JEE PYQ: Gravitation Question 43
Question 43 - 2019 (10 Jan Shift 1)
A satellite is moving with a constant speed $v$ in circular orbit around 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 ejection, the kinetic energy of the object is:
(1) $2mv^2$
(2) $mv^2$
(3) $\frac{1}{2}mv^2$
(4) $\frac{3}{2}mv^2$
Show Answer
Answer: (2)
Solution
For the object to just escape: total energy $= 0$. $\frac{1}{2}mv_{\text{esc}}^2 - \frac{GMm}{r} = 0$. Since orbital velocity $v = \sqrt{GM/r}$, we get $v_{\text{esc}} = v\sqrt{2}$. The ejection is from the satellite moving at $v$, so by momentum/energy analysis, KE $= \frac{1}{2}mv_{\text{esc}}^2 = \frac{1}{2}m(2v^2) = mv^2$.
BETA
