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. \(\frac{12}{2.5}~\text{h}\)

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. \(\frac{GMm}{R+h}\)

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. \(t' = 2T'\)

A body weighs \(200\) N on the surface of the earth. How much will it weigh halfway down the centre of the earth?

The work done to raise a mass \(m\) from the surface of the earth to a height \(h\), which is equal to the radius of the earth, is:
1. \(\frac{3}{2}mgR\)
2. \(mgR\)
3. \(2mgR\)
4. \(\frac{1}{2}mgR\)

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. \(48~\text{N}\)

The escape velocity from the Earth's surface is \(v\). The escape velocity from the surface of another planet having a radius, four times that of Earth and the same mass density is: 

A particle of mass \(m\) is projected with a velocity, \(v=kV_{e} ~(k<1)\) from the surface of the earth. The maximum height, above the surface, reached by the particle is: (Where \(V_e=\) escape velocity, \(R=\) radius of the earth)