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A satellite of mass $m$ revolves around the earth of radius $R$ at a height $x$ from its surface. If $g$ is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is
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Verified Answer
The correct answer is:
$\left(\frac{g R^2}{R+x}\right)^{1 / 2}$
$\left(\frac{g R^2}{R+x}\right)^{1 / 2}$
For the satellite, the gravitational force provides the necessary centripetal force i.e.
$$
\begin{aligned}
& \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{(\mathrm{R}+\mathrm{X})^2}=\frac{\mathrm{Mv}_0^2}{(\mathrm{R}+\mathrm{X})} \text { and } \frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}^2}=\mathrm{g} \\
& \therefore \mathrm{v}_0=\left(\frac{g \mathrm{R}^2}{\mathrm{R}+\mathrm{X}}\right)^{1 / 2} \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{(\mathrm{R}+\mathrm{X})^2}=\frac{\mathrm{Mv}_0^2}{(\mathrm{R}+\mathrm{X})} \text { and } \frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}^2}=\mathrm{g} \\
& \therefore \mathrm{v}_0=\left(\frac{g \mathrm{R}^2}{\mathrm{R}+\mathrm{X}}\right)^{1 / 2} \\
&
\end{aligned}
$$
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