JEE Main 2025 — 23 January, Evening Shift — Question 68

(i) If earth is assumed to be stationary

N

orbital velocity $\mathrm{v}_{0}=\sqrt{\frac{\mathrm{GM}}{4 \mathrm{R} / 3}}=\sqrt{\frac{3 \mathrm{GM}}{4 \mathrm{R}}}$

Angular momentum of satellite $=\frac{M}{2} v_{0} \frac{4 R}{3}$

$=\frac{\mathrm{M}}{2} \cdot \sqrt{\frac{3 \mathrm{GM}}{4 \mathrm{R}}} \cdot \frac{4 \mathrm{R}}{3}$

$=M \sqrt{\frac{\mathrm{GMR}}{3}}$

$\mathrm{x}=3$

(ii) Since mass of satellite is comparable to the mass of earth.

1

$\frac{\text { G.M. } \frac{M}{2}}{\left(\frac{4 R}{3}\right)^{2}}=\frac{M}{2} \omega^{2} \cdot \frac{8 R}{9}$

$\omega=\sqrt{\frac{81 \mathrm{GM}}{128 \mathrm{R}^{3}}}$