JEE Advanced 2025 Paper 2 Physics Question 12 Solution

For the geostationary satellite, the time period $T_1 = 24$ hours.
According to Kepler's third law, $T^2 \propto r^3$ . $\frac{T_1}{T_2} = \left(\frac{r_1}{r_2}\right)^{3/2} = (1.21)^{3/2} = (1.1)^3 = 1.331$
Thus, $T_2 = \frac{T_1}{1.331} = \frac{24}{1.331}$ hours.
Angular velocities are $\omega_1 = \frac{2\pi}{T_1}$ and $\omega_2 = \frac{2\pi}{T_2}$ . Since they orbit in opposite directions, the relative angular velocity is: $\omega_{rel} = \omega_1 + \omega_2 = 2\pi \left(\frac{1}{T_1} + \frac{1}{T_2}\right)$
The relative time period $T_{rel} = \frac{2\pi}{\omega_{rel}} = \frac{1}{\frac{1}{T_1} + \frac{1}{T_2}} = \frac{T_1 T_2}{T_1 + T_2}$ .
Substituting $T_2 = \frac{24}{1.331}$ and $T_1 = 24$ : $T_{rel} = \frac{24 \cdot \frac{24}{1.331}}{24 + \frac{24}{1.331}} = \frac{24}{1.331 + 1} = \frac{24}{2.331}$
Comparing with $\frac{24}{p}$ , we get $p = 2.331$ . Final Answer: 2.331 (Acceptable range 2.3 to 2.4)

Question 12