Question

A satellite of mass $m$ revolves around the earth of radius $R$ at a height x from its surface. If $g$ is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is

• A. $\sqrt{gx}$
• B. $\sqrt{\frac{gR}{R-x}}$
• C. $\sqrt{\frac{g{R}^2}{R-x}}$
• D. $\sqrt{\frac{g{R}^2}{R+x}}$

Answer 1

The correct option is D $\sqrt{\frac{g{R}^2}{R+x}}$

We have:

$\frac{m{v}^2}{R+x}=\frac{GMm}{{(R+x)}^2}\Rightarrow v=\sqrt{\frac{GM}{R+x}}=\sqrt{\frac{GM}{R^2}\times \frac{R^2}{R+x}}=\sqrt{\frac{g{R}^2}{R+x}}$

because the gravitational force provides the centripetal force for revolution.