Question

A satellite is revolving in a circular orbit at a height ' $h$' from the earth's surface (radius of earth $R ; h < < R$). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth’s gravitational field, is close to : (Neglect the effect of atmosphere.)

• A. $\sqrt{2\,gR}$
• B. $\sqrt{g\,R}$
• C. $\sqrt{g\,R / 2 }$
• D. $\sqrt{gR} ( \sqrt{2} - 1 ) $

Answer 1

Answer: (D)

Orbital velocity $v = \sqrt{\frac{GM}{R + h}} = \sqrt{\frac{GM}{R}}$ as h < < R
Velocity required to escape
$ \frac{1}{2} mv'^{2} = \frac{GMm}{R+h} ; v' = \sqrt{\frac{2GM}{R+h}} = \sqrt{\frac{2GM}{R}} \left(h < < R\right)$
$\therefore $ Increase in velocity
$ v' - v = \sqrt{\frac{2GM}{R}} - \sqrt{\frac{GM}{R}} = \sqrt{2gR} - \sqrt{gR} = \sqrt{gR} \left( \sqrt{2} - 1 \right) $