A remote sensing satellite of the earth revolves in a circular orbit at a height of \(0.25 \times10^6~\text{m}\) above the surface of the earth. If Earth’s radius is \(6.38\times10^6~\text{m}\) and \(g=9.8~\text{ms}^{-2},\) then the orbital speed of the satellite is:
1. \(7.76~\text{kms}^{-1}\)
2. \(8.56~\text{kms}^{-1}\)
3. \(9.13~\text{kms}^{-1}\)
4. \(6.67~\text{kms}^{-1}\)

Subtopic:  Orbital velocity |
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A satellite \(S\) is moving in an elliptical orbit around the Earth. If the mass of the satellite is very small as compared to the mass of the earth, then:

1. The angular momentum of \(S\) about the centre of the earth changes in direction, but its magnitude remains constant.
2. The total mechanical energy of \(S\) varies periodically with time.
3. The linear momentum of \(S\) remains constant in magnitude.
4. The acceleration of \(S\) is always directed towards the centre of the earth.
Subtopic:  Orbital velocity |
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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 average distance \(r\) between the sun and planet i.e. \(T^2 = Kr^3\), 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 \(F = \frac{GMm}{r^2},\) here \(G\) is the gravitational constant. The relation between \(G\) and \(K\) is described as:
1. \(GK = 4\pi^2\)
2. \(GMK = 4\pi^2\)
3. \(K =G\)
4. \(K = \frac{1}{G}\)

Subtopic:  Kepler's Laws |
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Two spherical bodies of masses \(M\) and \(5M\) and radii \(R\) and \(2R\) are released in free space with initial separation between their centres equal to \(12R.\) If they attract each other due to gravitational force only, then the distance covered by the smaller body before the collision is:

1. \(2.5R\) 2. \(4.5R\)
3. \(7.5R\) 4. \(1.5R\)

Subtopic:  Newton's Law of Gravitation |
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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\(m=5.98\times 10^{24}~\text{kg})\) have to be compressed to be a black hole?
1. \(10^{-9}~\text{m}\)
2. \(10^{-6}~\text{m}\)
3. \(10^{-2}~\text{m}\)
4. \(100​~\text{m}\)

Subtopic:  Escape velocity |
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Dependence of intensity of gravitational field \((\mathrm{E})\) of the earth with distance \((\mathrm{r})\) from the centre of the earth is correctly represented by: (where \(\mathrm{R}\) is the radius of the earth)

1. 2.
3. 4.
Subtopic:  Gravitational Field |
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A body of mass \(m\) is taken from the Earth’s surface to the height equal to twice the radius \((R)\) of the Earth. The change in potential energy of the body will be: 

1. \(\frac{2}{3}mgR\) 2. \(3mgR\)
3. \(\frac{1}{3}mgR\) 4. \(2mgR\)
Subtopic:  Gravitational Potential Energy |
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An infinite number of bodies, each of mass \(2~\text{kg}\) are situated on the \(x\text-\)axis at distances \(1 ~\text m, ~2~\text m, ~4~\text m, ~8~\text m,......\)respectively, from the origin. The resulting gravitational potential due to this system at the origin will be:
1.  \(-\frac{8}{3}{G}\) 2. \(-\frac{4}{3} {G}\)
3.  \(-4 {G}\) 4. \(-{G}\)
Subtopic:  Gravitational Potential |
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The radius of a planet is twice the radius of the Earth. Both have almost equal average mass densities. If \(v_P\) and \(v_E\) are escape velocities of the planet and the earth, respectively, then:
1. \(v_P = 1.5 v_E\)
2. \(v_P = 2v_E\)
3. \(v_E = 3 v_P\)
4. \(v_E = 1.5v_P\)
Subtopic:  Escape velocity |
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A particle of mass \(m\) is kept at rest at a height \(3R\) from the surface of the Earth, where \(R\) is the radius of the Earth and \(M\) is the mass of the Earth. The minimum speed with which it should be projected, so that it does not return, is:
(where \(g\) is the acceleration due to gravity on the surface of the Earth)
1. \(\left(\frac{{GM}}{2 {R}}\right)^{\frac{1}{2}} \) 2. \(\left(\frac{{g} R}{4}\right)^{\frac{1}{2}} \)
3. \( \left(\frac{2 g}{R}\right)^{\frac{1}{2}} \) 4. \(\left(\frac{G M}{R}\right)^{\frac{1}{2}}\)
Subtopic:  Escape velocity |
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