Question

A satellite is moving at a constant speed $V$ in a circular orbit about the earth. An object of mass $m$ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is

• A. $\frac{1}{2}m{V}^2$
• B. $m{V}^2$
• C. $\frac{3}{2}m{V}^2$
• D. $2m{V}^2$

Answer 1

The correct option is B $m{V}^2$

For earth, escape velocity at a given distance is calculated by the formula,

$v_e=\sqrt{\frac{2GM}{R}}$

We need the orbital height, which we can get from,

Satellite motion, circular;

$v=\sqrt{\frac{GM}{R}}$

$v=$velocity in m/s

$m$ is mass od central body in kg

$R$ is radius in orbit in meter.

$v^2=\sqrt{\frac{GM}{R}}=DR=\frac{GM}{v^2}$

Plugging that into the escape velocity equations, we get,

$v_e=\sqrt{\frac{2GM}{\frac{GM}{v^2}}}=\frac{v}{\sqrt{2}}$

$K.E=\frac{1}{2}m{v}^2=\frac{1}{2}m(\frac{v}{\sqrt{2}}{)}^2=m{v}^2$