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

Newtonian Gravitation And Planetary Motion Long Answer Type Questions

Question 1. How will the value of the acceleration due to gravity be affected if

  1. The earth stops rotating
  2. Does the Earth rotate faster?

Answer:

Due to the rotation of the earth, the outward centrifugal force acting on bodies on the earth’s surface and directed away from the centre of the earth decreases the weight of these bodies slightly. Hence, there is a small apparent change in g.

Apparent weight = mg’ = mg- mω² Rcos²θ; g’ = effective acceleration due to gravity, g = actual value of the acceleration due to gravity, ω = angular velocity of the earth, R = radius of the earth, θ = latitude of the place.

∴ g’ = g – ω²Rcos²θ.

  1. If the earth stops rotating, ω = 0 and g’ = g. Thus the effective value of g increases by ω²Rcos²θ.
  2. If the earth rotates faster, the value of co will increase. This will cause a further decrease in the apparent value of g. It is to be noted that changes in the speed of rotation of the earth about its axis do not influence the value of g at the poles.

Example 2. What will be the effect on the value of the acceleration due to gravity on the earth’s surface, if the radius of the earth suddenly reduces to half its present value and the mass remains constant?
Answer:

Let the mass of the earth be M and its radius be R.

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

When the radius changes to latex]\frac{R}{2}[/latex] and the mass remains constant, acceleration due to gravity

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

∴ g’ = 4g.

Thus, the acceleration due to gravity would be 4 times the present value.

Question 3. If the mass and the radius of the earth suddenly reduce to half their present values,

  1. How many hours will be there in a day and
  2. How many days will constitute a year?

Answer:

1. The time taken by the earth to spin once about its axis is the measure of a day. If the earth is taken as a uniform sphere, its moment of inertia about the axis of spinning, I = \(\frac{2}{5}\)MR² (M = mass of the earth and R = radius of the earth). Let the angular velocity of the earth be ω. When both the mass and radius of the earth become half the present values, a moment of inertia and angular velocity change to I’ and ω’.

∴ I’ = \(I^{\prime}=\frac{2}{5}\left(\frac{M}{2}\right)\left(\frac{R}{2}\right)^2=\frac{1}{8} \times \frac{2}{5} M R^2=\frac{1}{8} I\)

From the law of conservation of angular momentum \(I \omega=I^{\prime} \omega^{\prime} \text { or, } I \omega=\frac{1}{8} I \omega^{\prime} \text { or, } \omega^{\prime}=8 \omega\)

As angular velocity increases to 8 times the present value, the length of a day will be \(\frac{24}{8}\) or 3 h.

2. The time taken by the earth to revolve once around the sun (1 year) does not depend on the mass or radius of the earth. But as each day is now of 3 h duration and the total time in a year remains unchanged, it will now contain 365 x 8 = 2920 days.

Example 4. Explain why a person in an artificial satellite, orbiting the earth, feels himself to be weightless.
Answer:

Let the artificial satellite be in an orbit of radius r with respect to the centre of the earth. The necessary centripetal force for keeping the man of mass m moving in the orbit is supplied by the resultant of the gravitational force of the earth mg’ on the person and the upward reaction R of the plane of the satellite on the person.

That is mg’ – R = \(\frac{m v^2}{r}\)….(1)

The necessary centripetal force, to keep the satellite of mass M moving, is supplied by the gravitational attraction of the earth on the satellite.

Hence, Mg’ = \(\frac{m v^2}{r}\) or, g’ = \(\frac{v^2}{r}\)…..2)

From (1) and (2) we have, mg’ – R = mg’ or, R = 0

As the upward reaction force is zero, the person feels weightless.

Question 5. The sun and the earth both exert a gravitational force of attraction on any object on the earth’s surface. The two gravitational forces are in opposite directions at noon and in the same direction at midnight. Will an object weigh more at midnight?
Answer:

The Earth is a planet in the solar system. We know that there is no effect of the gravitational pull of the earth on any object in an artificial satellite, i.e., the object becomes weightless.

Similarly, any object on the surface of the earth is weightless with respect to the sun, i.e., there is no effect of the sun’s gravitational force on the object. As a result, the weight of an object (which is equal to the gravitational pull of the earth experienced by the body) remains the same both at noon and at midnight.

Question 6. Does the moon have any weight?
Answer:

Any planet or satellite orbiting due to the effect of the gravitational force of the respective star or planet is weightless. Hence, the moon does not have any weight with respect to the Earth.

Question 7. If the value of the universal gravitational constant G starts decreasing very slowly with time, what will be the effect on the motion of the moon around the Earth? Explain clearly.
Answer:

Let the earth be at O and the moon, while revolving around the earth, reach point A. When the value of G remains unchanged, the moon will continue its motion along path AB, that is to say, its orbit will remain unchanged.

Newtonian Gravitation And Planetary Motion Value Of Gravitational Constant

On the other hand, if suddenly the value of G falls to zero, the earth will have no attraction on the moon and the moon will fly off along AC, tangential to the orbit. If the value of G decreases slowly with time, the path of the moon will be towards AD, i.e., along some direction between AB and AC. As a result, the moon will gradually move away from the Earth and will take a spiral path as shown.

Question 8. Select the correct option stating the reason for your choice. A satellite is revolving around the earth in a circular path. It has

  1. Constant velocity
  2. Constant acceleration
  3. Variable acceleration.

Answer:

An object continuously changes the direction of its velocity while moving in a circular path. Hence, the velocity is not a constant. Also, during circular motion, there is an acceleration along the radius of the orbit directed towards the centre.

The magnitude of this acceleration is constant, but its direction changes continuously. Hence, the acceleration is not constant. The acceleration of the satellite keeps changing, and therefore, option 3 is correct.

Question 9. Select the correct option stating the reason for your choice. If the mass of the earth remains the same, but its diameter is decreased by 1%, the value of the acceleration due to gravity on the surface of the earth

  1. Remains the same
  2. Decreases
  3. Increases.

Answer:

The value of acceleration due to gravity on the earth’s surface is given by g = \(\frac{G M}{R^2}\). Thus, if M remains constant and diameter, i.e., radius R decreases, the value of g will increase. Hence, option 3 is correct.

Question 10. Between the earth and the moon, which one has a greater escape velocity? Give reason. Or, Is the value of the escape velocity from the earth’s surface the same as that from the surface of the moon?
Answer:

Escape velocity from a planet or satellite \(v_e=\sqrt{2 g R}\)  where g = acceleration due to gravity on the surface of that planet or satellite and R = radius of that planet or satellite. Values of g and R for the earth are much greater than that for the moon. Hence, the escape velocity from the earth’s surface is greater.

Question 11. Two satellites move around the earth at the same height. The mass of one satellite Is twice that of the other. Which satellite has a greater velocity?
Answer:

The orbital speed of a satellite in a fixed orbit at a height h from the earth’s surface is \(v=\sqrt{\frac{G M}{R+h}}\), where M = mass of the earth, R = radius of the earth. Clearly, orbital speed does not depend on the mass of the orbiting body. Hence, both move with the same velocity.

Question 12. Acceleration due to gravity on the surface of a planet is 196 cm · s-2. If It Is safe to jump from a height of 2 m on the Earth, what is the safe height for taking a jump on that planet?
Answer:

If a man jumps from a height h and v is the velocity attained on reaching the earth’s surface, then v = \(\sqrt{2 g h}\)

Let the safe height for taking a jump on the other planet be h’. In that case, the velocity attained on reaching the planet’s surface is,

v’ = \(\sqrt{2 g^{\prime} h^{\prime}} .\)

For the problem v’ = v

∴ gh = g’h’ or, h’ = \(\frac{g}{g \prime}\) h

Given g = 9.8 m · s-2 , h = 2m and g’ = 196 cm · s-2 = 1.96 m · s-2

∴ h’ = \(\frac{9.8 \times 2}{1.96}=10 \mathrm{~m}\)

Hence, it is quite safe to take a jump from a maximum height of 10 m from the surface of the planet.

Question 13. A freely falling body has no weight explains.
Answer:

Suppose, a body of mass m resting on an imaginary plane is falling down with an acceleration due to gravitational attraction. If the force of reaction of the plane on the body be R (upwards), mg – R = ma or, R = m(g-a)

For a freely falling body a = g and thus R – m(g- g) = 0.

∴ The reaction force acting on the body is zero and thus its apparent weight is zero.

Question 14. A body is released in space from an artificial satellite moving in a fixed orbit. What happens to the body?
Answer:

The released body will revolve around the earth in the same orbit and with the same velocity as that of the It should be noted that if a small body is projected with an initial velocity from an orbiting satellite out into space, there will be a change in the orbit of the body. Using the principle of conservation of angular momentum, in this case, it can be shown that there will be a small change in the orbit of the satellite.

Question 15. Why are tidal waves not observed in a large lake?
Answer:

There is no change in the total volume of water in a lake or sea, due to the gravitational attraction of the moon or the sun. In that portion of the sea surface, where the gravitational attraction is high, sea water accumulates causing high tide. Sea water flows from the area of minimum attraction to the area of maximum attraction and low tide occurs at the place where the attraction is minimum.

However, irrespective of the size, no lake is very large; the gravitational pull of the moon practically remains the same at any two places on the lake’s surface at a time. As there is no scope for the flow of water from one place to another in a lake, tidal waves are not observed even on the surfaces of large lakes.

Question 16. What is the condition for setting up an artificial satellite at a height h from the earth’s surface? (Radius of the earth = R, mass of the earth = M)
Answer:

Suppose height of the artificial satellite from the earth’s surface is h.

Hence, the radius of the orbit is (R+ h). If v is the orbital speed, the centripetal force necessary to keep the satellite in that orbit is \(\frac{m v^2}{(R+h)}\) (m = mass of the satellite).

If the mutual force of gravitation between the earth and the satellite can supply this centripetal force, the satellite would be in the right Gravitational force between the satellite and the earth
is \(\frac{G M m}{(R+h)^2}\)

Hence, the necessary condition is \(\frac{m v^2}{R+h}=\frac{G M m}{(R+h)^2} \text { or, } v=\sqrt{\frac{G M}{R+h}}\)

Question 17. At what velocity will a rocket escape the or, gravitational pull of the Earth?
Answer:

To be free from the gravitational pull of the earth, a rocket need not have any minimum velocity. When the fuel in the rocket bums, a reaction force acts on the rocket. When the average reaction force exceeds the average gravitational attraction of the earth on the rocket, it escapes the earth’s gravitational field.

Question 18. What will be the feeling of an astronaut inside a satellite about his weight when the satellite is in the process of being launched by a rocket?
Answer:

The velocity of the satellite is increased rapidly with the help of the rocket as the satellite is launched from the earth’s surface. The upward acceleration of the satellite sometimes increases up to 15 times the value of the acceleration due to gravity.

If the mass of the astronaut = m, the upward reaction of the plane of the satellite on the astronaut = R and acceleration of the satellite a = 15g, then R- mg = ma = m – 15g or, R = 16mg This is the apparent weight of the astronaut. He feels 16 times heavier than his actual weight.

Question 19. A piece of matter of mass m is thrown up vertically from the earth’s surface and it rises up to a height R. (Radius of the earth = R.) What is the initial velocity of the piece of matter? Show that the increase in potential energy of the piece of matter = \(\frac{1}{2}\)mgR.
Answer:

The potential energy of a body of mass m on the surface of the earth, i.e., at a distance R from the centre of the earth = –\(\frac{G M m}{R}\).

If the initial upward velocity is u, its kinetic energy = \(\frac{1}{2}\)mu². Thus, the total energy of the body on the surface of the earth = \(\frac{1}{2}\)mu² – \(\frac{G M m}{R}\).

At a height R from the earth’s surface, which is at a distance 2R from the centre of the earth, potential energy = –\(\frac{G M m}{2R}\). As per the given condition, kinetic energy at that height = 0. Hence, the total energy = –\(\frac{G M m}{2R}\)

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

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

or, initial velocity \(u=\sqrt{\frac{G M}{R}}\)

Also, as per the law of conservation of energy, an increase in potential energy = a decrease in kinetic energy

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

Question 20. The force of gravitation on all objects Is directly proportional to their masses, yet a heavy body does not fall faster than a light body. Why?
Answer:

If the force of gravity on a body is F and the mass of the body is m, then F ∝ m

or, F = km [ k = proportionality constant]

or, \(\frac{F}{m}\) = k ….(1)

Suppose mass m gains an acceleration of an under the action of force F. Hence, from Newton’s second law of motion,

F = ma or, \(\frac{F}{m}\) = a…….(2)

From (1) and (2), a = k

Hence, acceleration produced on any mass is a constant, so all bodies fall to the earth with equal speed.

Question 21. What will be the change In the mutual force of gravitation between two bodies when the mass of each body as well as the distance separating them is doubled?
Answer:

In the first case, the force of gravitation \(F_1=\frac{G m_1 m_2}{r^2}\) where m1 and m2 are masses of the two bodies and r is the distance separating them.

In the second case, the force of gravitation \(F_2=G \cdot \frac{2 m_1 \cdot 2 m_2}{(2 r)^2}=G \cdot \frac{m_1 m_2}{r^2}\)

∴ F2 = F1, i.e., there will be no change in the gravitational force.

Question 22. Show that the orbital speed of an artificial satellite increases as the energy of the satellite decreases.
Answer:

Total energy ofan artificial satellite = –\(\frac{G M m}{2r}\) and its kinetic energy = +\(\frac{G M m}{2r}\) . In this case, M = mass of the earth, m = mass of the artificial satellite and r = distance of the satellite from the centre of the earth.

When the total energy of the satellite starts decreasing, i.e., the value of –\(\frac{G M m}{2r}\) starts falling, the value of the kinetic energy +\(\frac{G M m}{2r}\)starts increasing. As the orbital speed is directly proportional to the square root of the kinetic energy, it also increases.

Question 23. The value of the acceleration due to gravity at a height h from the surface of the earth is g1 and that at a depth h below the earth’s surface is g2. Show that(radius of earth R>>h) \(\frac{g_2}{g_1}=1+\frac{h}{R}\).
Answer:

It is known \(g_1=1-\frac{2 h}{R}\) and \(g_2=1-\frac{h}{R}\)

∴ \(\frac{g_2}{g_1}=\frac{1-\frac{h}{R}}{1-\frac{2 h}{R}}=\left(1-\frac{h}{R}\right)\left(1-\frac{2 h}{R}\right)^{-1}\)

= \(\left(1-\frac{h}{R}\right)\left(1+\frac{2 h}{R}\right)[because R \gg h]\)

= \(1-\frac{h}{R}+\frac{2 h}{R}[because R \gg h]\)

= \(1+\frac{h}{R}\)

Question 24. If the acceleration due to gravity at a height h from the surface of the earth is the same as that at a depth h below the surface of the earth, prove that h = \(\frac{\sqrt{5}-1}{2} R\) (R = radius of the earth).
Answer:

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

Acceleration due to gravity at a height h from the earth’s surface \(g_1=\frac{G M}{(R+h)^2}\)

Hence, \(\frac{g_1}{g}=\frac{R^2}{(R+h)^2}\)

Also, taking the earth as a uniform sphere of radius R, acceleration due to gravity at a depth h

⇒ \(g_2=g \frac{(R-h)}{R}\)

∴ \(\frac{g_1}{g_2}=\frac{R^3}{(R+h)^2(R-h)}\)

If \(g_1=g_2\)

⇒ \(R^3=\left(R^2-h^2\right)(R+h)=R^3+R^2 h-h^2 R-h^3 \)

or, \(h^2+R h-R^2=0\)

∴ h = \(\frac{-R \pm \sqrt{R^2+4 R^2}}{2}=\frac{(-1 \pm \sqrt{5}) R}{2}\)

As a negative value of h is not acceptable, h= \(\frac{\sqrt{5}-1}{2} R\)

Question 25. The force of attraction of the earth on an apple is of the same magnitude as the force of attraction of the apple on the earth, yet the apple moves towards the earth, but the earth does not move towards the apple. Why?
Answer:

Suppose the earth of mass M attracts an apple of mass m with a force F towards it. As per the law of gravitation, the apple also attracts the earth with the same force F towards it.

Acceleration of the apple due to the earth’s pull

= \(\frac{\text { force on apple }}{\text { mass of apple }}=\frac{F}{m}\)

Acceleration of the earth due to the attraction of the apple = \(\frac{\text { force on the earth }}{\text { mass of the earth }}=\frac{P}{M}\)

Hence, \(\frac{\text { acceleration of the apple }}{\text { acceleration of the earth }}=\frac{\frac{F}{m}}{\frac{F}{M}}=\frac{M}{m}\)

As the mass of the apple is negligible compared to that of the earth, \(\frac{M}{m}\)>>1.

That is to say, acceleration of the apple >> acceleration of the earth. Practically, the apple moves towards the earth. The motion of the earth towards the apple is too small to be noticed.

Question 26. In which case will the decrease in the value of the acceleration due to gravity be greater (with respect to that on the earth’s surface)—at 1 km above the earth’s surface or 1 km below the earth’s surface?
Answer:

Let the radius of the earth = R

Value of acceleration due to gravity on the earth = g

Decrease in g at a height h = \(\frac{2 h g}{R}\)…..(1)

and decrease in g at a depth h = \(\frac{h g}{R}\)….(2)

Comparing the two equations, it can be concluded that the decrease in the acceleration due to gravity at a height of 1 km above the earth’s surface will be greater.

Question 27. Due to some reason, if the average distance of the earth from the sun decreases, will the length of a year Increase or decrease? Give reasons for your answer.
Answer:

If the average distance of the earth from the sun is r and the time period of revolution of the earth around the sun is T, then from Kepler’s law T² ∝ r³. Hence, with the decrease in r, the span of a year (T) would decrease.

Question 28. What happens to a satellite if its orbital speed is greater than the escape velocity corresponding to that orbit?
Answer:

In any orbit, the orbital speed of an artificial satellite is less than the escape velocity from the planet. Now, for some reason, if the orbital speed exceeds the escape velocity, the gravitational attraction of the planet will not be able to keep the satellite in its orbit. The satellite will move out of the gravitational field of the planet following a spiral path.

Question 29. Why is the span of a year smaller for a planet closer: to the sun?
Answer:

As per Kepler’s law (time period of revolution)² ∝ (radius of the orbit)³ Radius of the orbit for a planet closer to the sun is smaller. Hence, the time period, i.e., the span of a year is also smaller.

Question 30. Suppose there existed a planet that went around the sun eight times as fast as the Earth. What would be its orbital size as compared to that of the Earth?
Answer:

If the period of revolution of the earth around the sun is T, the period of the planet is \(\frac{T}{8}\). If the semi-major axes of their elliptical orbits are r1 and r2 respectively, then from Kepler’s law,

⇒ \(\left(\frac{T}{T / 8}\right)^2=\left(\frac{r_1}{r_2}\right)^3\)

or, \(\frac{r_1}{r_2}=8^{2 / 3}=4, \text { or, } r_2=\frac{r_1}{4}\)

So, the orbital size of the planet is 1/4 th that of the earth.

Question 31. The force of gravitation between two spherical masses M and m, placed in air, is F. The space between the two spheres is filled with a liquid of relative density 3. What will be the change in the value of the gravitational force acting between the spheres?
Answer:

The force of gravitation does not depend on the properties of the medium in which the two bodies are placed. Hence, the force in the second case is also F.

Question 32. The conservation of what physical quantity does Kepler’s second law refer to?
Answer:

Kepler’s second law refers to the law of conservation of angular momentum.

According to Kepler’s second law, the area swept out by the line joining the Earth and the sun in unit time is a constant.

Area swept out in unit time by the line

= \(\frac{1}{2}\)rω x r =  \(\frac{1}{2}\)r²ω = k (constant)

In this case,

r = distance of the earth from the sun

ω = angular velocity of the earth around the sun

If the linear speed of the earth is v and its mass is m, the angular momentum of the earth

= mv x r = mω² = mx 2k = constant

Thus, the angular momentum is conserved.

Question 33. What is the appearance of a communicating satellite from its plane of projection?
Answer:

Communication satellites are actually geostationary satellites. This artificial satellite revolves around the earth with an angular velocity equal to that of the diurnal motion of the earth (and follows the same spinning direction), i.e., it revolves around the earth in 24 hours.

Hence, the angular velocity of the satellite becomes zero with respect to the angular velocity of the earth. Hence, when observed from the plane of projection, i.e., from the earth’s surface, the satellite will appear to be stationary in the sky.

Question 34. Is it possible to place an artificial satellite in an orbit such that it is always visible in the sky of Lucknow or New Delhi? Give the reason.
Answer:

Only a geostationary satellite, which revolves around the earth on the equatorial plane, may always be visible directly overhead from someplace on the equator. Lucknow and New Delhi are not on the equator (nearly at 25°N latitude).

So, if a geostationary satellite is on this side of the earth above the equator, it would always be visible from Lucknow or New Delhi in an oblique direction towards the south, not directly overhead.

Question 35. The gravitational potential energy of a body on the surface of the earth is -6.4 x 106 joule. Explain the statement.
Answer:

The meaning of the statement is that 6.4 x 106 joule of energy would be required to send away the body outside the gravitational field of the earth.

Question 36. If T is the period of revolution of an artificial satellite orbiting very close to the earth’s surface and ρ is the density of the earth, then show that ρT² is a universal constant.
Answer:

The period of revolution of an artificial satellite orbiting very close to the Earth’s surface is given by

T =  \(2 \pi \sqrt{\frac{R}{g}}\); R = radius of the earth

g = acceleration due to gravity on the earth’s surface

We know that the average density of the earth is given by \(\rho=\frac{3 g}{4 \pi G R} \text { or, } \frac{R}{g}=\frac{3}{4 \pi G \rho}\)

∴ T \(=2 \pi \sqrt{\frac{3}{4 \pi G \rho}}\)

or, \(T^2=\frac{4 \pi^2 \times 3}{4 \pi G \rho} or, \rho T^2=\frac{3 \pi}{G}\) (a universal constant)

Question 37. Three particles, each of mass m, are placed at the vertices of an equilateral triangle. What is the force acting on a particle of mass 2m placed at the I centroid D of the triangle?
Answer:

The positions of the particles are shown.

Let DA = DB = DC = x.

Newtonian Gravitation And Planetary Motion Three Particles Are plated At Equlateral Triangle

The gravitational forces on the particle at D, due to the particles at A, B, and C are equal, with value \(\frac{G(2 m)(m)}{t^2}\) directed as shown.

By symmetry the resultant force on 2m =0.

Question 38. Acceleration due to gravity decreases as we go below the surface of the earth. What will be the nature of the graph between the change of acceleration due to gravity and the depth below the surface of the earth?
Answer:

The acceleration due to gravity at a depth d below the surface of the earth is given by, \(g^{\prime}=g\left(1-\frac{d}{R}\right)=g-g \frac{d}{R} \text { or, } g-g^{\prime}=g \frac{d}{R}\)

Since g and R are constants, g-g’ ∝ d

So the graph between (g-g’) and d will be a straight line.

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