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Physics · Gravitation

JEE Advanced 2025 — Paper 2 — 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=1.21r2r_{1}=1.21 r_{2}. The time period of the second satellite as measured

from the geostationary satellite is 24p\frac{24}{p} hours. The value of pp is \qquad

Answer: 2.33

Numerical answer — enter this value.

Step-by-step solution

T∝r3/2\mathrm{T} \propto \mathrm{r}^{3 / 2}

T2 T1=(r2r1)3/2\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{3 / 2}

ω2ω1=(r1r2)3/2=(1.21)3/2\frac{\omega_{2}}{\omega_{1}}=\left(\frac{r_{1}}{r_{2}}\right)^{3 / 2}=(1.21)^{3 / 2}

ω2=ω1(1.331)\omega_{2}=\omega_{1}(1.331)

(ω2+ω1)t0=2π\left(\omega_{2}+\omega_{1}\right) t_{0}=2 \pi

t0=2πω2+ω1=2π(43+1)ω1=6π7ω1\mathrm{t}_{0}=\frac{2 \pi}{\omega_{2}+\omega_{1}}=\frac{2 \pi}{\left(\frac{4}{3}+1\right) \omega_{1}}=\frac{6 \pi}{7 \omega_{1}}

t0=6π2π( TGSS)(7)\mathrm{t}_{0}=\frac{6 \pi}{2 \pi} \frac{\left(\mathrm{~T}_{\mathrm{GSS}}\right)}{(7)}

t0=3×24 hours 7=24p\mathrm{t}_{0}=\frac{3 \times 24 \text { hours }}{7}=\frac{24}{\mathrm{p}} hours p=73=2.33\mathrm{p}=\frac{7}{3}=2.33

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2025
Paper
Paper 2
Subject
Physics
Chapter
Gravitation
Topic
Planetary Motion & Binary Star System (Kepler's Law)

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