JEE Advanced · Physics · 9. Gravitation
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 \(\frac{24}{p}\) hours. The value of \(p\) is
- A 0.34
- B 0.55
- C 2.34
- D 2.3
Answer & Solution
Correct Answer
(C) 2.34
Step-by-step Solution
Detailed explanation

\(\mathrm{T} \propto \mathrm{r}^{3 / 2}\)
\(\frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^{3 / 2}\)
\(\frac{\omega_2}{\omega_1}=\left(\frac{r_1}{r_2}\right)^{3 / 2}=(1.21)^{3 / 2}\)
\(\begin{aligned}
& \omega_2=\omega_1(1.331) \qquad ...(i)\\
& \left(\omega_2+\omega_1\right) t_0=2 \pi \qquad ...(ii)
\end{aligned}\)
\(t_0=\frac{2 \pi}{\omega_2+\omega_1}=\frac{2 \pi}{\left(\frac{4}{3}+1\right) \omega_1}=\frac{6 \pi}{7 \omega_1}\)
\(\mathrm{t}_0=\frac{6 \pi}{2 \pi} \frac{\left(\mathrm{~T}_{\mathrm{GSS}}\right)}{(7)}\)
\(\mathrm{t}_0=\frac{3 \times 24 \text { hours }}{7}=\frac{24}{\mathrm{p}} \text { hours }\)
\(\mathrm{p}=\frac{7}{3}=2.33\)
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