Question

A satellite is moving with 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

$m{v}^2$

Escape velocity and orbital velocity

orbital velocity is given by $v_o=\sqrt{\frac{GM}{r}}$

and escape velocity is given by $v_e=\sqrt{\frac{2GM}{r}}$

where $G$ is gravitational constant, $M$ is mass of body and $r$ is distance from centre of mass

From the above relations we have,

$v_e=\sqrt{2}{v}_o$

The body ejected from satellite will have velocity $v_o=v$ initially,

so the kinetic energy is $\frac{1}{2}m{v_e}^2=\frac{1}{2}m{\left(\sqrt{2}{v}_o\right)}^2=m{v}^2$

Hence, option B is correct.