Question

The orbital velocity of a satellite at a height $h$ above the surface of earth is $v$. The value of escape velocity from the same location is given by:

• A. $\sqrt{2}v$
• B. $v$
• C. $\frac{v}{\sqrt{2}}$
• D. $\frac{v}{2}$

Answer 1

Answer: (A)

The orbital velocity $(v_o)$ of a satellite at a height h above the surface of earth is
$v_o=\sqrt{\frac{G{M}_e}{R_e+h}\dots }(i)$
The escape velocity $\left(v_e\right)$ is
$v_e=\sqrt{\frac{2G{M}_e}{R_e+h}}\dots (ii)$
From Eqs. (i) and (ii), we get
$v_e=\sqrt{2}.v_o$
Given, $v_o=v$
$\therefore v_e=\sqrt{2}v$