Gravitation

JEE Main level, single correct, medium

204 of 223

Question 204 of 223

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 :
(\left(\right. Given == Radius of geo-stationary orbit for earth is 4.2×104 km4.2 \times 10^4 \mathrm{~km} )

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Answer: (B) 1.05×104 km1.05 \times 10^4 \mathrm{~km}

T=2πr3/2GMT1 T2=(r1r2)3/2(M2M1)1/2624=(r1)3/2(4.2×104)3/2(MM/4)1/2r1=1.05×104 km\begin{aligned} & \mathrm{T}=\frac{2 \pi \mathrm{r}^{3 / 2}}{\sqrt{\mathrm{GM}}} \\ & \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^{3 / 2}\left(\frac{\mathrm{M}_2}{\mathrm{M}_1}\right)^{1 / 2} \\ & \frac{6}{24}=\frac{\left(\mathrm{r}_1\right)^{3 / 2}}{\left(4.2 \times 10^4\right)^{3 / 2}}\left(\frac{\mathrm{M}}{\mathrm{M} / 4}\right)^{1 / 2} \\ & \mathrm{r}_1=1.05 \times 10^4 \mathrm{~km}\end{aligned}

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