An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the escape velocity from the earth of radius R. It is at height h=R above the earths surface. If the satellite is stopped suddenly in its orbit and allowed to fall freely on the earth, the speed with which it will hit the surface of the earth will be

Potential energy of the satellite in its orbit is

$E_1=-\frac{GmM}{r}=-\frac{GmM}{r} \left(\therefore r=2R\right)$

The kinetic energy is zero because the satellite is stopped. Potential energy of the satellite on the surface of the earth is 

$E_2=-\frac{GmM}{R}$

$$\begin{gathered} \therefore Loss of PE=E_1-E_2=-\frac{GmM}{2R}-\left(-\frac{GmM}{R}\right) \\ =\frac{GmM}{2R} \end{gathered}$$

This is converted into kinetic energy. If v is the speed with which the satellite hits the surface of the earth, then from the law of conservation of energy, we have 

$$\begin{gathered} \frac{1}{2} m{v}^2=\frac{GmM}{2R} \\ or v^2=\frac{GM}{R}=gR \left(\therefore g=\frac{GM}{R^2}\right) \end{gathered}$$

Hence the correct choice is (a).