JEE MAIN - Physics (2024 - 5th April Evening Shift - No. 12)
A satellite revolving around a planet in stationary orbit has time period 6 hours. The mass of planet is one-fourth the mass of earth. The radius orbit of planet is :
(Given $$=$$ Radius of geo-stationary orbit for earth is $$4.2 \times 10^4 \mathrm{~km}$$)
$$1.68 \times 10^5 \mathrm{~km}$$
$$1.4 \times 10^4 \mathrm{~km}$$
$$8.4 \times 10^4 \mathrm{~km}$$
$$1.05 \times 10^4 \mathrm{~km}$$
Explanation
$$\begin{aligned}
& \frac{T_1}{T_2}=\left(\frac{r_1}{r_2}\right)^{3 / 2} \sqrt{\frac{m_2}{m_1}} \\
& \Rightarrow \quad \frac{24}{6}=\left(\frac{4.2 \times 10^4}{r_2}\right)^{3 / 2} \sqrt{\frac{m / 4}{m}} \\
& \Rightarrow r_2=1.05 \times 10^4 \mathrm{~km}
\end{aligned}$$
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