Q 11-07-072JEE MainJEE Main 2024 (9 Apr, Shift 1)Medium
An astronaut takes a ball of mass $m$ from earth to space. He throws the ball into a circular orbit about earth at an altitude of $318.5\ \text{km}$. From earth's surface to the orbit, the change in total mechanical energy of the ball is $x\dfrac{GM_em}{21R_e}$. The value of $x$ is (take $R_e = 6370\ \text{km}$):
Answer: (D) 11
Orbit radius $r = 6370 + 318.5 = 6688.5\ \text{km} = \dfrac{21}{20}R_e$.
On the surface (at rest): $E_1 = -\dfrac{GM_em}{R_e}$. In orbit: $E_2 = -\dfrac{GM_em}{2r} = -\dfrac{10}{21}\dfrac{GM_em}{R_e}$.
$$\Delta E = \frac{GM_em}{R_e}\left(1 - \frac{10}{21}\right) = \frac{11\,GM_em}{21R_e} \;\Rightarrow\; x = 11$$
Solution by Sreeraj P, M.Sc Physics