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← JEE Advanced 2025 · Paper 2
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JEE Advanced 2025 · Paper 2, Physics question 12

A geostationary satellite above the equator is orbiting around the earth at a fixed distance r1r_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 r2r_2 from the center of the earth, such that r1=1.21 r2r_1 = 1.21\,r_2. The time period of the second satellite as measured from the geostationary satellite is 24p\dfrac{24}{p} hours. The value of pp is _____.

Answer and worked solution

Official answer: 2.3

Period of the second satellite. By Kepler's third law,
T2=24(r2r1)3/2=241.331 h.T_2=24\left(\frac{r_2}{r_1}\right)^{3/2}=\frac{24}{1.331}\ \text{h}.

Relative motion. The two orbit in opposite directions, so their angular velocities add:
ωrel=2π24(1+1.331)  ⟹  Trel=242.331 h.\omega_{rel}=\frac{2\pi}{24}(1+1.331)\implies T_{rel}=\frac{24}{2.331}\ \text{h}.

So p=2.331p=2.331.