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 closed to(Neglect the effect of atmosphere.)

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

Answer 1

Answer: (D)

$\sqrt{gR}(\sqrt{2}-1)$

For a satellite in orbit, the velocity can be evaluated from the following equation,

$\frac{GMm}{r^2}=\frac{m{v}_o^2}{r}$

$v_o=\sqrt{\frac{GM}{r}}$

Here, $r=R+h$

For satellite to escape, total energy of satellite must be zero. Let escape velocity of satellite at height h be $v_e$.

$E=PE+KE=0$

$\frac{-GMm}{r}+\frac{1}{2}m{v}_e^2=0$

$v_e=\sqrt{\frac{2GM}{r}}$

Difference in velocities is given by

$\Delta v=v_e-v_o$

$\Delta v=\sqrt{\frac{GM}{r}}(\sqrt{2}-1)$

$\Delta v\approx \sqrt{\frac{GM}{R}}(\sqrt{2}-1)$ as $h<<R$

But $g=\frac{GM}{R^2}$

$\therefore \Delta v\approx \sqrt{gR}(\sqrt{2}-1)$