Question

Derive the expression for the critical velocity of a satellite revolving close to the surface of the Earth in terms of acceleration due to gravity.

Answer 1

1. When the satellite is revolving close to the surface of the Earth, the height is very small as compared to the radius of the Earth.

2. Hence the height can be neglected and radius of the orbit is nearly equal to R, i.e., R + h ≈ R

3. The critical speed of the satellite then becomes,

v_c = \(\sqrt{\frac{GM}{R}}\)

4. G is related to acceleration due to gravity by the relation,

g = \(\frac{GM}{R^2}\)

∴ GM = gR²

5. Thus, critical speed in terms of acceleration due to gravity, neglecting the air resistance, can be obtained as,

v_c = \(\sqrt{\frac{gR^2}{R}}=\sqrt{gR}\)