Question

If an artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth, the height of the satellite above the surface of the earth is : (Radius of earth $=R$)

• A. $2R$
• B. $\frac{R}{2}$
• C. $R$
• D. $\frac{R}{4}$

Answer 1

The correct option is C $R$

For a satellite moving in circular orbit at a height $h$ from Earth's surface, centripetal force is given by
$F_c=\frac{m{v}^2}{R_e+h}$
This force is provided by the Gravitational force of Earth, i.e.
$F_g=\frac{G{M}_{e}m}{(R_e+h{)}^2}$
$\therefore \frac{m{v}^2}{R_e+h}=\frac{G{M}_{e}m}{(R_e+h{)}^2}$
$\Rightarrow v^2=\frac{G{M}_e}{R_e+h}\Rightarrow v=\sqrt{\frac{G{M}_e}{R_e+h}}$
Escape velocity is given by,
$v_e=\sqrt{\frac{2G{M}_e}{R_e}}$
It is given that the speed of the satellite is half of the escape speed.
$\therefore {\left[\frac{1}{2}\left(\sqrt{\frac{2G{M}_e}{R_e}}\right)\right]}^2=\frac{G{M}_e}{R_e+h}$
$\Rightarrow \frac{G{M}_e}{2R_e}=\frac{G{M}_e}{R_e+h}$
$\Rightarrow 2R_e=R_e+h\Rightarrow h=R_e$