Question

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 $\left(h\right)$ of the satellite above the earth's surface is (Take radius of earth as $R_e)$.

• A. $h=R_e^2$
• B. $h=R_e$
• C. $h=2R_e$
• D. $h=4R_e$

Answer 1

Answer: (B)

$h=R_e$

The escape velocity from earth is given by
$v_e=\sqrt{2g{R}_e}.....\left(i\right)$
The orbital velocity of a satellite revolving around earth is given by
$v_0=\frac{\sqrt{G{M}_e}}{(R_e+h)}$
where, $M_e=$ mass of earth, $R_e=$ radius of earth, $h=$ height of satellite from surface of earth.
By the relation $G{M}_e=g{R}_e^2$
So, $v_0=\frac{\sqrt{g{R}_e^2}}{(R_e+h)}.....\left(ii\right)$
Dividing equation (i) by (ii), we get
$\frac{v_e}{v_0}=\frac{\sqrt{2(R_e+h)}}{\left(R_e\right)}$
Given, $v_0=\frac{v_e}{2}$
$\frac{2v_e}{v_e}=\frac{\sqrt{2(R_e+h)}}{R_e}$
Squaring on both side, we get
$4=\frac{2(R_e+h)}{R_e}$

or $R_e+h=2R_e$.

$⟹h=R_e$