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A satellite is revolving in a circular orbit at a height h from the carth surface, such that $\mathrm{h}< < \mathrm{R}$ where $\mathrm{R}$ is the radius of the earth. Assuming that the effect of earth"s atmosphere can be neglected the minimum increase in the speed required so that the satellite could escape from the gravitational field of earth is
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The correct answer is:
$\sqrt{g R}(\sqrt{2}-1)$
For a satellite orbiting close to the earth, orbital velocity is given by
$v_{0}=\sqrt{g(R+h)}=\sqrt{g R}$
Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)$ is
$v_{e}=\sqrt{2 g(R+h)}=\sqrt{2 g R} \quad[\because h < c R$
$\Delta v=v_{e}-v_{0}=(\sqrt{2}-1) \sqrt{g R}$
$v_{0}=\sqrt{g(R+h)}=\sqrt{g R}$
Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)$ is
$v_{e}=\sqrt{2 g(R+h)}=\sqrt{2 g R} \quad[\because h < c R$
$\Delta v=v_{e}-v_{0}=(\sqrt{2}-1) \sqrt{g R}$
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