What is the depth at which the value of acceleration due to gravity becomes \(\frac{1}{{n^{th}}}\) time it's value at the surface of the earth? (radius of the earth = \(\mathrm{R}\))  
1. \(R \over n^2\)
2. \(R~(n-1) \over n\)
3. \(Rn \over (n-1)\)
4. \(R \over n\)

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)

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 the earth to be a sphere of uniform density, its acceleration due to gravity acting on a body:

Two planets are in a circular orbit of radius \(R\) and \(4R\) about a star. At a specific time, the two planets and the star are in a straight line. If the period of the closest planet is \(T,\) then the star and planets will again be in a straight line after a minimum time: