A body weighs \(72~\text{N}\) on the surface of the earth. What is the gravitational force on it at a height equal to half the radius of the earth?
1. \(32~\text{N}\)
2. \(30~\text{N}\)
3. \(24~\text{N}\)
4. \(48~\text{N}\)

Assuming that the gravitational potential energy of an object at infinity is zero, the change in potential energy (final - initial) of an object of mass \(m\) when taken to a height \(h\) from the surface of the earth (of radius \(R\) and mass \(M\)), is given by:
1. \(-\frac{GMm}{R+h}\)
2. \(\frac{GMmh}{R(R+h)}\)
3. \(mgh\)
4. \(\frac{GMm}{R+h}\)

The time period of a geostationary satellite is \(24~\text{h}\) at a height \(6R_E\) \((R_E\) is the radius of the earth) from the surface of the earth. The time period of another satellite whose height is \(2.5R_E\)${}_{}$ from the surface will be:
1. \(6\sqrt{2}~\text{h}\)
2. \(12\sqrt{2}~\text{h}\)
3. \(\frac{24}{2.5}~\text{h}\)
4. \(\frac{12}{2.5}~\text{h}\)

A mass falls from a height '\(h\)' and its time of fall '\(t\)' is recorded in terms of time period \(T\) of a simple pendulum. On the surface of the earth, it is found that \(t=2T\). The entire setup is taken on the surface of another planet whose mass is half of that of the Earth and whose radius is the same. The same experiment is repeated and corresponding times are noted as '\(t\)' and '\(T\)'. Then we can say:
1. \(t' = \sqrt{2}T\)
2. \(t'>2T'\)
3. \(t'<2T'\)
4. \(t' = 2T'\)

If $v_e$ is the escape velocity and $v_o$ is the orbital velocity of a satellite for orbit close to the earth's surface, then these  are related by:

Which one of the following plots represents the variation of a gravitational field on a particle with distance \(r\) due to a thin spherical shell of radius \(R?\)
(\(r\) is measured from the centre of the spherical shell)

1.		2.
3.		4.

A particle of mass \(\mathrm{m}\) is thrown upwards from the surface of the earth, with a velocity \(\mathrm{u}\). The mass and the radius of the earth are, respectively, \(\mathrm{M}\) and \(\mathrm{R}\). \(\mathrm{G}\) is the gravitational constant and \(\mathrm{g}\) is the acceleration due to gravity on the surface of the earth. The minimum value of \(\mathrm{u}\) so that the particle does not return back to earth is:
1. \(\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}^2}} \)
2. \(\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \)
3.\(\sqrt{\frac{2 \mathrm{gM}}{\mathrm{R}^2}} \)
4. \(\sqrt{ \mathrm{2gR^2}}\)

A particle of mass M is situated at the centre of a spherical shell of the same mass and radius a. The magnitude of the gravitational potential at a point situated at a/2 distance from the centre will be:

1. $-\frac{\mathrm{GM}}{\mathrm{a}}$

2. $-\frac{2\mathrm{GM}}{\mathrm{a}}$

3. $-\frac{3\mathrm{GM}}{\mathrm{a}}$

4. $-\frac{4\mathrm{GM}}{\mathrm{a}}$

The dependence of acceleration due to gravity 'g' on the distance 'r' from the centre of the earth, assumed to be a sphere of radius R of uniform density, is as shown in figure below:
   
The correct figure is:
1. a

2. b

3. c

4. d