The time period of revolution of a satellite close to planet's surface is 80 minutes. The time period of…

The time period of revolution of a satellite close to planet's surface is 80 minutes. The time period of another satellite which is at a height of 3 times the radius of the planet from surface is
  1. 64 minutes
  2. 640 minutes
  3. 320 minutes
  4. 240 minutes

Solution

By Kepler's period law, $\begin{aligned} & \mathrm{T}^2 \propto \mathrm{r}^3 \\ & \therefore\left(\frac{\mathrm{~T}_2}{\mathrm{~T}_1}\right)^2=\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^3=\left(\frac{\mathrm{R}+3 \mathrm{R}}{\mathrm{R}}\right)^3=64 \\ & \Rightarrow \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=8 \Rightarrow \mathrm{~T}_2=8 \mathrm{~T}_1=8 \times 80=640 \mathrm{~min} \end{aligned}$

Asked in: AP EAMCET 2024 (23 May Shift 1)

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