Question

If $v_0$ be the orbital velocity of a satellite in a circular orbit close to the earth's surface and $v_e$ is the escape velocity from the earth, then relation between the two is

• A. $v_o=v_e$
• B. ${\upsilon }_e=\sqrt{3}{\upsilon }_o$
• C. ${\upsilon }_e=\sqrt{2}{\upsilon }_o$
• D. $v_e=2v_o$

Answer 1

Answer: (C)

Orbital velocity of satellite $=v_0$
and escape velocity from the earth $=v_e$.
We have orbital velocity $\left(v_{\circ }\right)=\sqrt{gR}$
and escape velocity $\left(v_e\right)=\sqrt{2gR}$
Therefore ratio of orbital velocity to escape velocity
$\frac{v_o}{v_e}=\sqrt{\frac{gR}{2gR}}=\frac{1}{\sqrt{2}}$ or
$v_e=\sqrt{2}{v}_o.$