Question

The orbital velocity of an artificial satellite in a circular orbit just above the earths surface is $V_0$. The orbital velocity of a satellite orbiting at an altitude of half of the radius, is

• A. $ \frac{3}{2}v_{0} $
• B. $ \frac{2}{3}v_{0} $
• C. $ \sqrt{\frac{2}{3}}v_{0} $
• D. $ \sqrt{\frac{3}{2}}v_{0} $

Answer 1

Answer: (C)

The orbital velocity of a satellite orbiting in a circular orbit at a height $h$ above the earth's surface is

$v_{0}=\sqrt{\frac{G M_{e}}{R+h}}$
where $R$ is radius of earth, $G$ the gravitational constant and $M_{e}$ the mass of earth.
Given, $h=\frac{R}{2}$
$\therefore v_{0}'=\sqrt{\frac{G M_{e}}{R+\frac{R}{2}}} $
$=\sqrt{\frac{G M_{e}}{\frac{3}{2} R_{e}}}=\sqrt{\frac{2}{3}} \cdot v_{0}$