JEE Advanced 2025 Paper 2 · Q12 · Kepler's Laws
A geostationary satellite above the equator is orbiting around the earth at a fixed distance $r_1$ from the center of the earth. A second satellite is orbiting in the equatorial plane in the opposite direction to the earth's rotation, at a distance $r_2$ from the center of the earth, such that $r_1 = 1.21\,r_2$. The time period of the second satellite as measured from the geostationary satellite is $24/p$ hours. The value of $p$ is __________.
Reveal answer + step-by-step solution
Correct answer:2.33
Solution
Kepler's third law: $T \propto r^{3/2}$. $T_1$ (geostationary) $= 24$ h. $T_2 = 24 \times (r_2/r_1)^{3/2} = 24/(1.21)^{3/2} = 24/1.331$ (since $1.21 = 1.1^2 \to (1.1)^3 = 1.331$). Sat 2 moves opposite to earth's rotation; sat 1 (geostationary) co-rotates with earth (period $24$ h). Their angular velocities are oppositely directed in the inertial frame, so relative angular velocity $= \omega_1 + \omega_2$. Effective period as seen from sat 1: $T_{\text{rel}} = 2\pi/(\omega_1 + \omega_2) = (T_1 T_2)/(T_1 + T_2) = 24 \times (24/1.331)/(24 + 24/1.331) = (24/1.331)/(1 + 1/1.331) = 24/(1.331 + 1) = 24/2.331$. Comparing with $24/p$ gives $p = 2.331 \approx 7/3 \approx 2.33$.
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