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 $(h)$ of the satellite above the earth’s surface is (Take radius of earth as $R_e$)

• A. $h = R^2_e$
• B. $h = R_{e}$
• C. $h = 2R_e$
• D. $h = 4R_e$

Answer 1

Answer: (B)

The escape velocity from earth is given by
$v_{e}=\sqrt{2 g R_{e}}$ ...(i)
The orbital velocity of a satellite revolving around earth is given by
$v_{0}=\frac{G M_{e}}{\left(R_{e}+h\right)}$
where, $M_{e}=$ mass of earth,
$R_{e}=$ radius of earth,
$h =$ height of satellite from surface of earth.
By the relation $G M_{e}=q R_{\rho}^{2}$
so, $v_{0} \frac{\sqrt{g R_{e}^{2}}}{\left(R_{e}+h\right)}$ ...(ii)
Dividing equation $(i)$ by $(ii)$, we get
$\frac{v_{e}}{v_{0}} \frac{\sqrt{2\left(R_{e}+h\right)}}{R_{e}}$
Squaring on both side, we get
$4=\frac{2\left(R_{e}+h\right)}{R_{e}}$
or $R_{e}+h=2 R_{e}$ i.e., $h=R_{e}$