A geo-stationary satellite is orbiting earth at a height of 5R above surface of earth, R being radiu

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7.Gravitation

normal

A geo-stationary satellite is orbiting the earth at a height of $5R$ above surface of the earth, $R$ being the radius of the earth. The time period of another satellite in hours at a height of $2R$ from the surface of earth is

$6 \sqrt 2$

$\frac{6}{{\sqrt 2 }}$

$5$

$10$

Solution

$T = \frac{{2\pi {r^{3/2}}}}{{\sqrt {Gm} }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,T \propto {r^{3/2}}\left| \begin{gathered}
Time\,period\,of \hfill \\
Gss – 24\,hours \hfill \\
\end{gathered} \right.$

$\Rightarrow \frac{\mathrm{T}_{1}}{\mathrm{T}_{2}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{3 / 2} \Rightarrow \frac{24}{\mathrm{T}_{2}}=\left(\frac{6 \mathrm{R}}{2 \mathrm{R}}\right)^{3 / 2}$

$\Rightarrow \mathrm{T}_{2}=6 \sqrt{2}$ hours

Standard 11

Physics

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A geo-stationary satellite is orbiting the earth at a height of $5R$ above surface of the earth, $R$ being the radius of the earth. The time period of another satellite in hours at a height of $2R$ from the surface of earth is

Solution

Similar Questions

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