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, $h=\frac{R}{2}$
$\therefore v_0^{\prime }=\sqrt{\frac{G{M}_e}{R+\frac{R}{2}}}$
$=\sqrt{\frac{G{M}_e}{\frac{3}{2}{R}_e}}=\sqrt{\frac{2}{3}}\cdot v_0$