WBCHSE Class 11 Physics Newtonian Gravitation And Planetary Motion Short Answer Questions

Newtonian Gravitation And Planetary Motion Short Answer Type Questions

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:

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:

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

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:

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

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:

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:

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.

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:

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.

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:

⇒ \(g_1: g_2=5: 2\)

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.

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:

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.

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:

⇒ \(v_e=\sqrt{\frac{2 G M}{R}}=C or, R=\frac{2 G M}{C^2}\) ; 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:

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: 

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.

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:

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:

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:

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

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