Question

An artificial satellite is moving around earth in a circular orbit with speed equal to one fourth the escape speed of a body. From the surface of earth the hight of satellite above earth's surface is (R = radius of earth)

Answer 1

We know that,

Escape velocity = $\sqrt{2gR}$

Given, Orbital speed of satellite = $\frac{\sqrt{2gR}}{4}$

By force balance we have,

Gravitational force = Centripetal force

$G\frac{Mm}{x^2}=\frac{m{v}^2}{x}$

$⟹x=\frac{GM}{v^2}=\frac{GM}{2gR}\times 16=\frac{8GM}{\frac{GM}{R^2}}\times R=8R$

Height of satellite above earth's surface is $8R-R=7R$.