earth’s surface in a circular orbit. Find the orbital speed of the satellite. The radius of the earth =6400 km and g = 980 cm · s^-2.
Solution:

Given

An artificial satellite is orbiting around the earth at a height of 3400 km above the earth’s surface in a circular orbit.

Radius of the earth, R = 6400 km = 64 x 10⁷ cm

Distance of the artificial satellite from the centre of the earth, r = 6400 + 3400 = 9800 km = 98 x 10⁷ cm

Hence, the orbital speed of the satellite, \(v=R \sqrt{\frac{g}{r}}=64 \times 10^7 \sqrt{\frac{980}{98 \times 10^7}}\)

= 6.4 x 10⁵ cm · s^-1 = 6.4 km · s^-1

Example 5. A satellite at a height of 700 km is revolving around the earth in a circular orbit Find the velocity of the satellite with respect to the earth’s surface. (Radius of the earth R = 6300 km, g= 9.8 m · s^-2)
Solution:

Given

A satellite at a height of 700 km is revolving around the earth in a circular orbit

The radius of the earth (RF) = 6300 km = 63 x 10⁵ m.

Distance of the satellite from the centre of the earth, r = 6300 + 700 = 7000 km = 7 x 10⁶ m

Hence, the velocity of the artificial satellite, \(\nu=R \sqrt{\frac{g}{r}}=63 \times 10^5 \sqrt{\frac{9.8}{7 \times 10^6}}=74.543 \times 10^2 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

= \(7.454 \mathrm{~km} \cdot \mathrm{s}^{-1}\)

Example 6. A small satellite revolves around a plane of density 10 g · cm^-3. The radius of the orbit of the satellite is slightly more than the radius of the planet. Find the time period of rotation of the satellite. G = 6.68 x 10^-8 CGS unit.
Solution:

Given

A small satellite revolves around a plane of density 10 g · cm^-3. The radius of the orbit of the satellite is slightly more than the radius of the planet. Find the time period of rotation of the satellite. G = 6.68 x 10^-8 CGS unit.

Let the mass of the planet = M, radius = R and density = ρ.

∴ \(M=\frac{4}{3} \pi R^3 \rho\)

As per the question, the radius of the orbit of the satellite ≅ R. If its mass = m and orbital speed = v, centripetal force = \(\frac{m v^2}{R}\)

The mutual force of gravitation between the planet and satellite provides this centripetal force.

∴ \(\frac{m v^2}{R}=\frac{G M m}{R^2}\)

or, \(v^2=\frac{G M}{R}=\frac{G}{R} \frac{4}{3} \pi R^3 \rho=\frac{4}{3} \pi G \rho R^2\)

or, \(\nu=R \sqrt{\frac{4}{3} \pi G \rho}\)

Time period = \(\frac{\text { circumference of orbit }}{\text { orbital speed }}=\frac{2 \pi R}{\nu}\)

= \(\frac{2 \pi}{\sqrt{\frac{4}{3} \pi G \rho}}=\sqrt{\frac{3 \pi}{G \rho}}=\sqrt{\frac{3 \times \pi}{6.68 \times 10^{-8} \times 10}}\)

= \(3756 \mathrm{~s}=1 \mathrm{~h} 2 \mathrm{~min} 36 \mathrm{~s} .\)

Kinetic Energy And Potential Energy Of An Artificial Satellite: If an artificial satellite of mass m revolves around the earth in a circular orbit of radius r with an orbital speed v, then \(\frac{m v^2}{r}=\frac{G M m}{r^2} \text { or, } v^2=\frac{G M}{r}\)

Hence, the kinetic energy of the satellite,

K = \(\frac{1}{2} m v^2=\frac{G M m}{2 r}\)….(1)

The potential energy of an object of mass m at a distance r from another mass M,

U = \(-\frac{G M m}{r}\)…..(2)

The value of this potential energy is negative as the force of gravitation is an attractive force. It is seen from equations (1) and (2) that the potential energy of a satellite revolving around the earth is double its kinetic energy, but negative.

Thus, the total energy of the satellite in orbit, \(\)…..(3)

Due to friction in the earth’s atmosphere or due to collision with meteors and meteorites in space, the energy of a
satellite (-\(\frac{G M m}{2 r}\)) decreases, that is, the magnitude of \(\frac{G M m}{2 r}\) increases.

• Hence, the value of r decreases, which means that the satellite starts shifting towards the earth’s surface. As a result, the kinetic energy (\(\frac{G M m}{2 r}\)) starts increasing. In brief, as the distance of a satellite from the surface of the earth decreases, its orbital speed and kinetic energy increase, but the total energy decreases.
• This is possible because a decrease in the potential energy of a satellite is twice the increase in kinetic energy when it transfers from one orbit to another. Thus, for any artificial satellite to be working in its orbit for a long time (for example, a geostationary satellite), energy has to be supplied from outside to compensate for its loss of energy. The use of solar energy for this purpose is indispensable.
• It should be noted that for an elliptical orbit, the expression for the total energy of the artificial satellite is –\(\frac{G M m}{2 4}\), where a is the semimajor axis of the ellipse.

Binding Energy Of A Satellite: The energy required by a satellite to leave its orbit around the earth and escape to infinity is called its binding energy.

The total energy of a satellite is \(\frac{G M m}{2 r}\). In order to escape to infinity, it must be supplied an extra energy equal to +\(\frac{G M m}{2 r}\), so that, its total energy E becomes zero. Hence binding energy of a satellite is \(\frac{G M m}{2 r}\).

The plot of energies of a satellite versus orbit radius is shown.

Unit 6 Gravitation Chapter 1 Newtonian Gravitation And Planetary Motion

Kinetic Energy Of Artificial Satellites Numerical Examples

Real-Life Examples of Artificial Satellites

Example 1. Find the velocity with which a body can be projected vertically upwards so that it can reach a height equal to the radius of the earth. The radius of the earth = 6400 km, g = 980 cm · s^-2.
Solution:

Given

The radius of the earth = 6400 km, g = 980 cm · s^-2.

Initial distance of the body from the centre of the earth = radius of the earth = R.

Final distance of the body from the centre of the earth = R+R = 2 R

Let the mass of the earth = M; the mass of the body = m; velocity of projection from the earth’s surface = v, Hence, its kinetic energy on the earth’s surface = \(\frac{1}{2} m v^2\)

At a height R above the earth’s surface, the body stops momentarily and then falls. Hence, at that height R, kinetic energy = 0.

Potential energy on the earth’s surface = –\(\frac{G M m}{R}\)

∴ Total energy of the body on the earth’s surface = kinetic energy + potential energy = \(\frac{1}{2} m v^2\)–\(\frac{G M m}{R}\)

Again, potential energy at height R = –\(\frac{G M m}{2R}\)

and its total energy at that height = 0 – \(\frac{G M m}{2R}\) = –\(\frac{G M m}{2R}\)

From the law of conservation of energy \(\frac{1}{2} m v^2-\frac{G M m}{R}=-\frac{G M m}{2 R}\)

or, \(\frac{1}{2} m v^2=\frac{G M m}{R}-\frac{G M m}{2 R}=\frac{G M m}{2 R}\)

= \(\frac{G M}{R^2} \cdot \frac{m R}{2}=g \cdot \frac{m R}{2}=\frac{1}{2} m g R\)

∴ \(v^2=g R\)

or,  \(v=\sqrt{g R}=\sqrt{980 \times 64 \times 10^7}=79.2 \times 10^4 \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

= \(7.92 \times 10^5 \mathrm{~cm} \cdot \mathrm{s}^{-1}=7.92 \mathrm{~km} \cdot \mathrm{s}^{-1}\)

Alternative Method: Let the radius of the earth be R, the potential energy of the body on the earth’s surface be 0, and the acceleration due to gravity at a height h above the earth’s surface be g’. Hence, the potential energy at height h is mg’h. For a further increase dh in height, let the increase in potential energy be dW.

∴ dW = \(m g^{\prime} d h=m g \frac{R^2}{(R+h)^2} d h\left[because g^{\prime}=\frac{R^2}{(R+h)^2} \cdot g\right]\)

Hence, the total increase in the potential energy for an increase in height R from the surface of the earth,

⇒ \(\int_0^W d W=\int_0^R m g \frac{R^2}{(R+h)^2} d h\)

or, \( W=m g R^2\left[-\frac{1}{R+h}\right]_0^R=m g R^2\left(-\frac{1}{R+R}+\frac{1}{R}\right)\)

= \(\frac{m g R^2}{2 R}=\frac{1}{2} m g R\)

Let the kinetic energy of the body on the earth’s surface = \(\frac{1}{2}m v^2\).

As per the question, the kinetic energy at a height R from the earth’s surface is 0.

Hence, from the law of conservation of energy, \(\frac{1}{2} m v^2=\frac{1}{2} m g R\) or, \(\nu^2=g R\)

or, \(v=\sqrt{g R}=7.92 \times 10^5 \mathrm{~cm} \cdot \mathrm{s}^{-1}=7.92 \mathrm{~km} \cdot \mathrm{s}^{-1}\)

Example 2. Two bodies of masses M and m were initially at infinite distance from each other and they started approaching each other due to their mutual force of gravitation. Prove that their velocity of approach becomes \(\sqrt{\frac{2 G}{r}(M+m)}\) when they are at a distance of r from each other.
Solution:

Given

Two bodies of masses M and m were initially at infinite distance from each other and they started approaching each other due to their mutual force of gravitation.

At an infinite distance from each other, both bodies did not have any potential energy. Also, their velocities were zero.

Hence, the initial momentum of both the bodies = 0.

Total initial energy = kinetic energy + potential energy = 0 + 0 = 0.

At a distance r from each other, let the velocity of the body of mass m be v and that of mass M be V.

Hence, their velocity of approach (c) = v+V

As the velocities v and V are in opposite directions, the total momentum at this stage = mv – MV

Hence, from the law of conservation of momentum

0 = mv – MV or, V = \(\frac{m}{M}\)v

∴ c = v + V = v + \(\frac{m}{V}\)v = v(1+\(\frac{m}{M}\))

Hence, the kinetic energy of the bodies when they are at a distance r from each other.

= \(\frac{1}{2} m \nu^2+\frac{1}{2} M \nu^2-\frac{1}{2} m v^2+\frac{1}{2} M \cdot \frac{m^2}{M^2} \cdot \nu^2\)

= \(\frac{1}{2} m \nu^2\left(1+\frac{m}{M}\right)=\frac{1}{2} m \cdot \frac{v^2\left(1+\frac{m}{M}\right)^2}{1+\frac{m}{M}}=\frac{1}{2} m \cdot \frac{c^2}{\frac{M+m}{M}}\)

= \(\frac{1}{2} \cdot \frac{M m c^2}{M+m}\)

Also, potential energy at a distance r (due to gravitation) = \(-\frac{G M m}{r}\)

Hence, from the law of conservation of energy, \( 0=\frac{1}{2} \frac{m M c^2}{M+m}-\frac{G M m}{r}\)

or, \(\frac{1}{2} \frac{m M c^2}{M+m}=\frac{G M m}{r} \text { or, } c^2=\frac{2 G}{r}(M+m)\)

c = \(\sqrt{\frac{2 G}{r}(M+m)}\)

Example 3. An artificial satellite revolves around the Earth in a circular orbit Its velocity is half the value of the escape velocity from the Earth,

1. What is the height of the satellite from the earth’s surface?
2. If its revolution around the earth is stopped and the satellite is allowed to fall freely towards the earth, what will be the velocity with which it will strike the earth’s surface? (Radius of the earth = 6.4 x 10⁶ m;  g = 9.8 m · s^-2)

Solution:

1. Orbital speed of the satellite = \(u=\frac{1}{2} \times \text { escape velocity }\) = \(\frac{1}{2} \times \sqrt{2 g R}=\sqrt{\frac{g R}{2}}\)

Again, if it is at a height h above the earth’s surface,

u = \(R \sqrt{\frac{g}{R+h}}=\sqrt{\frac{g R^2}{R+h}}\)

∴ \(\sqrt{\frac{g R}{2}}=\sqrt{\frac{g R^2}{R+h}} \text { or, } \frac{g R}{2}=\frac{g R^2}{R+h} \text { or, } R+h=2 R\)

∴ h = \(R=6.4 \times 10^6 \mathrm{~m}=6400 \mathrm{~km} .\)

2. When the satellite stops revolving and falls freely towards the earth, its initial velocity of fall = 0. Hence, its initial kinetic energy = 0.

Since at this stage, the distance of the satellite from the centre of the earth is r = R+ h = 2R, its initial potential energy = \(\frac{G M m}{2R}\) (M = mass of the earth, m= mass of the satellite).

Hence, the total initial energy of the satellite = \(0-\frac{G M m}{2 R}=-\frac{G M m}{2 R}\)

Let the velocity of the satellite be v, just before touching the earth’s surface.

So its kinetic energy = \(\frac{1}{2} mv^2\). At the same time, its distance from the centre of the earth is r = R and its potential energy = \(-\frac {G M m}{R}\)

Hence, total energy of the satellite = \(\frac{1}{2} mv^2\)\(-\frac{G M m}{R}\)

From the law of conservation of energy, \(-\frac{G M m}{2 R}=\frac{1}{2} m v^2-\frac{G M m}{R}\)

or, \(\frac{1}{2} m v^2=\frac{G M m}{R}-\frac{G M m}{2 R}=\frac{G M m}{2 R}\)

or, \(v^2=\frac{G M}{R}=\frac{g R^2}{R}=g R\)

or, \(v=\sqrt{g R}=\sqrt{9.8 \times 6.4 \times 10^6}\)

= \(7.9 \times 10^3 \mathrm{~m} \cdot \mathrm{s}^{-1}=7.9 \mathrm{~km} \cdot \mathrm{s}^{-1} .\)

Example 4. Calculate the kinetic energy, potential energy and total energy of a geostationary satellite of mass 100 tonnes. The radius of the earth is 6400 km and the radius of the orbit of the satellite is 42400 km.
Solution:

Given

The radius of the earth is 6400 km and the radius of the orbit of the satellite is 42400 km.

100 tone = 100 x 10³ kg = 10⁵ kg. The radius of the earth = 6400 km = 6.4 x 10⁶ m and the distance of the geostationary satellite from the centre of the earth, r = 42400 km = 4.24 x 10⁷ m.

Hence, kinetic energy

= \(\frac{G M m}{2 r}=\frac{1}{2} m \frac{G M}{R^2} \cdot \frac{R^2}{r}\)

= \(\frac{m g R^2}{2 r}=\frac{10^5 \times 9.8 \times\left(6.4 \times 10^6\right)^2}{2 \times 4.24 \times 10^7}\)

= \(4.73 \times 10^{11} \mathrm{~J}\)

Potential energy = \(-\frac{G M m}{r}=-2 \times \frac{G M m}{2 r}\)

= \(-2 \times \frac{10^5 \times 9.8 \times\left(6.4 \times 10^6\right)^2}{2 \times 4.24 \times 10^7}\)

= \(-9.47 \times 10^{11} \mathrm{~J}\)

Total energy =\(-\frac{G M m}{2 r}=-4.73 \times 10^{11} \mathrm{~J} \text {. }\)

Weightlessness In Artificial Satellites: It is known that the weight of a body of mass m is mg, i.e., the earth pulls the body towards its centre with this force. When this body is placed on a non-accelerating plane, it exerts a force of mg on that plane.

• The plane also exerts an upward reaction force on the body R(= mg). This upward reaction force is the apparent weight of the body. The body feels only this upward reaction force as its weight. Hence, any person or body cannot feel its own weight while standing on a plane if the plane does not offer any upward reaction; this means that the person or body feels weightless.
• Let the plane along with the body have an acceleration vertically upwards. In this case, the upward reaction of the plane does not become equal to the weight. Because of this a passenger in a lift with an acceleration feels his weight to be increased or decreased.
• There are a few cases in which the apparent weight of a body is not related to the existence of the plane (on which the body is placed). But it is convenient to measure the apparent weight of a body taking into consideration the existence of such a plane.

Weight Experienced By An Astronaut Before The Spaceship Is Parked In Its Orbit: The velocity of the spaceship is increased rapidly by the rocket immediately after its launch from the surface of the earth. Sometimes the value of its upward acceleration may be 15 times the value of the acceleration due to gravity. If the mass of the astronaut is m, the upward reaction of the spaceship on the man is R, and the acceleration of the spaceship is 15g, then

R – mg – ma = m · 15g or, R = 16mg

This is the apparent weight of the astronaut. Thus he feels himself to be 16 times heavier than his actual weight.

Weight Experienced By An Astronaut When The Spaceship Is Moving In Its Orbit:

Let r = radius of the circular orbit of the spaceship with respect to the centre of the earth,

g’ = acceleration due to gravity at the position of the spaceship,

v = instantaneous velocity of the spaceship,

M = mass of the spaceship, m = mass of the astronaut,

R = upward reaction of the floor of the spaceship on the astronaut

For the rotation of the spaceship or of the astronaut around the earth, the force acting towards the centre of the earth supplies the necessary centripetal force. So, for the spaceship and for the astronaut, respectively,

⇒ \(M g^{\prime}=\frac{M v^2}{r}, \text { or } g^{\prime}=\frac{v^2}{r}\)

and \(m g^{\prime}-R=\frac{m v^2}{r} \text { or, } m \frac{v^2}{r}-R=\frac{m v^2}{r}\)

Thus, R = 0

The astronaut cannot feel his own weight as the reaction force is zero. He floats in the spaceship. This is called the weightlessness of man or any object in an artificial satellite.

Similarly, if it is assumed that the spaceship is resting on an imaginary plane and the motion of the spaceship along with its plane is considered, the spaceship also appears to be weightless.

• As the plane is imaginary, it can be concluded that all objects (planet, satellite, etc.) orbiting under the action of gravitation do not experience any force directed towards the centre of the orbit, and therefore, all such objects are weightless.
• An astronaut feels much heavier than his actual weight at the time of launching the spaceship, but he feels weightless as soon as the spaceship is placed in its orbit.
• The phenomenon of weightlessness is widely used in many cases of scientific and technological requirements. For example, crystal formation on the earth’s surface due to gravitation is imperfect, but perfect crystals can be developed in orbiting spaceships easily and such crystals have immense importance in making transistors and other such electronic devices.
• Again, oil and water cannot really be mixed on the earth’s surface, but in a spaceship, in the weightless condition, it is possible to make a homogeneous mixture of oil and water. In addition, a pure alloy can be produced by cooling a homogeneous mixture of molten metals in a spaceship.
• As in the case of materials kept in an artificial satellite, objects on the moon’s surface also have no weight due to the gravitational pull of the Earth. But nobody is weightless on the moon’s surface as the attraction of the moon acts on the body.
• This weight is about| th of its weight on the earth. It is to be noted that bodies in an artificial satellite are also subject to the force of attraction of the satellite or spaceship. But this force is too small to be felt.
• The gravitational force of attraction of the earth on an artificial satellite or on an astronaut can never be zero. If it had been zero, the necessary centripetal force could not have been supplied and the satellite would not have been able to revolve around the earth in a circular orbit.

Thus the statement that a body is weightless implies that only the reaction force acting on it is zero, but the gravitational force is not zero.

Unit 6 Gravitation Chapter 1 Newtonian Gravitation And Planetary Motion

Momentum Of Satellite Numerical Examples

Example 1. Two particles of equal mass revolve in a circular path of radius R due to their mutual force of attraction. Find the velocity of each particle.
Solution:

Given

Two particles of equal mass revolve in a circular path of radius R due to their mutual force of attraction.

At any time, during the revolution, the two particles stay at the two ends of any diameter of the circular path. The mutual force of gravitation acts along the diameter.

Considering the motion of any one of the particles, \(\frac{G m \times m}{(2 R)^2}=\frac{m v^2}{R} \text { or, } v^2=\frac{G m}{4 R} \text { or, } \nu=\sqrt{\frac{G m}{4 R}} \text {. }\)

Example 2. The acceleration due to gravity at two places are g and g’ respectively. A body is dropped from the same height at both places. In the second place, the required time to touch the ground is t seconds less than that in the first place, while the velocity attained in reaching the ground is higher by a value of v than that in the first place. Show that gg’ = \(\frac{v^2}{t^2}\).
Solution:

Given

The acceleration due to gravity at two places are g and g’ respectively. A body is dropped from the same height at both places. In the second place, the required time to touch the ground is t seconds less than that in the first place, while the velocity attained in reaching the ground is higher by a value of v than that in the first place.

Let the time taken by the body to fall at the first place be T and the velocity with which the body touches the ground be V.

Hence, the time taken to reach the ground and the corresponding velocity are (T- t) and V+ v in the second place. Let the body fall from a height h in each case. Considering the motion in the first place,

h = \(\frac{1}{2} g T^2 \text { or, } T=\sqrt{\frac{2 h}{g}}\)…(1)

and \(V^2=2 g h or, V=\sqrt{2 g h}\)….(2)

Similarly, for motion at the second place, h = \(\frac{1}{2} g^{\prime}(T-t)^2 \text { or, } T-t=\sqrt{\frac{2 h}{g^{\prime}}}\)

and \((V+v)^2=2 g^{\prime} h or, V+v=\sqrt{2 g^{\prime} h}\)

Subtracting (3) from (1) \(t=\sqrt{\frac{2 h}{g}}-\sqrt{\frac{2 h}{g^{\prime}}}=\sqrt{2 h}\left(\frac{1}{\sqrt{g}}-\frac{1}{\sqrt{g}}\right)\)

Subtracting (2) from (4) \(v=\sqrt{2 g^{\prime} h}-\sqrt{2 g h}=\sqrt{2 h}\left(\sqrt{g^{\prime}}-\sqrt{g}\right)\)

∴ \(\frac{v^2}{t^2}=\frac{2 h\left(\sqrt{g^{\prime}}-\sqrt{g}\right)^2}{2 h\left(\frac{\sqrt{g^{\prime}}-\sqrt{g}}{\sqrt{g g^{\prime}}}\right)^2}=g g^{\prime} .\)

Example 3. A satellite of mass m revolves around the earth in a circular orbit of radius r. Find the angular momentum of the satellite with respect to the centre of the orbit.
Solution:

Given

A satellite of mass m revolves around the earth in a circular orbit of radius r.

The angular momentum of the satellite about the centre of the orbit is L = mvr, where v is the orbital speed of the satellite. For motion in a fixed orbit, centripetal force = force of attraction due to gravity

or, \(\frac{m v^2}{r} = \frac{G M m}{r^2}\)

or, \(m v^2 r=G M m \text { or, }(m v r)^2=G M m^2 r\)

or,\(m v r=m \sqrt{G M r}\)

∴ L = \(m \sqrt{G M r}\)

Example 4. A satellite is orbiting in a circular path around the earth close to its surface. What additional velocity Is to be imparted to the satellite so that it escapes the Earth’s gravitational pull? The radius of the earth = 6400 km and g = 9.8 m · s^-2.
Solution:

Given

A satellite is orbiting in a circular path around the earth close to its surface.

The orbital speed of a satellite in an orbit close to the surface of the earth \(v=\sqrt{g R}\) escape velocity \(v_e=\sqrt{g R}\)

∴Required additional velocity

= \(v_e-v=\sqrt{2 g R}-\sqrt{g R}\)

= \((\sqrt{2}-1) \sqrt{g R}=0.41 \times \sqrt{9.8 \times 6400 \times 1000}\)

= \(3.25 \times 10^3 \mathrm{~m} \cdot \mathrm{s}^{-1}=3.25 \mathrm{~km} \cdot \mathrm{s}^{-1} \).

Example 5. Two satellites A and B of the same mass revolve around the Earth in circular orbits. They are at heights of R and 3R respectively from the surface of the earth (R = radius of the earth). Find the ratio of their kinetic energies and potential energies.
Solution:

Given

Two satellites A and B of the same mass revolve around the Earth in circular orbits. They are at heights of R and 3R respectively from the surface of the earth (R = radius of the earth).

The height of satellite A from the centre of the earth = R + R = 2R and the height of satellite B from the centre of the earth = R + 3R = 4R. If the mass of each satellite is m and the mass of the earth is M,

potential energy of satellite A, U_A = –\(\frac{G M m}{2 R}\)

and potential energy of satellite B, U_B = –\(\frac{G M m}{4 R}\)

∴ \(\frac{U_A}{U_B}=\frac{G M m / 2 R}{G M m / 4 R}=\frac{2}{1}\)

If the orbital speeds of the two satellites are v₁ and v₂, then the kinetic energy of satellite A, K_A = \(\frac{1}{2} m v_1^2\)

and kinetic energy of B, K_B = \(\frac{1}{2} m v_2^2\)

A, \( v_1=\sqrt{\frac{G M}{2 R}} \text { and } v_2=\sqrt{\frac{G M}{4 R}}\)

∴ \(\frac{K_A}{K_B}=\frac{v_1^2}{v_2^2}=\frac{G M / 2 R}{G M / 4 R}=\frac{2}{1} .\)

Example 6. Two satellites S₁ and S₂ are orbiting around the Earth in circular orbits in the same direction. Time period for the two satellites is 1h and 8h respectively. The radius of the orbit of satellite S₁ is 10⁴ km. If satellites S₁ and S₂ are on the same side of the earth, find the linear and angular speeds of S₂ with respect to S₁.
Solution:

Given

Two satellites S₁ and S₂ are orbiting around the Earth in circular orbits in the same direction. Time period for the two satellites is 1h and 8h respectively. The radius of the orbit of satellite S₁ is 10⁴ km. If satellites S₁ and S₂ are on the same side of the earth,

If the radii of satellites S₁ and S₂ are r₁ and r₂ and their respective time periods are T₁ and T₂, then

⇒ \(T_1^2 \propto r_1^3 \text { and } T_2^2 \propto r_2^3\)

∴ \(\frac{T_1^2}{T_2^2}=\frac{r_1^3}{r_2^3} \text { or, }\left(\frac{1}{8}\right)^2=\left(\frac{10^4}{r_2}\right)^3\)

or, \(\frac{1}{64}=\left(\frac{10^4}{r_2}\right)^3 \text { or, }\left(\frac{1}{4}\right)^3=\left(\frac{10^4}{r_2}\right)^3\)

∴ \(r_2=4 \times 10^4 \mathrm{~km}\)

The speed of satellite \(S_1\), \(v_1=\frac{2 \pi r_1}{T_1}\) = \(\frac{2 \pi \times 10^4}{1}\)

= \(2 \pi \times 10^4 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

and the speed of satellite \(S_2\), \(v_2=\frac{2 \pi r_2}{T_2}=\frac{2 \pi \times 4 \times 10^4}{8}\)

= \(\pi \times 10^4 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

∴ Linear speed of \(S_2\) with respect to \(S_1\)

= \(\left|v_2-v_1\right|=\pi \times 10^4 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

and angular speed of \(S_2\) with respect to \(S_1\),

ω = \(\frac{\left|v_2-v_1\right|}{r_2-r_1}=\frac{\pi \times 10^4}{4 \times 10^4-10^4}\)

= \(\frac{\pi}{3} \mathrm{rad} \cdot \mathrm{s}^{-1}.\)

Escape Velocity

WBBSE Class 11 Escape Velocity Notes

When an object is thrown vertically upwards, it moves up to a certain height and then comes down to the earth’s surface due to the gravitational pull of the earth.

At the maximum height, the velocity of the object becomes zero. The higher the projection velocity, the greater the height to which the object rises before its velocity becomes zero.

• If the velocity of projection continuously increases, a state is reached when the velocity of the object (for a finite height) does not become zero.
• This implies that it does not return to the earth’s surface. At this state, the object recedes from the earth’s surface and the force of gravity also decreases to almost a negligible value.
• Hence, it can be said that the body reaches an infinite distance from the earth. The minimum velocity of projection required for this to happen is called the escape velocity.
• It is to be noted that this concept of escape velocity is not only applicable to objects projected from the Earth’s surface but also valid for objects projected from the surface of other celestial bodies.

Read and Learn More: Class 11 Physics Notes

Escape Velocity Formula and Derivation

Escape Velocity Definition: The minimum velocity required to project a body from the surface of the earth or a planet or a satellite such that it can escape the gravitational attraction of the earth or the planet or the satellite is called escape velocity.

Receding from the gravitational attraction of a planet or satellite implies that the projected body does not return to the planet or satellite any more.

Let the mass of the earth = M and its radius = R. The force of gravity on a body of mass m at a distance x from the
centre of the earth = \(\frac{G M m}{x^2}\)

For a small displacement dx of the body against this force of gravity, work done = \(\frac{G M m}{x^2}\)dx

Hence, the work done to displace the body from the earth’s surface to an infinite distance

= \(\int_R^{\infty} \frac{G M m}{x^2} d x=G M m\left[-\frac{1}{x}\right]_R^{\infty}=\frac{G M m}{R}\)

If the escape velocity from the earth is v_e, the initial kinetic energy of the body is \(\frac{1}{2} m v_e^2\). When this energy does the work as calculated above, the body recedes from the earth’s gravitational attraction.

∴ \(\frac{1}{2} m v_e^2=\frac{G M m}{R} \text { or, } v_e^2=\frac{2 G M}{R}\)

or, \(v_e=\sqrt{\frac{2 G M}{R}}\)…(1)

The acceleration due to gravity on the Earth’s surface g= \(\frac{GM}{R^2}\) or, GM= gR²

∴ \(v_e=\sqrt{2 g R}\)…..(2)

Equation (2) gives the value of the escape velocity for a body from the earth’s surface. Note that, the escape velocity does not depend on the mass of the body. This means that for a small and a massive body, the value of the escape velocity is the same.

It is known that the acceleration due to gravity on the earth’s surface g = 9.8 m · s^-2 and the radius of the earth R = 6400 km = 6.4 x 10⁶m.

∴ \(v_e=\sqrt{2 \times 9.8 \times 6.4 \times 10^6}\)

= 11200 m · s^-1 = 11.2 km· s^-1

Factors Affecting Escape Velocity

Hence, if a body of any mass is projected upwards from the surface of the earth with a velocity of 11.2 km · s^-1, it moves out of the gravitational field of the earth, i.e., the body does not return to the earth’s surface.

Equation (2) can also be used for calculating the escape velocity from the surface of any planet or satellite by putting the corresponding values of g and R. For example, the value of g on the surface of the moon is \(\frac{1}{6}\)th of the value on the surface of the earth and the moon’s radius is \(\frac{1}{4}\)th that of the earth.

Putting these values in equation (2), the value of the escape velocity from the moon’s surface can be evaluated and the value is about 2.4 km · s^-1. This value is about \(\frac{1}{5}\)th the value of the escape velocity from the earth’s surface.

Applications of Escape Velocity in Space Exploration

Scarcity Of Light Gases In The Earth’s Atmosphere: When the Earth was formed, the Earth’s atmosphere was filled with lighter gases such as hydrogen and helium along with heavier gases like oxygen, nitrogen, etc. At present, there is practically no existence of lighter gases. The scarcity of lighter gases can be explained in the light of escape velocity.

• From the kinetic theory of gases, it is known that the rms speed of hydrogen at STP is nearly 1.6 km · s^-1. During the formation stage, the earth was warmer and the rms speed of hydrogen had a value close to 5 km · s^-1. Obviously, many hydrogen molecules had velocities more than or equal to the escape velocity (11km · s^-1).
• Hence, gradually with the passage of time, most of the hydrogen molecules escaped the gravitational field of the Earth and went into space. The same was the fate of helium. On the other hand, heavy molecules have very low rms velocity of thermal motion.
• As the value is much lower than the value of the escape velocity, heavy gas molecules remained confined to the earth’s atmosphere. Hence, at present, the Earth’s atmosphere is primarily made up of heavier gases like oxygen, nitrogen and carbon dioxide.
• In the case of the moon or Mercury, owing to their low masses and shorter radii, the value of the escape velocity from the surface is low. As the rms speeds of both the lighter and heavier molecules are comparable to the escape velocity, these gas molecules could escape.

Hence, there is no atmosphere on the moon or on Mercury. On the other hand, the escape velocity for Jupiter or Saturn is of much higher value because of the higher mass and size. Thus, neither lighter nor heavier gas molecules could escape the gravitational attraction of these planets. So, there is an abundance of hydrogen and helium gas on these planets.

Unit 6 Gravitation Chapter 1 Newtonian Gravitation And Planetary Motion

Escape Velocity Numerical Examples

Short Answer Questions on Escape Velocity
Real-Life Examples of Escape Velocity

Example 1. What is the escape velocity of a meteorite situated 1800 km above the surface of the earth? Given, the radius of the earth = 6300 km and the acceleration due to gravity on the earth’s surface = 9.8 m · s^-2.
Solution:

Given, the radius of the earth = 6300 km and the acceleration due to gravity on the earth’s surface = 9.8 m · s^-2.

Radius of the earth (R) = 6300 km = 63 x 10⁵ m

Distance of the meteorite from the centre of the earth (r) = 6300 + 1800 = 8100 km = 81 x 10⁵ km

Acceleration due to gravity on the earth’s surface, g = \(\frac{G M}{R^2}\);

acceleration due to gravity at the position of the meteorite, g’ = \(\frac{G M}{r^2}\).

∴ \(\frac{g}{g^{\prime}}=\frac{r^2}{R^2} \text { or, } g^{\prime}=\frac{g R^2}{r^2}=g\left(\frac{R}{r}\right)^2\)

Escape velocity of the meteorite, \(v_e^{\prime}=\sqrt{2 g^{\prime} r}=\sqrt{2 g\left(\frac{R^2}{r^2}\right)} r=R \sqrt{\frac{2 g}{r}}\)

= \(63 \times 10^5 \times \sqrt{\frac{2 \times 9.8}{81 \times 10^5}}\)

= \(98 \times 10^2 \mathrm{~m} \cdot \mathrm{s}^{-1}=9.8 \mathrm{~km} \cdot \mathrm{s}^{-1} .\)

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Example 2. A man can jump up to a height of 1.5 m on the earth’s surface. What should be the radius of a planet having the same average density as that of the Earth so that man can come out of the gravitational field of that planet in one jump? The radius of the earth is 6400 km.
Solution:

Given

The radius of the earth is 6400 km.

A man can jump up to a height of 1.5 m on the earth’s surface.

The man can jump up to a height h on the earth’s surface.

Hence, his initial kinetic energy = his potential energy at height h.

Thus, \(\frac{1}{2}\)mv² = mgh [m = mass of the man, v = initial velocity]

or, v = \(\sqrt{2 g h}\)….(1)

The man can jump on the surface of any planet with this velocity v. If v equals the escape velocity from a planet, the man can go out of the gravitational pull of that planet. Let the acceleration due to gravity for a planet be g’, its radius Rf and average density ρ, then

g’ = \(\frac{4}{3} \pi G R^{\prime} \rho\)

As \(\frac{4}{3} \pi\)G and ρ are constants, g’ ∝ R’.

Hence, \(\frac{g^{\prime}}{g}=\frac{R^{\prime}}{R} \text { or, } g^{\prime}=g \cdot \frac{R^{\prime}}{R}\)………(2) [R = radius of the earth]

Escape velocity for the planet is \(v=\sqrt{2 g^{\prime} R^{\prime}}\)…..(3)

From (1) and (3), \(\sqrt{2 g h}=\sqrt{2 g^{\prime} R^{\prime}} \text { or, } g h=g^{\prime} R^{\prime}\)

or. \(g^{\prime}=g \cdot \frac{h}{R^{\prime}}\)….(4)

From (2) and (4), \(\frac{g R^{\prime}}{R}\)=\(\frac{g h}{R^{\prime}}\)

or, \(R^{\prime}=\sqrt{h R}=\sqrt{1.5 \times\left(6400 \times 10^3\right)}\)

= \(3098 \mathrm{~m}=3.098 \mathrm{~km} .\)

Variations In Acceleration Due To Gravity

Acceleration due to gravity is not a constant. It changes due to the following reasons:

1. The earth is an oblate spheroid; hence, the value of R is different at different points on the earth’s surface. Thus, the values of g at different places are different.
2. The value of g differs at different heights from the earth’s surface.
3. The value of g differs at different depths below the earth’s surface.
4. Due to the diurnal motion of the earth, the value of g is different at different points on the surface of the earth.

Effect Of The Earth’s Oblate Spheroidal Shape: The earth is not a perfectly spherical body. It is oblate spheroidal in shape. The Arctic and the Antarctic regions are slightly depressed and the equatorial region is a little bulging. Hence, the equatorial radius is greater than the polar radius by about 21 km.

We know that, g ∝  \(\frac{1}{r^2}\), i.e., the value of g is inversely proportional to the square of the distance of any point on the earth’s surface from the centre of the earth. Now, as the polar region is closer to the centre, the value of g is greater than that in the equatorial region.

Read and Learn More: Class 11 Physics Notes

Variation With Altitude: Let the value of the acceleration due to gravity on the earth’s surface be g; the value of the acceleration due to gravity at a point P situated at height h from the earth’s surface be g’ and the distance of the earth’s surface from the centre of the earth O, i.e., the radius of the earth be R.

Gravity Variation with Depth Below Earth’s Surface

Hence, the distance of P from O = R+h. As the magnitude of the acceleration due to gravity is inversely proportional to the square of the distance of the point from the centre of the earth,

⇒ \(\frac{g^{\prime}}{g}=\frac{R^2}{(R+h)^2}\)

= \(\frac{1}{\left(1+\frac{h}{R}\right)^2}=\left(1+\frac{h}{R}\right)^{-2}=\left(1-\frac{2 h}{R}\right)\)

[higher terms in the above binomial expansion are neglected as the value of h is usually too small in comparison to that R]

∴ \(g^{\prime}=g\left(1-\frac{2 h}{R}\right)\)….(2)

This is the relationship between the value of acceleration due to gravity at a height h from the earth’s surface and that on the earth’s surface. Equation (2) shows that g’ decreases with the increase in h, which means that the acceleration due to gravity decreases with height from the earth’s surface.

Now, g – g’ = decrease in the acceleration due to gravity at a height h from the surface of the earth

= \(\frac{2 h g}{R}\)….(3)

Note that equation (2) is applicable when h<<R, but when h is comparable to R, equation (1) has to be applied.

WBBSE Class 11 Variations in Acceleration Due to Gravity Notes

Variation With Depth: Let the radius of the earth be R and a mass m be kept at a point P situated at a depth h from the surface of the earth.

The point O is the centre of the earth. A sphere is imagined with its centre at O and of radius (R-h). The surface of this imaginary sphere divides the Earth into two parts:

1. An inner spherical core of radius (R-h) and
2. An outer spherical shell of thickness h. It can be proved that the body of mass m, being placed just outside the inner core, experiences a gravitational pull due to this inner core only. Being inside the spherical shell, the mass m does not experience any gravitational pull due to the shell. Hence, to determine the gravitational pull, only the inner spherical core is to be taken into consideration.

The average density of the earth, \(\rho=\frac{\text { mass of the earth }}{\text { volume of the earth }}=\frac{M}{\frac{4}{3} \pi R^3}=\frac{3 M}{4 \pi R^3}\)

Volume of the inner spherical core =\( \frac{4}{3} \pi(R-h)^3\)

Hence, mass of the inner spherical core, \(M^{\prime}=\frac{4}{3} \pi(R-h)^3 \rho=\frac{4}{3} \pi(R-h)^3 \cdot \frac{3 M}{4 \pi R^3}=\frac{M(R-h)^3}{R^3}\)

Hence, the gravitational force acting on the mass m

F = \(\frac{G M^{\prime} m}{(R-h)^2}=\frac{G M m}{R^3}(R-h)\)

So, if the acceleration due to gravity at P, i.e., at a depth h be g’, then \(g^{\prime}=\frac{F}{m}=\frac{G M}{R^3}(R-h)=\frac{G M}{R^2} \frac{(R-h)}{R}=g\left(1-\frac{h}{R}\right)\)…(4)

In this case, g = \(\frac{G M}{R^2}\) = acceleration due to gravity on the surface of the earth.

From equation (4), we conclude that,

1. The acceleration due to gravity from the centre of the earth to its surface is directly proportional to the distance (R-h) of the place from the centre of the earth. Thus the value of the acceleration due to gravity below the earth’s surface (g’) is less than that on the surface of the earth, and as depth h increases, the value of g’ decreases.
2. At the centre of the earth, g’ = 0, because h = R. Hence, the value of the acceleration due to gravity at the centre of the earth is zero. In fact, when an object is kept at the centre of the earth, it is subjected to gravitational forces acting uniformly from all directions and the resultant of these forces is zero. Hence, the acceleration due to gravity is also zero.
3. g-g’ = decrease in the acceleration due to gravity at a depth h from the earth’s surface = \(\frac{h g}{R}\)…(5)

Understanding Acceleration Due to Gravity Variations

It is seen from equations (2) and (4) that the value of the acceleration due to gravity decreases with height above the surface of the earth, as well as with depth below the surface of the earth. Hence, the acceleration due to gravity is maximum on the surface of the earth.

Comparing equations (3) and (5), it can be said that the acceleration due to gravity at a height h from the surface of the earth is less than that at a depth h, i.e., g decreases with height faster rate than with depth.

The graph shows the variation of g with an increase in distance from the centre of the earth. From the centre of the earth to the earth’s surface, the value of g is directly proportional to the distance and is represented by a straight line starting from the origin. After this, the value of g decreases with distance along the curved line PQ.

Effect Of The Earth’s Diurnal Motion: For its diurnal motion, the Earth rotates about its NS axis once in 24 hours. Therefore, except the objects at the poles, all objects on the surface of the earth rotate in a circular path. The centres of these circular paths lie on the earth’s NS axis.

Taking the surface of the earth as a reference frame, it can be stated that during circular motion, a centrifugal force acts on an object situated on the earth’s surface and partially balances the gravitational force on the body acting towards the centre of the earth.

This causes an apparent decrease in the weight of the body. As a result, the effective value of acceleration due to gravity is less compared to its actual value. The actual value means that of the acceleration due to gravity in the absence of the diurnal motion of the earth.

Let the radius of the earth be R, and the latitude of a point A on the surface of the earth be θ. A mass m is kept at A. The distance of the mass from the axis of rotation NS is r = Rcosθ. As the earth rotates with angular velocity ω about its own axis NS, in a circular path of radius r, the centrifugal force on the mass along the radius r acting away from the centre of the earth is mω²r.

Effects of Altitude on Acceleration Due to Gravity

Also, if g is the actual value of the acceleration due to gravity at A, the force of attraction directed towards the centre of the earth on the body = mg. At the same time, the component of the centrifugal force, directed away from the centre of the earth at that point = mω²rcosθ = mω²Rcos²θ. Hence, the resultant force on the mass m towards the centre of the earth,

F = \(m g-m \omega^2 R \cos ^2 \theta \quad \text { or, } F=m g\left(1-\frac{\omega^2 R}{g} \cos ^2 \theta\right)\)

This is the apparent weight or the apparent force of gravity on the mass m at A. Hence, the apparent value of the acceleration due to gravity at A is

g’ = \(\frac{F}{m}=g\left(1-\frac{\omega^2 R}{g} \cos ^2 \theta\right)\)…(6)

The above equation shows that the higher the value of cosθ, the lesser the value of the acceleration due to gravity will be at that place.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Effect Of The Earth’s Diurnal Motion Special cases:

1. At the poles θ = 90°; cosθ= 0

Hence, g’ = g. Therefore, the acceleration due to gravity does not decrease in the polar regions, owing to the diurnal motion of the earth.

2. At the equator θ = 0°; cosθ = 1

Hence, g’ \(=g\left(1-\frac{\omega^2 R}{g}\right) .\). Thus, due to the diurnal motion of the earth, the acceleration due to gravity is the lowest in the equatorial region. Putting R = 6.4 x 10⁶ m, g = 9.8m · s^-2 and \(\omega=\frac{2 \pi}{24 \times 3600} \mathrm{rad} \cdot \mathrm{s}^{-1}\), we get \(\frac{\omega^2 R}{g}=3.454 \times 10^{-3}\) and the value of the acceleration due to gravity in the equatorial region, g’ = g(1-3.454 x 10^-3) = 0.996546g.

Thus, due to the diurnal motion of the earth, the acceleration due to gravity in the equatorial region is 99.6546% of that in the polar region. So, the value decreases only by 0.3454%.

So, owing to the earth’s oblate spheroidal shape and diurnal motion, the acceleration due to gravity is the highest at the poles and the lowest at the equator.

Unit 6 Gravitation Chapter 1 Newtonian Gravitation And Planetary Motion

Variations In Acceleration Due To Gravity Numerical Examples

Short Answer Questions on Gravity Variation

Example 1. What will be the value of the acceleration due to gravity at a point 3 km above the earth’s surface? The diameter of the earth = 12800 km and g = 980 cm · s^-2 on the surface of the earth.
Solution:

Radius of the earth R = \(\frac{12800}{2}\) = 6400 km.

As the height of the point (h = 3 km) is negligibly small in comparison to R, the acceleration due to gravity at that height,

g’ = \(g\left(1-\frac{2 h}{R}\right)=980\left(1-\frac{2 \times 3}{6400}\right)=980 \times \frac{6394}{6400}\) = 979.08 cm · s^-2.

Example 2. An object of mass m is raised up to a small height h above the earth’s surface. If the acceleration due to gravity on the earth’s surface is g, prove that the weight of the object will decrease by \(\frac{2mgh}{R}\) with respect to that on the earth’s surface. R = radius of the earth.
Solution:

Acceleration due to gravity at a small height h above the earth’s surface is given by

g’ = \(g\left(1-\frac{2 h}{R}\right)\)

Hence, weight at that height, \(m g^{\prime}=m g\left(1-\frac{2 h}{R}\right)=m g-\frac{2 m g h}{R}\)

or, \(m g-m g^{\prime}=\frac{2 m g h}{R}\)

So, the decrease in weight with respect to that on the earth’s surface = \(\frac{2mgh}{R}\)

Example 3. At what depth below the surface of the earth, will the acceleration due to gravity decrease by 1% with respect to that on the earth’s surface? The earth can be taken as a uniform sphere of radius 6400 km.
Solution:

If the acceleration due to gravity on the earth’s surface is g, then at a depth h below the earth’s surface,

g’ = \(g\left(1-\frac{h}{R}\right)=g-\frac{g h}{R}\)

∴ Decrease in value, g – g’ = \(\frac{g h}{R}\)

As per the question, \(\frac{g h}{R}\) = 0.01 g

or, h = 0.01 x R = 0.01 x 6400 = 64 km..

Example 4. The acceleration due to gravity at a height h above the surface of the earth is the same as its value at a t depth d below the earth’s surface. Find the relationship between d and h.
Solution:

Acceleration due to gravity at a height h, \(g_1=g\left(1-\frac{2 h}{R}\right)\)

Also, the value of acceleration due to gravity at a depth d, \(g_2=g\left(1-\frac{d}{R}\right)\)

Given \(g_1=g_2 or, g\left(1-\frac{2 h}{R}\right)=g\left(1-\frac{d}{R}\right) or, d=2 h\).

Example 5. If the present radius of the earth is doubled keeping the mass unchanged, how will the weight of a body on the surface of the earth change?
Solution:

Weight of a body of mass m on the earth’s surface, W = mg = \(\frac{G M m}{R^2}\), where M and R are the mass and the radius of the earth.

When mass remains constant and radius doubles, then weight \(W_1=\frac{G M m}{(2 R)^2}=\frac{G M m}{4 R^2}\)

∴ \(\frac{W_1}{W}=\frac{G M m}{4 R^2} \cdot \frac{R^2}{G M m}=\frac{1}{4}\)

Hence, the new weight would be \(\frac{1}{4}\) th of the present weight.

Example 6. What is the difference in the values of acceleration due to gravity at the polar region and at the equator due to the diurnal motion of the Earth? The radius of the earth = 6400 km
Solution:

Angular velocity of diurnal motion of the earth \(\omega=\frac{2 \pi}{24 \times 60 \times 60} \mathrm{rad} \cdot \mathrm{s}^{-1}\)

Radius of the earth = 6400 km = 6.4 x 10⁸ cm

Acceleration due to gravity at latitude θ, g’ = \(g\left(1-\frac{\omega^2 R}{g} \cos ^2 \theta\right)\)

= \(g-\omega^2 R \cos ^2 \theta\)

In the polar regions θ = 90°, i.e., cosθ = 0.

Therefore, the acceleration due to gravity is g₁ = g.

At the equator θ = 0° and cosθ = 1; thus the acceleration due to gravity is g₂ = g – ω²R.

So, the difference in the values of acceleration due to gravity in these two regions

⇒ \(g_1-g_2=\omega^2 R=\frac{(2 \pi)^2 \times 6.4 \times 10^8}{(24 \times 60 \times 60)^2}=3.38 \mathrm{~cm} \cdot \mathrm{s}^{-2} \text {. }\)

Mathematical Formulas for Gravity Variation

Example 7.  For what value of the angular velocity of the earth, the acceleration due to gravity in the equatorial region would have been zero? The average density of the earth’s material = 5.5 g · cm^-3 and G = 6.67 x 10^-8 CGS unit. Find the ratio of this calculated value and the present value of the angular velocity of the earth.
Solution:

The acceleration due to gravity at the equator g’ = g – ω²R

For g’ to be zero, the value of ω₀ should be such that \(\)

As the average density of the earth is ρ and its radius is R, the mass of the earth M = \(\frac{4}{3} \pi R^3\)

Also, g = \(\frac{G M}{R^2}=\frac{G}{R^2} \cdot \frac{4}{3} \cdot \pi R^3 \rho=\frac{4 \pi G R \rho}{3}\)

Hence, the required angular velocity, \(\omega_0=\sqrt{\frac{g}{R}}=\sqrt{\frac{4 \pi G \rho}{3}}\)

∴ \(\omega_0=\sqrt{\frac{4 \times \pi \times 6.67 \times 10^{-8} \times 5.5}{3}}\)

= \(1.24 \times 10^{-3} \mathrm{rad} \cdot \mathrm{s}^{-1}\)

Present angular velocity of the earth \(\omega =\frac{2 \pi}{24 \times 60 \times 60}=7.27 \times 10^{-5} \mathrm{rad} \cdot \mathrm{s}^{-1}\)

∴ \(\frac{\omega_0}{\omega}=\frac{1.24 \times 10^{-3}}{7.27 \times 10^{-5}}=17 \text { (approx) }\)

Hence, if the angular velocity of the diurnal motion of the earth had been 17 times its present value, the acceleration due to gravity at the equator would be zero.

Alternative Method: From known values of ω, R and g, \(\frac{\omega^2 R}{\boldsymbol{g}} \approx \frac{1}{289} \text { or, } \omega=\sqrt{\frac{g}{289 R}} \text {. }\)

If the acceleration due to gravity is zero at the equator, then the required angular velocity, \(\omega_0=\sqrt{\frac{g}{R}}\)

∴ \(\frac{\omega_0}{\omega}=\frac{\sqrt{g / R}}{\sqrt{g /(289 R)}}=\sqrt{289}=17 .\)

Example 8. If the diurnal motion of the earth stops for some reason, what would be the percentage change in the weight of a body at the equator? The radius of the earth = 6400 km and g = 9.8 m · s^-2.
Solution:

Due to the diurnal motion of the earth, the apparent weight of a body at the equator W’ = W-mω²R

When the earth stops rotating, ω = 0 and hence, W’ = W. In this case, the increase in weight of the object,

ΔW=W-W’ = W-(W- mω²R) = mω²R

∴ Percentage increase in weight

= \(\frac{\Delta W}{W} \times 100=\frac{m \omega^2 R}{m g} \times 100=\left(\frac{2 \pi}{T}\right)^2 \times \frac{R}{g} \times 100\)

= \(\frac{4 \pi^2}{(24 \times 60 \times 60)^2} \times \frac{6400 \times 10^3}{9.8} \times 100=0.3454 \% .\)

Example 9. Find the percentage decrease in weight of a body, when taken 16 km below the surface of the earth. Take the radius of the earth as 6400 km.
Solution:

Here, R = 6400 km, x = 16 km.

The value of acceleration due to gravity at a depth x, \(g^{\prime}=g\left(1-\frac{x}{R}\right)=g\left(1-\frac{16}{6400}\right)\)

∴ \(g-g^{\prime}=\frac{1}{400} g=2.5 \times 10^{-3} \mathrm{~g}\)

If m is the mass of the body, then mg and mg’ are the weight of the body on the surface of the earth and at a depth of 16 km below the surface of the earth respectively. Then, the percentage decrease in the weight of the body

= \(\frac{m g-m g^{\prime}}{m g} \times 100\)

= \(\frac{g-g^{\prime}}{g} \times 100=\frac{2.5 \times 10^{-3}}{g} g \times 100=0.25 \%\)

Gravity

WBBSE Class 11 Acceleration Due to Gravity Notes

Gravity Definition: The force of attraction exerted by the earth on the objects on or near the surface of the earth is called gravity.

Because of gravity, a body, when released from a height, starts moving towards the earth’s surface. Clearly, gravity is a form of gravitation.

Let the mass of the earth be M and its radius be R. At a distance of r from the centre of the earth, let there be a body of mass m. Taking the earth as an isotropic, uniform and homogeneous sphere, its entire mass may be considered to be concentrated at its centre. According to the law of gravitation, the force of attraction on the body due to the earth is

F = \(\frac{G M m}{r^2}\)….(1)

From Newton’s second law of motion, it can be stated that the body of mass m, being acted upon by a force F, gets an acceleration a in the direction of F, i.e., towards the centre of the earth. Acceleration produced in a body due to gravity is called acceleration due to gravity. This is denoted by g.

Read and Learn More: Class 11 Physics Notes

Acceleration Due To Gravity: Acceleration produced in a freely falling body under the action of gravity is called acceleration due to gravity.

In the case of a mass m, using Newton’s second law of motion, we have, F = mg

Hence, comparing with equation (1) \(m g=\frac{G M m}{r^2} \text { or, } g=\frac{G M}{r^2}\)….(2)

Both G and M are constants, hence, \(g \propto \frac{1}{r^2}\)…(3)

In other words, the acceleration due to gravity at a point near the surface of the earth is inversely proportional to the square of the distance between the body and the centre of the earth.

It is easily understood from equation (2) that the value of the acceleration due to gravity at a place does not depend on the mass of the body. At any fixed place, the value of acceleration due to gravity remains the same for a light or a heavy body.

Thus points equidistant from the centre of the earth have the same value of g. For any point on the surface of the earth r = R. Hence, from equation (2), the value of the acceleration due to gravity on the earth’s surface is

g = \(\frac{G M}{R^2}\)…(4)

This is the relation between acceleration due to gravity g and the gravitational constant G.

If the average density of the earth is p, then M = \(\frac{4}{3}\)πR³ρ

∴ g = \(\frac{G \cdot \frac{4}{3} \pi R^3 \rho}{R^2}=\frac{4}{3} \pi G R \rho\)

or, ρ = \(\frac{3 g}{4 \pi R G}\)……(5)

This is the relation between the gravitational constant G and the average density of the earth.

Using equations (4) and (5), or by other experiments, the value of the acceleration due to gravity on the surface of the earth can be obtained. The average value of g on the surface of the earth is:

Key Concepts of Acceleration Due to Gravity

CGS System: g = 980.6 cm · s^-2

SI: 9.806 m · s^-2 ≅ 9.8 m · s^-2

Acceleration Of The Body And Acceleration Of The Earth: Let us take the example of two bodies as shown. The earth of mass M pulls a body of mass m towards its centre with a force F. As per the law of gravitation, the mass m also pulls the earth towards it with the same force F. (This is an example of the action and reaction forces being equal and opposite.)

Hence, from Newton’s second law of motion,

acceleration of the body = \(\frac{\text { force on the body }}{\text { mass of the body }}=\frac{F}{m}\)

also acceleration of the earth = \(\frac{\text { force on the earth }}{\text { mass of the earth }}=\frac{F}{M}\)

Therefore, \(\frac{\text { acceleration of the body }}{\text { acceleration of the earth}} =\frac{F / m}{F / M}=\frac{M}{m}\)

As the masses of ordinary objects, situated on or near the surface of the earth, are negligible compared to the mass of the earth, \(\frac{M}{m}\) >>1.

Hence, \(\frac{\text { acceleration of the body }}{\text { acceleration of the earth }}\) >>1

∴ acceleration of a body >> acceleration of the earth

Hence, the acceleration of a body is many times greater than the acceleration of the earth. Practically, it is the body that moves towards the earth; the acceleration of the earth towards the body is so small that it cannot be felt.

Newtonian Gravitation And Planetary Motion

Gravity Numerical Examples

Short Answer Questions on Acceleration Due to Gravity

Example 1. What will be the percentage, increase or decrease in the value of acceleration due to gravity on the surface of the earth, if the radius of the earth decreases by 1 % and the mass of the earth remains unchanged?
Solution:

Acceleration due to gravity on the surface of the earth, for the present value of the earth’s radius R, is g = \(\frac{G M}{R^2}\).

Let R’ = radius of the earth when it decreases by 1%, i.e., R’ = 0.99R.

At that time, the value of the acceleration due to gravity on the earth’s surface is g’ = \(\frac{G M}{R^2}\)

∴ \(\frac{g^{\prime}}{g}=\frac{R^2}{R^{\prime 2}}=\frac{R^2}{(0.99 R)^2}=\frac{1}{(0.99)^2}=1.02\)

Hence, g’ = 1.02g = g + 0.02g

Hence, in this case, the value of acceleration due to gravity on the earth’s surface increases by 2%.

Alternative Method:

We know, g = \(\frac{G M}{R^2}\)

or, \(\frac{d g}{d R}=-G M \frac{2}{R^3}=-2 \frac{G M}{R^2} \cdot \frac{1}{R}=-\frac{2 g}{R}\)

∴ \(\frac{d g}{g}=-2 \frac{d R}{R}\)

Radius decreased by 1% implies that, \(\frac{d R}{R}\) x 100 = -1

∴ Percentage change in g = \(\frac{d g}{g}\) x 100 = -2\(\frac{d R}{R}\) x 100 = 2

Hence, the value of g increases by 2%.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Example 2. The mass of the moon is \(\frac{1}{80}\)th of the mass of the earth and its radius is \(\frac{1}{4}\)th of that of the earth. Compare the accelerations due to gravity on the Earth’s and the Moon’s surfaces.
Solution:

Given

The mass of the moon is \(\frac{1}{80}\)th of the mass of the earth and its radius is \(\frac{1}{4}\)th of that of the earth.

Let the mass of the earth be M and its radius R; the acceleration due to gravity on the earth’s surface is g = \(\frac{G M}{R^2}\).

Similarly, taking the mass and radius of the moon as m and r respectively, acceleration due to gravity on the moon’s surface is g’ = \(\frac{G M}{r^2}\).

∴ \(\frac{g^{\prime}}{g}=\frac{m}{M} \cdot \frac{R^2}{r^2}=\frac{m}{M} \cdot\left(\frac{R}{r}\right)^2=\frac{1}{80} \times\left(\frac{4}{1}\right)^2=\frac{1}{5} .\)

Hence, the value of the acceleration due to gravity on the moon’s surface is \(\frac{1}{5}\)th of that on the surface of the earth.

Variation of Acceleration Due to Gravity with Height

Mass And Average Density Of The Earth

Mass: Let R = radius of the earth, M = its mass and ρ = average density.

From equation (4), M = \(\frac{g R^2}{G}\)

In this case, g = 9.8 m · s^-2; R = 6400 km = 6.4 x 10⁶ m and G = 6.7 x 10^-11 N · m² · kg^-2.

∴ M \(=\frac{9.8 \times\left(6.4 \times 10^6\right)^2}{6.7 \times 10^{-11}}=6.0 \times 10^{24} \mathrm{~kg}\)

This is the measure of the mass of the earth.

Average Density: From equation (5) get ρ = \(\frac{3 g}{4 \pi R G}\)

Substituting the values of g, R and G, we obtain ρ = 5500 kg · m^-3 = 5.5 g · cm^-3.

The density of seawater and of soil on the upper crust of the earth is 1 g · cm^-3 and 2.7 g · cm^-3 respectively. Thus, the value of the average density of the earth (5.5 g · cm^-3) indicates that the earth is not a sphere of uniform mass density.

The density of materials in the inner layers is definitely high. In fact, the density at the central core is about 13 g · cm^-3. It is, therefore, understood that the inner layers of the earth’s crust consist of heavy materials like metals.

Gravitational Field And Gravitational Potential And Gravitational Potential Energy

WBBSE Class 11 Gravitational Field Notes

The space surrounding a body where gravitational attraction can be felt due to the mass of the body, i.e., the space in which the body exerts a gravitational force on any other body, is called its gravitational field.

Actually, the gravitational field due to any object is extended up to infinity. However, the space beyond a certain distance, where the gravitational force becomes negligible in reality, may not be considered as the gravitational field of the body.

Gravitational Field: Along with the identification of the space that acts as the gravitational field of the body, the measurement of the field at every point of that space is also important. The physical quantity that defines the field at any point is called the gravitational field intensity, which in many cases, is mentioned only as gravitational field.

Gravitational Field Intensity: The gravitational force acting on a body of unit mass, placed at any point in a gravitational field, is called the gravitational field intensity of that point.

Read and Learn More: Class 11 Physics Notes

Calculation Of Gravitational Field Due To A Point Mass: Let a particle of mass M is placed at point O and a second particle of mass m is placed at point P which is at a distance r apart. The mass M creates a field \(\vec{E}\) at the position of mass m which results in a force on the mass m. The force due to the field \(\vec{E}\) is,

⇒ \(\vec{E}\) = m\(\vec{E}\)

But the force \(\vec{F}\) on the mass m due to the mass M is, \(F=\frac{G M m}{r^2} \text {, acting along } \overrightarrow{P O}\)

Thus, the gravitational field at P is, \(E=\frac{G M}{r^2} \text { along } \overrightarrow{P O}\)

In vector notation, it can be written as, \(\vec{E}=-\frac{G M}{r^2} \hat{r}\)

where \(\hat{r}\) is the unit vector along \(\overrightarrow{O P} \text {. }\).

The negative sign characterises the attractive nature of the gravitational field. It is to be noted that

Understanding Gravitational Field Strength

1. Both force and acceleration are vector quantities, and so the gravitational field intensity is also a vector quantity.
2. A gravitational force can be felt at any point even due to an assembly of point masses instead of a single point mass. In that case, the gravitational field intensities due to all the masses are to be added by the vector addition method to get the total intensity,
3. If the gravitational field intensity at a point, i.e., the force on a unit mass placed at that point is E, the force that will act on a mass m, placed at that point, is F = mE.

Gravitational Field Intensity Formula

Gravitational Field Units:

• CGS System: cm · s^-2
• SI: m · s^-2

Gravitational Potential: The work done by an external agent to bring a unit mass from infinity to a point in the gravitational field is called the gravitational potential at that point.

Gravitational Potential Units:

CGS System: erg · g^-1

SI: J · kg^-1

At an infinite distance, where no gravitational field exists, the gravitational potential is taken to be zero. To bring a unit mass from that point to a point in the gravitational field, work is done by the gravitational force alone.

As displacement occurs in the same direction as that of force, by convention, work done is positive. We know that change in potential or potential energy is, therefore, negative. So it can be said that, as the gravitational force is always attractive, the gravitational potential at any point in a gravitational field is negative.

Expression For The Gravitational Potential: Let the point O be the centre of a body of mass M, taken as the origin. The gravitational potential at a point P, in the gravitational field of the mass M, is to be determined,

Let r = distance OP, \(\hat{r}\) = unit vector along OP,

so that, \(\overrightarrow{O P}=\vec{r}=r \hat{r}\)

Now, \(\vec{E}=-\frac{G M}{r^2} \hat{r}\) = gravitational field intensity,

and \(d \vec{r}=-d r \hat{r}\) = an infinitesimal displacement from infinity towards point P.

So, the work done to displace a unit mass by \(d \vec{r}\) is,

∴ \(\vec{E} \cdot d \vec{r}=-\frac{G M}{r^2} \hat{r} \cdot(-d r \hat{r})=\frac{G M}{r^2} d r\) (because \(\hat{r} \cdot \hat{r}=1\))

Applications of Gravitational Fields in Physics

The gravitational potential (V) at the point P = work done to bring a unit mass from infinity to the point P, at a distance r from M.

So, V = \(\int_{\infty}^r \vec{E} \cdot d \vec{r}=G M \int_{\infty}^r \frac{d r}{r^2}=G M\left[-\frac{1}{r}\right]_{\infty}^r\)

i.e., V=\(-\frac{G M}{r}\)

The gravitational field intensity and the potential are related as \(\vec{E}=-\frac{d V}{d r} \hat{r}\).

It is to be noted that

1. As work is a scalar quantity, gravitational potential is also a scalar quantity.
2. If V is the gravitational potential at a point, then the potential energy of a body of mass m kept at that point = mV.

Gravitational Potential Energy Definition: The gravitational potential energy of a body at a point is defined as the amount of work done in bringing the body from infinity to that point in the gravitational field.

Gravitational Potential Energy Unit:

• CGS System: erg
• SI: joule

Whenever two material bodies are kept at a finite distance apart, the system possesses potential energy. It is because some work must have been done to bring them at finite distance apart from infinity.

Potential energy is positive, if the force between the two bodies is repulsive and negative if the force between them is attractive in nature. In the case of two material bodies, this potential energy is called gravitational potential energy.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Expression For The Gravitational Potential Energy: Let us consider that the earth is a uniform sphere of radius R and mass M. We have to calculate the gravitational potential energy of the body of mass m at a point A, such that OA = r

By definition, the gravitational potential energy of the body at point A,

U = work done in bringing the body from infinity to point A. Suppose, at any instant of time, the body is situated at the point P, such that OP = x.

Difference Between Gravitational Field and Gravitational Force

Then the gravitational force on the body at the point P is given by, F = \(\frac{G M m}{x^2}\)

Now, let the body be displaced by an infinitesimally small distance dx. The small amount of work done is given by \(\)

Therefore, the total work done in bringing the body from infinity to point A,

W = \(\int_{\infty}^r \frac{G M m}{x^2} d x=G M m \int_{\infty}^r x^{-2} d x=-G M m\left[\frac{1}{x}\right]_{\infty}^r\)

= \(-G M m\left(\frac{1}{r}-\frac{1}{\infty}\right)=-\frac{G M m}{r}\)

This work is stored in the body as gravitational potential energy.

∴ \(W=U=\frac{-G M m}{r}\)

It should be noted that, as the distance r increases, the gravitational potential energy becomes less negative, i.e., increases and it becomes zero, i.e., maximum, when r = ∞

Newtonian Gravitation And Planetary Motion – Gravitational Numerical Examples

Example 1. Two small but heavy spheres, each of mass M are on a horizontal plane separated by a distance r. Find the gravitational potential at the mid-point of the line joining the centres of the two spheres.
Solution:

Gravitational potential at the mid-point of the line and joining the two centres, \(V=\frac{-G M}{r / 2}+\frac{-G M}{r / 2}=-\frac{2 G M}{r}-\frac{2 G M}{r}=-\frac{4 G M}{r}.\)

Example 2. At the vertices of an equilateral triangle of side a, three particles each of mass m are kept. Find the gravitational potential and the gravitational field at the centroid of the triangle.
Solution:

Here AB = BC = CA = a; perpendiculars drawn from the points A, B and C on respective opposite sides meet at the centroid O and proceed to bisect the sides BC, CA and AB respectively.

Let AO = BO = CO = r

From ΔABD, AD = a sin60° = \(\frac{\sqrt{3}}{2} a\)

Also, AO = \(\frac{2}{3}\) AD

∴ \(r=\frac{2}{3} \times \frac{\sqrt{3}}{2} a=\frac{a}{\sqrt{3}} .\)

Hence the gravitational potential at the centroid O is \(V=-\frac{G m}{r}+\frac{-G m}{r}+\frac{-G m}{r}=-\frac{3 G m}{\frac{a}{\sqrt{3}}}=-\frac{3 \sqrt{3} G m}{a}\)

Let the gravitational field intensities at O due to masses at A, B and C be F_A (directed from O to A), F_B (directed from O to B) and F_C (directed from O to C) respectively.

∴ \(F_A=\frac{G m}{r^2}=F_B=F_C\)

Resolving the intensities along and perpendicular to the direction of AO, we get,

⇒ \(F_A-F_B \cos 60^{\circ}-F_C \cos 60^{\circ}=F_A-\frac{F_B}{2}-\frac{F_C}{2}=0\) (because \(F_B=F_C=F_A\))

and \(0+F_C \sin 60^{\circ}-F_B \sin 60^{\circ}=\frac{F_{C^{\sqrt{3}}}}{2}-\frac{F_B \sqrt{3}}{2}=0\) (because \(F_B=F_C\))

Hence, the gravitational field intensity at the centroid of the triangle = 0.

Short Answer Questions on Gravitational Field

Example 3. Find the potential energy of a system of four particles of equal masses placed at the comers of a square of side l. Also, obtain the potential at the centre of the square.
Solution:

Four particles, each of mass m, are placed at the four corners of a square ABCD of side l as shown.

Here AO = CO = BO = DO = \(\frac{l}{\sqrt{2}}\)

We know that the gravitational potential energy associated with two particles of masses m₁ and m₂ at a distance r is -G\(\frac{m_1 m_2}{r}\)

Here, masses at the four corners are at a distance l from one another and the two diagonal pairs are at a distance \(\frac{l}{\sqrt{2}}+\frac{l}{\sqrt{2}}=\sqrt{2} l\)

So, the gravitational potential energy of the whole system,

⇒ \(E_p=4\left(-\frac{G m^2}{l}\right)+2\left(-G \frac{m^2}{\sqrt{2} l}\right)=-2 G \frac{m^2}{l}\left(2+\frac{1}{\sqrt{2}}\right)\)

= \(-5.41 \frac{G m^2}{l}\)

Gravitational potential at the centre (O) of the square is \(V=4\left(-\frac{G m}{\frac{l}{\sqrt{2}}}\right)=-4 \sqrt{2} \frac{G m}{l}\)

Real-Life Examples of Gravitational Fields

Example 4. Four particles of masses m, 2m, 3m and 4m are placed at the four corners of a square of side a. Find the gravitational force on a particle of mass m placed at the centre of the square.
Solution:

Given

Four particles of masses m, 2m, 3m and 4m are placed at the four corners of a square of side a.

Let O be the centre of the square ABCD of side a.

∴ BD = \(\sqrt{a^2+a^2}=\sqrt{2} a\)

Here, OA = OB = OC = OD = \(\frac{B D}{2}=\frac{a}{\sqrt{2}}=x \text { (say). }\)

Suppose, the particle of mass m placed at O experiences gravitational forces F₁, F₂, F₃, and F₄ due to the particles placed at A, B, C, and D respect

Now, \(F_1=G \frac{m \times m}{x^2}\) along \(\overrightarrow{O A}\)

⇒ \(F_2=G \frac{m \times 2 m}{x^2} \text { along } \overrightarrow{O B}\)

⇒ \(F_3=G \frac{m \times 3 m}{x^2} \text { along } \overrightarrow{O C}\)

⇒ \(F_4=G \frac{m \times 4 m}{x^2} \text { along } \overrightarrow{O D}\)

∴ The resultant of \(F_1\) and \(F_3\),

⇒ \(F^{\prime}=G \frac{m \times 3 m}{x^2}-G \frac{m \times m}{x^2}=G \frac{2 m^2}{x^2} \text { along } \overrightarrow{O C}\)

and the resultant of \(\mathrm{F}_2\) and \(\mathrm{F}_4\), \(F^{\prime \prime}=G \frac{m \times 4 m}{x^2}-G \frac{m \times 2 m}{x^2}=G \frac{2 m^2}{x^2} \text { along } \overrightarrow{O D}\)

Here, \(F^{\prime}\) and \(F^{\prime \prime}\) are equal in magnitude and at right angles to each other. The resultant of \(F^{\prime}\) and \(F^{\prime \prime}\),

F= \(\sqrt{F^{\prime 2}+F^{\prime \prime 2}}\) = \(F^{\prime}\sqrt{2}\)

= \(\frac{G .2 m^2}{x^2} \times \sqrt{2}=\frac{G .2 m^2}{\frac{a^2}{2}} \times \sqrt{2}=4 \sqrt{2} \frac{G m^2}{a^2}\)

Example 5. What will be the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth?
Solution:

Let M be the mass of the earth. Then, the gravitational potential energy of the object of mass m at the surface of the earth,

⇒ \(U_i=-\frac{G M m}{R}\)

and the potential energy of the object, when taken to a height equal to R,

⇒ \(U_f=-\frac{G M m}{R+R}=-\frac{G M m}{2 R}\)

∴ Potential energy gain = \(U_f-U_i\)

= \(-\frac{G M m}{2 R}-\left(-\frac{G M m}{R}\right)=\frac{G M m}{2 R}\)

Motion Of Planets And Satellites

WBBSE Class 11 Motion of Planets and Satellites Notes

Newton’s law of gravitation can be applied to arrive at Kepler’s laws theoretically. Using Newton’s law of gravitation, it can also be shown that the revolution of an object, natural or artificial (like a satellite), around another object (say, a planet) is similar to the motion of planets revolving around the sun.

• This means that satellites also follow Kepler’s laws within the parameter that the mass of the revolving object should be negligible compared to the mass of the object around which it revolves as in the case of the solar system.
• It should be noted that the eccentricities of the orbits of the planets in the solar system are very law, i.e., the orbits are almost circular. As a circle is a special case of ellipse (for circles, e = 0 ), Kepler’s laws are equally applicable for circular orbits. For the case of calculation, in the following cases, the orbits are taken as circular.
• All of Kepler’s laws are valid for circular and elliptical orbits, i.e., for orbits with 0 ≤ e < 1. As parabolas and hyperbolas are open conic sections, neither Kepler’s first law nor his third law applies to them. For example, a comet that makes a single pass by the sun follows a parabolic or hyperbolic path.

Orbital Speed And Period Of Revolution Of Planets: Let a planet of mass m revolve in a circular orbit of radius r, with the sun of mass M₀ at the centre of the orbit.

Orbital Speed: The speed With which a planet moves in its orbit is called its orbital speed(v). In the case of a circular orbit, the value of this orbital speed is a constant. The centripetal force needed for this circular motion = \(\frac{m v^2}{r}\).

Read and Learn More: Class 11 Physics Notes

The mutual gravitational force of attraction between a planet and the sun provides this centripetal force. GM₀m The force of gravitation, in this case, is \(\frac{G M m}{r^2}\).

Hence, \(\frac{m v^2}{r}=\frac{G M_0 m}{r^2}\)

or, \(v^2=\frac{G M_0}{r}\)

or, \(v=\sqrt{\frac{G M_0}{r}}\)

It should be noted that in the case of an elliptical orbit orbital speed (v) is not constant. If the distance of a planet from the sun is r, at an instance v ∝ \(\frac{1}{r}\).

Understanding Kepler’s Laws of Planetary Motion

Orbital Angular Velocity: The angular velocity of the planet revolving in its orbit is called its orbital angular velocity (ω). This is related to orbital speed (v) as ω = \(\frac{v}{r}\). This ω varies in an elliptical orbit as r and v change, but in a circular orbit, the value of ω is a constant.

From equation (1) ω = \(\frac{\nu}{r}=\sqrt{\frac{G M_0}{r^3}}\)….(2)

Time Period Of Revolution: The time taken by a planet to revolve once around the sun is called its time period of revolution (T).

In time T, the planet performs one complete revolution around the sun and thus moves through a distance 2πr.

Hence, orbital speed \(v=\frac{2 \pi r}{T}\)

or, \(T=\frac{2 \pi r}{v}=2 \pi r \sqrt{\frac{r}{G M_0}}\) [from equation (1)]

∴ T = \(2 \pi \sqrt{\frac{r^3}{G M_0}}\)….(3)

Hence, \(T^2=4 \pi^2 \frac{r^3}{G M_0}\)

or, \(T^2= constant \times r^3\)….(4)

Orbital Motion of Satellites Explained

This is Kepler’s third law. From this law, it can be stated that the orbital time period is longer for the planets which are farther from the sun.

Mercury, the planet nearest to the sun, revolves around the sun in 88 days, whereas Neptune, the farthest planet in the solar family, revolves in 165 years. The International Astronomical Union has classified Pluto as a dwarf planet and excluded it from the solar planet family.

It can be noted from equations (1), (2) and (3) that the orbital speed, the orbital angular velocity and the time period of revolution of a planet do not depend on the mass of the planet m, though each quantity depends on the mass of the sun M₀.

Hence, from the knowledge of distance r of a planet from the sun and its time period T, the mass of the sun M₀ can be calculated using equation (3), but the mass of the planet m cannot be determined.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Clearly, the above discussion is not only applicable to planets moving around the sun, but also applicable to any object orbiting the sun, or even in the case of satellites moving around their respective planets.

Motion Of Planets And Satellites Numerical Examples

Short Answer Questions on Planetary Motion

Example 1. The distance of the dwarf planet Pluto from the sun I is 40 times the average distance of the Earth from the sun. How much time does Pluto take to revolve around the sun?
Solution:

Given

The distance of the dwarf planet Pluto from the sun I is 40 times the average distance of the Earth from the sun.

Let T₁ = time period of Pluto,

T₂ = time period of the earth,

r₁ = distance of Pluto from the sun and

r₂ = distance of the earth from the sun

From Kepler’s law, \(\frac{T_1^2}{T_2^2}=\frac{r_1^3}{r_2^3} \text { or, } \frac{T_1^2}{(1)^2}=\left(\frac{40 r_2}{r_2}\right)^3=40^3\)

∴ \(T_1^2=40^3 \text { or, } T_1=253 \mathrm{y} \text { (approx.). }\)

Example 2. Assuming that the earth moves around the sun in a circular orbit, find the orbital speed (in km · h^-1) and the orbital angular velocity of the earth. The radius of the earth’s orbit = 1.5 x 10⁸ km.
Solution:

Given

Assuming that the earth moves around the sun in a circular orbit,

The radius of the earth’s orbit = 1.5 x 10⁸ km.

Time period of the earth ( T) = 365 d = 365 x 24 h

The radius of the orbit (r) = 1.5 x 10⁸ km

∴ Orbital speed \((\nu)=\frac{2 \pi r}{T}=\frac{2 \times \pi \times 1.5 \times 10^8}{365 \times 24}\)

= \(1.076 \times 10^5 \mathrm{~km} \cdot \mathrm{h}^{-1} \)

Orbital angular velocity \((\omega)=\frac{\nu}{r}=\frac{2 \pi}{T}=\frac{2 \times \pi}{365 \times 24}\)

= \(7.17 \times 10^{-4} \mathrm{rad} \cdot \mathrm{h}^{-1} \text {. }\)

Example 3. If the distance of the earth from the sun suddenly decreases to half its present value, then how many days will make a year?
Solution:

Let the present distance of the earth from the sun = r, time period (T) = 365 d.

If distance decreases to \(\frac{r}{2}\), the corresponding time period is T₁(say).

Following Kepler’s law, \(\frac{T^2}{r^3}=\frac{T_1^2}{r_1^3} \text { or, } T_1^2=T^2 \cdot \frac{r_1^3}{r^3}\)

or, \(T_1=T \sqrt{\frac{r_1^3}{r^3}}\) = \(365 \sqrt{\frac{\left(\frac{r}{2}\right)^3}{r^3}}\) = \(365 \sqrt{\frac{1}{8}}=\frac{365}{2 \sqrt{2}}=129 \mathrm{~d}\)

Example 4. The mean orbital radius of the Earth around the sun is 1.5 x 10⁸ km. Calculate the mass of the sun if G = 6.67 x 10^-11 N · m² · kg^-2.
Solution:

Given

The mean orbital radius of the Earth around the sun is 1.5 x 10⁸ km.

Here r = 1.5 x 10⁸ km = 1.5 x 10¹¹ m

T = 365 days = 365 x 24 x 3600 s

Since, centripetal force required = gravitational force between the earth and the sun.

∴ \(\frac{m v^2}{r}=\frac{G M m}{r^2} \text { or, } \frac{m}{r}\left(\frac{2 \pi r}{T}\right)^2=\frac{G M m}{r^2}\)

or, \(M=\frac{4 \pi^2 r^3}{G T^2}=\frac{4 \times 9.87 \times\left(1.5 \times 10^{11}\right)^3}{6.67 \times 10^{-11} \times(365 \times 24 \times 3600)^2}\)

∴ M = \(2.01 \times 10^{30} \mathrm{~kg}\)

Some Useful Data On The Sun, The Earth And The Moon:

Mass Of The Sun: Let the mass of the sun = M₁, the distance of a planet from the sun = r and the time period of the planet around the sun = T.

∴ \(M_0=\frac{4 \pi^2 r^3}{G T^2}\)….(1)

In the case of the earth, the average distance between the sun and the earth r = 1.5 x 10⁸ km (approx.) = 15 x 10¹⁰ m. The time period of revolution of the earth = 365 d (approx.) = 365 x 24 x 3600 s

Substituting the values in equation (1), \(M_0=\frac{4 \times \pi^2 \times\left(15 \times 10^{10}\right)^3}{6.7 \times 10^{-11} \times(365 \times 24 \times 3600)^2}\)

= \(2 \times 10^{30} \mathrm{~kg} \text { (approx.) }\)

Mass Of The Earth: Mass of the earth can be obtained when the motion of the moon around the earth is considered. The average distance between the moon and the earth, r = 3.84 x 10⁸ m

The time period of revolution of the moon, T = 27.3 d = 27.3 x 24 x 60 x 60 s

Hence, from equation (1), the mass of the earth

M = \(\frac{4 \times \pi^2 \times\left(3.84 \times 10^8\right)^3}{6.7 \times 10^{-11} \times(27.3 \times 24 \times 3600)^2}\)

= \(6 \times 10^{24} \mathrm{~kg} \text { (approx.) }\)

The mass of the earth can be calculated following an easier method if the value of the acceleration is due to gravity.

Gravitational Force and Planetary Motion

Gravitational Attraction Between The Sun And The Earth: The mass of the sun, M0 = 2 x 10³⁰ kg; the mass of the earth, M = 6 x 10²⁴ kg; the average distance between the sun and the earth, r = 15 x 10¹⁰ m.

Hence, the mutual force of attraction between the sun and the earth,

F = \(\frac{G M_0 M}{r^2}\)

= \(\frac{6.67 \times 10^{-11} \times\left(2 \times 10^{30}\right) \times\left(6 \times 10^{24}\right)}{\left(15 \times 10^{10}\right)^2}\)

= \(3.56 \times 10^{22} \mathrm{~N} .\)

Gravitational Attraction Between The Earth And The Moon: The mass of the Earth, M = 6 x 10²⁴ kg; the mass of the Moon, m = 7.33 x 10²² kg; the average distance between the Earth and the Moon, r = 3.84 x 10⁸ m.

Hence, the gravitational force of attraction between the two,

F = \(\frac{G M m}{r^2}\)

= \(\frac{6.67 \times 10^{-11} \times\left(6 \times 10^{24}\right) \times\left(7.33 \times 10^{22}\right)}{\left(3.84 \times 10^8\right)^2}\)

= \(1.99 \times 10^{20} \mathrm{~N}\)

Hence, \(\frac{\text{gravitational force between the sun and the earth}}{\text{gravitational force between theearth and the moon}}\)

= \(\frac{3.56 \times 10^{22} \mathrm{~N}}{1.99 \times 10^{20} \mathrm{~N}}=179 .\)

Newton’s Laws Of Gravitation

WBBSE Class 11 Newton’s Law of Gravitation Notes

Sir Isaac Newton analysed these laws of planetary motion and formulated the nature of the force of attraction between the sun and the planets and at the same time established a rule for the measurement of this force. This force of attraction is called gravitation, which acts not only between the sun and the planets but also between any two particles in the universe.

Newton’s Law of Gravitation Definition: The force of attraction between any two particles in the universe is called gravitation.

The rule for the measurement of this force of attraction is known as Newton’s law of gravitation or the law of universal gravitation.

Newton’s Law of Gravitation Statement: Any two particles in the universe attract each other along a straight line joining them. The magnitude of this force of attraction is directly proportional to the product of the masses of the particles and inversely proportional to the square of the linear distance between them.

Let two particles of masses m₁ and m₂ be kept at a distance r from each other. If the the force of attraction between them is F, then as per Newton’s law of gravitation F ∝ m₁m₂, when r remains constant and \(\), when m₁ and m₂ remain unchanged.

Read and Learn More: Class 11 Physics Notes

Hence, \(F \propto \frac{m_1 m_2}{r^2} \text { or, } F=G \frac{m_1 m_2}{r^2}\)….(1)

G is called the universal gravitational constant or the gravitational constant. Equation (1) is the mathematical representation of the law of gravitation.

Applications of Newton’s Law of Gravitation

The masses m₁ and m₂ used in Newton’s law of gravitation are called gravitational masses. On the other hand, the mass referred to in Newton’s second law of motion is called inertial mass. Though the concept of these two different masses is derived from two different laws, experiments and observations show that the two masses are identical physical quantities.

Hence, it is not necessary to mention these two masses separately.

Vector Form Of The Law Of Gravitation: Let \(\vec{r_1}\) and \(\vec{r_2}\) be the position vectors of two particles of mass m₁ and m₂ respectively. So the position vector of m₂ with respect to m₁ and that of m₁ with respect to m₂ are respectively,

⇒ \(\vec{r}_{21}=\vec{r}_2-\vec{r}_1\)….(1)

and \(\vec{r}_{12}=\vec{r}_1-\vec{r}_2\)….(2)

Understanding Universal Law of Gravitation

Let the unit vector along \(m_2\) from \(m_1\) is \(\hat{r}_{21}\)

and \(\left|\vec{r}_{21}\right|=r=\left|\vec{r}_{12}\right|\).

Therefore, \(\hat{r}_{21}=\frac{\vec{r}_{21}}{r}\) and similarly, \(\hat{r}_{12}=\frac{\vec{r}_{12}}{r}\)

Hence, according to Newton’s law of gravitation, the force on \(m_2\) due to \(m_1\) is,

⇒ \(\vec{F}_{12}=-G \frac{m_1 m_2}{\left|\overrightarrow{r_{21}}\right|^2} \hat{r}_{21}=-G \frac{m_1 m_2}{r^3} \vec{r}_{21}\)….(4)

Similarly, the force on \(m_1\) due to \(m_2\) is, \(\vec{F}_{21}=-G \frac{m_1 m_2}{\left|\vec{r}_{12}\right|^2} \hat{r}_{12}=-G \frac{m_1 m_2}{r^3} \vec{r}_{12}\)

Equations (4) and (5) are the vector form of the law of gravitation.

Gravitational Force Formula and Derivation

Gravitational Force Between Two Parties Is Equal And Opposite: From equations (2) and (4) we get,

⇒ \(\vec{F}_{12}=-G \frac{m_1 m_2}{r^3}\left(\vec{r}_2-\vec{r}_1\right)=G \frac{m_1 m_2}{r^3}\left(\vec{r}_1-\vec{r}_2\right)\)

= \(G \frac{m_1 m_2}{r^3} \vec{r}_{12}\) [according to equation (3)]

= \(-\vec{F}_{21}\) [according to equaiton (5)]

Gravitational Constant (G) Explained

Gravitational Constant: On substitution of m₁ = m₂ = 1 and r = 1, equation (1) is reduced to G = F. This defines the universal gravitational constant.

Gravitational Constant Definition: The mutually operative force of attraction between two particles of unit mass kept a unit distance apart is called the gravitational constant.

Unit: From equation (1),

unit of G = \(=\frac{\text { unit of } F \times(\text { unit of } r)^2}{(\text { unit of } m)^2}\)

Hence, the units of G in different systems are as follows:

CGS System: dyn · cm² · g^-2

SI: N · m² · kg^-2

Dimension And Ahemative Units: From equation (1), the dimension of G = \(\frac{\text { dimension of } F \times(\text { dimension of } r)^2}{(\text { dimension of } m)^2}\)

= \(\frac{M L T^{-2} \times L^2}{\mathrm{M}^2}=\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\)
Following this analysis, we can write the alternative units of G in different systems.

CGS System: cm³ ·  g^-1 · s^-2

SI: m³ · kg^-1 · s^-2

The units stated earlier and those mentioned above are obviously identical.

Magnitude: The magnitude of the universal gravitational constant obtained from the results of different experiments is:

G = 6.67 x 10^-8 dyn · cm² · g^-2 = 6.67 x 10^-8 CGS unit

= 6.67 x 10^-11 N · m² · kg^-2 =6.67 x 10^-11 in SI

As per the definition of G, when two objects of mass 1 kg each are kept lm away from each other, they attract each other with a force of 6.67 x 10^-11 N.

Clearly, this is a force of very low magnitude. Hence, we cannot feel the gravitational force acting between objects used in our daily lives.

But, on the other hand, the force of attraction of the earth is considerable as the mass of the earth is quite immense.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Derivation Of Newton’s Law Of Gravitation From Kepler’s Law: Let a planet of mass m move around the sun in a circular orbit of radius r. Let M be the mass of the sun, and F is the centripetal force exerted by the sun on the planet.

We know that F = mrω²; but ω = \(\frac{2 \pi}{T}\) [where cω = angular velocity and T is the period of revolution of the planet around the sun]

Then, \(F=m r\left(\frac{2 \pi}{T}\right)^2=\frac{4 \pi^2 m r}{T^2}\)

According to Kepler’s third law, T² = Kr³, where K is constant for all planets.

∴ F = \(\frac{4 \pi^2 m r}{K r^3}=\frac{4 \pi^2 m}{K r^2}\)

Hence, F ∝ \(\frac{m}{r^2}\)

Therefore, the force F, exerted by the sun on the planet is directly proportional to the mass of the planet and inversely proportional to the square of the distance of the planet from the sun.

Since the force is mutual, so, the planet will also exert the same force (in magnitude) on the sun in the opposite direction. Thus, the force exerted by the planet on the sun will also be directly proportional to the mass M of the sun.

Therefore, F ∝ \(\frac{m M}{r^2}\)

This is Newton’s law of gravitation.

Universality Of The Law Of Gravitation: As per Newton’s law of gravitation, the magnitude of the force of attraction between two bodies depends on their masses and the distance between them. This force of attraction does not depend on

1. The state of the objects (solid, liquid, gas),
2. Their chemical composition,
3. Temperature,

The intermediate medium, or any other factor. The law of gravitation is equally applicable to earthly objects separated by small distances and celestial bodies in space, widely separated from one another.

It has been possible to explain the motion of planets around the sun satisfactorily using this law. For these reasons, the law of gravitation is taken as a universal law and the gravitational constant is called the universal gravitational constant.

But as per the theory of relativity propounded by Einstein,

1. The mass of a body depends on its velocity,
2. The distance measured between two objects depends on the velocity of the observer, i.e., The distance between two objects measured by an observer at rest will not be the same as that measured by an observer in motion,
3. Newton’s law of gravitation is not applicable in the case of interatomic distances (10^-9 m).

Because of the reasons stated above, Newton’s law of gravitation cannot be called a universal law any more. However, the universality of G has not been denied even in Einstein’s theory.

 Newtonian Gravitation And Planetary Motion

Newton’s Laws Of Gravitation Numerical Examples

Short Answer Questions on Gravitational Force

Example 1. If the distance between the centre of a gold sphere of radius 5 nun and of a lead sphere of radius 11.5 cm is 15cm and the force of gravitational attraction between them is 2.16 x 10^-4 dyn, find the value of G, the universal gravitational constant. Densities of gold and lead are 19.3 g · cm^-3 and 11.3 g · cm^-3 respectively.
Solution:

Given

If the distance between the centre of a gold sphere of radius 5 nun and of a lead sphere of radius 11.5 cm is 15cm and the force of gravitational attraction between them is 2.16 x 10^-4 dyn,

Mass of gold sphere, \(m_1=\frac{4}{3} \times \pi \times(0.5)^3 \times 19.3=10.1 \mathrm{~g}\)

Mass of lead sphere, \(m_2=\frac{4}{3} \times \pi \times(11.5)^3 \times 11.3=7.2 \times 10^4 \mathrm{~g}\)

Given, \(r=15 \mathrm{~cm}\)

∴ G = \(\frac{F r^2}{m_1 m_2}=\frac{2.16 \times 10^{-4} \times(15)^2}{10.1 \times 7.2 \times 10^4}\)

= \(6.68 \times 10^{-8} \mathrm{CGS} \text { unit }\)

Example 2. Three particles of the same mass are kept at the if vertices of an equilateral triangle. The mass of each particle is m and the length of an arm of the triangle is l. Due to the mutual gravitational force of attraction, the particles revolve along the circumcircle of the triangle. Find the velocity of each particle.
Solution:

Given

Three particles of the same mass are kept at the if vertices of an equilateral triangle. The mass of each particle is m and the length of an arm of the triangle is l. Due to the mutual gravitational force of attraction, the particles revolve along the circumcircle of the triangle.

ABC is an equilateral triangle. At its vertices, three particles A, B and C, each of mass m, are kept. The point of intersection of the E medians of ABC is O, which is also the centre of the circumcircle of the triangle.

Hence, radius of the circumcircle, r = OA = \(\frac{2}{3}\)AD = \(\frac{2}{3}\)AB sin60° = \(\frac{2}{3} \cdot l \cdot \frac{\sqrt{3}}{2}=\frac{l}{\sqrt{3}}\)

Hence, the centripetal force needed by each particle to revolve along the circumcircle with velocity v is

∴ \(F_1=\frac{m v^2}{r}=\frac{m v^2}{\frac{l}{\sqrt{3}}}=\frac{\sqrt{3} m v^2}{l}\)

Now the gravitational force acting on the particle at point A can be calculated. The force of gravitation on the particle at A due to the particle at B along AB,

F = \(\frac{G \cdot m \cdot m}{(A B)^2}=\frac{G m^2}{R}\)

Component of this force along AE = F cos 60° and that along AD = F sin60°

Again, gravitational attraction on the particle at A due to the particle kept at C is F along AC.

The component of this force along AH = F cos 60° and that along AD = F sin60°.

Clearly, components along AH and AE cancel each other. Hence, the resultant force F₂ on the particle kept at A = sum of the components along AD.

∴ F₂ = Fsin60° + F sin60° = 2 F sin60°

= \(2 F \cdot \frac{\sqrt{3}}{2}=\sqrt{3} F=\frac{\sqrt{3} G m^2}{R}\)

As per the question, this force of attraction due to gravitation supplies the necessary centripetal force.

Hence, \(F_1=F_2 \text { or, } \frac{\sqrt{3} m v^2}{l}=\frac{\sqrt{3} G m^2}{l^2}\)

or, \(\nu^2=\frac{G m}{l}\)

or, \(\nu=\sqrt{\frac{G m}{l}} \)

Example 3. A mass M is broken into two parts of masses m₁ and m₂. How are m₁ and m₂ related so that the force of gravitational attraction between the two parts is maximum?
Solution:

Given

A mass M is broken into two parts of masses m₁ and m₂.

Let m₁ = m, then m₂ = M – m.

The gravitational force between the two parts, when they are placed at a distance r apart, is,

F = \(G \frac{m(M-m)}{r^2}=\frac{G m M}{r^2}-\frac{G m^2}{r^2}\)

Differentiating with respect to m, we have, \(\frac{d F}{d m}=\frac{G M}{r^2}-\frac{2 G m}{r^2}\)

For F to be maximum, \(\frac{d F}{d m}=0\)

∴ \(\frac{G M}{r^2}=\frac{2 G M}{r^2} \text { or, } M=2 m \text { or, } m=\frac{M}{2}\)

∴ \(m_1=m_2=\frac{M}{2}\)

Real-Life Examples of Gravitational Forces

Example 4. Assuming the earth’s orbit around the sun to be circular, show that the area swept by its radius vector in unit time (areal velocity of the earth) is a constant.
Solution:

Suppose the earth moves in a circular orbit of radius r with the sun at the centre (O) of the orbit.

Let the radius vector OP, in an infinitesimal interval of time dt, describe an angle dθ at the centre. Hence, arc PQ = rθ. As the value of PQ is very small, the arc PQ can be taken to be a straight line (chord PQ).

∴ Area swept in time dt = area of triangle OPQ = \(\frac{1}{2} O P \times P Q=\frac{1}{2} r \cdot r d \theta=\frac{1}{2} r^2 d \theta\)

∴ Area swept per unit time = \(\frac{1}{2} r^2 \cdot \frac{d \theta}{d t}=\frac{1}{2} r^2 \omega\) = constant [as the orbit is circular, r and co of the earth should be constants]

∴ The area swept in unit time by the radius vector of the earth is a constant.

Example 5. How fast (in m²/s) is the area swept out by

1. The radius from the sun to Earth?
2. The radius from Earth to the Moon? Given the distance of the sun to earth = 1.496 x 10¹¹ m, the distance of the earth to the moon = 3.845 x 10⁸ m and the period of revolution of the moon = 27\(\frac{1}{3}\) days.

Solution:

1. The rate, at which the area is swept out by the radius from the sun to the earth, \(\frac{d A}{d t}=\frac{\pi r^2}{T}\)

Here, r = 1.496 x 10¹¹ m , T = 365 days = 365 x 24 x 60 x 60 s

∴ \(\frac{d A}{d t}=\frac{\pi \times\left(1.496 \times 10^{11}\right)^2}{365 \times 24 \times 60 \times 60}=2.23 \times 10^{15} \mathrm{~m}^2 \cdot \mathrm{s}^{-1}\)

2. Here, r = 3.845 x 10⁸ m, \(T=27 \frac{1}{3} \text { days }=\frac{82}{3} \times 60 \times 60 \times 24 \mathrm{~s}\)

∴ The rate, at which the area is swept out by radius from the earth to the moon,

⇒ \(\frac{d A}{d t}=\frac{\pi \times\left(3.845 \times 10^8\right)^2}{\frac{82}{3} \times 24 \times 60 \times 60}\)]

= \(1.97 \times 10^{11} \mathrm{~m}^2 \cdot \mathrm{s}^{-1}\)

Transmission Of Heat Very Short Answer Type Questions

Question 1. Name the quickest mode of transmission of heat.
Answer: Radiation

Question 2. Name the mode of transmission of heat in which the molecules of the material medium do not change their positions.
Answer: Conduction

Read And Learn More WBCHSE Class 11 Physics Very Short Question And Answers

Question 3. Name the mode of transmission of heat in which heat is transferred by actual displacement of the material molecules.
Answer: Convection

Question 4. Name the mode of transmission of heat in which the material molecules has no role.
Answer: Radiation

Question 5. In which mode of transmission, the medium is not heated at the time of heat transmission?
Answer: Radiation

Question 6. Name the mode, in which heat is always transmitted in straight paths.
Answer: Radiation

Question 7. Which metal has the highest thermal conductivity?
Answer: Silver

Question 8. Name a liquid which is a good conductor of heat,
Answer: Mercury

Question 9. Thermal conductivity of copper is 0,92 CCS unit, Express it in SI unit.
Answer: 386.4 J · m^-1 · K^-1 · s^-1

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

 

Question 10. On what factor does the coefficient of thermal conductivity of a conductor depend?
Answer: On the material

WBBSE Class 11 Transmission of Heat Very Short Answer Questions

Question 11. What is the value of k of an ideal conductor?
Answer: Infinfie

Question 12. State whether the gradient of temperature In case of conduction of heat is positive or negative.
Answer: Negative

Question 13. Of copper, iron, wood, and mercury which one has the highest thermal resistivity?
Answer: Wood

Question 14. Two metallic rods having equal lengths and equal cross-sectional areas are coated with wax uniformly. Now one end of each rod is kept at the same high temperature, The lengths of the rods up to which wax melts are 2 cm and 3cm respectively. What is the ratio of the coefficients of thermal conductivity of the rods?
Answer: 4:9

Question 15. On which properties of the material of a conductor does the heat conduction depend till the temperature of the conducting rod increases?
Answer: Coefficient of thermal conductivity, specific heat, and density

Question 16. On which properties of the material of a conductor does the heat conduction depend when the conducting rod is in a steady state?
Answer: Coefficient of thermal conductivity

Question 17. What do you mean by thermal steady state in heat conduction?
Answer: A constant temperature is attained by each point on the die conductor

Question 18. A combined strip is produced by riveting two plates. In case of heat conduction through it which property remains die same for die two plates?
Answer: The rate of heat conduction

Question 19. Almost all metals are bad conductors of heat. Is the statement true or false?
Answer: False

Question 20. What is the CGS unit of the coefficient of thermal conductivity? Answer: cal • cm^-1 • °C^-1 • s^-1

Question 21. What is die dimension of the coefficient of thermal conductivity? Answer: MLT^-3Θ^-1

Question 22. How does the rate of heel conduction change if the temperature difference between the two ends of a conducting rod be made twice the previous value?
Answer: Doubled

Question 23. If a conducting rod is covered by an insulating material then in a steady state the rate of beat conduction will ———
Answer: Remain unchanged]

Question 24. The ratio of both the lengths and diameters of two cylindrical rods is 1:2. What will be the ratio of their thermal resistances?
Answer: 2: 1

Real-Life Examples of Heat Transmission

Question 25. If the coefficient of thermal conductivity of the material of a conductor is k, density is ρ and specific heat is s, then what will be the thermal diffusivity of the material?
Answer: \(\left[\frac{k}{\rho s}\right]\)

Question 26. The coefficient of thermal conductivity of a rod is 4 unit. What will be its thermal resistivity?
Answer: 0.25 unit

Question 27. One end of a rod covered with an insulating material is kept at 100°C and the other end is kept at the room temperature of 30°C. What will be the temperature of the mid-point of the rod in steady state?
Answer: 65°C

Question 28. Two metallic rods of equal lengths but having coefficients of thermal conductivity k₁ and k₂ and areas of cross-section A₁ and A₂ respectively are taken. The temperatures of their ends are equal. If the rate of heat conduction in them is equal \(k_1 / k_2\) = _____
Answer: \(A_1 / A_2\)

Question 29. Two metallic strips of equal cross sections are riveted together to form a combined strip. If the thermal resistances of the individual strips are R₁ and R₂, then what will be the thermal resistance of the combined strip?
Answer: R₁ + R₂

Question 30. If the velocity of light in vacuum is c then what will be the velocity of radiant heat in it?
Answer: c

Question 31. The upper layer of the atmosphere remains cold compared to the earth’s surface during the daytime. What does it indicate?
Answer: Radiant heat does not heat the medium

Question 32. What type of wave is the radiant heat?
Answer: Electromagnetic wave

Question 33. Which part of electromagnetic radiation is the most active for thermal radiation?
Answer: Infrared

Question 34. In which way does visible light differ from radiant heat?
Answer: Wavelength

Question 35. Name the body which absorbs the whole part of incident radiation. Answer: Ideal black body

Question 36. What is the absorptive power of an ideal black body?
Answer: 1

Question 37. At what temperature does a body not radiate heat?
Answer: 0K

Step-by-Step Explanations for Very Short Heat Transfer Questions

Question 38. Newton’s law of cooling is the consequence of which law?
Answer: Stefan’s law

Question 39. How many times will the rate of heat absorbed by the earth be if the temperature of the sun is doubled?
Answer: 16 times

Question 40. If a sphere and a cube made of the same material and of equal masses be kept at 200°C in air then which one of them will cool faster?
Answer: The cube

Question 41. Through solid, heat cannot be transmitted by convection. Is the statement true or false?
Answer: True

Question 42. What is the approximate range of wavelengths of thermal radiation? Answer: 6000 Å to 10^-4 m

Question 43. The frequency of radiant heat is of the order of 1014 Hz. Is the statement true or false?
Answer: True

Question 44. From Kirchhoff’s law it is known that a good radiator is also a bad absorber. Is the statement true or false?
Answer: False

Question 45. Which type of body reflects all the incident radiant heat?
Answer: Ideal white body

Question 46. A star A looks reddish from the earth, while another star B looks white. What conclusion can be drawn from this observation?
Answer: The temperature of A is less

Question 47. What is the condition for which Newton’s law of cooling is applicable?
Answer: Low-temperature difference

Question 48. For which type of surface of a hot body the rate of heat radiation will be rapid?
Answer: Black and rough

Question 49. If an iron ball and an iron plate of the same mass are heated to the same temperature, then which of them will cool faster?
Answer: Plate

Question 50. An iron plate takes times t₁, t₂, and t₃ to cool down from 100°C to 90°C, 90°C to 80°C and 80°C to 70°C respectively. What will be the relation between t₁, t₂, and t₃?
Answer: t₁<t₂ <t₃

Thermal Conductivity and Its Importance Short Answers

Question 51. The temperature of a black body is increased from 7°C to 2870 C. How does the rate of its heat radiation change?
Answer: 16 times

Question 52. Which mode of transmission of heat depends on the force of gravity?
Answer: Connection

Question 53. Due to this mode of transmission of heat land breeze and sea breeze are formed in coastal regions.
Answer: Connection

Question 54. Name one greenhouse gas.
Answer: Carbon dioxide (CO₂)