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. $\frac{1}{2}\, mv^2$
• B. $mv^2$
• C. $\frac{1}{3}\, mv^2$
• D. $\frac{2}{3}\, mv^2$

Answer 1

Answer: (B)

Escape speed,
$v_{e}=\sqrt{2} \times$ orbital speed $=\sqrt{2}v$
$\therefore \quad$ Kinetic energy of the object at the time of
ejection $=\frac{1}{2} mv^{2}_{e}=\frac{1}{2} m\left(\sqrt{2}v\right)^{2}=mv^{2}$