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Periodic time of a satellite revolving above the earth's surface at a height equal to radius of the earth ' $R$ ' is [ $g=$ acceleration due to gravity]
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The correct answer is:
$4 \pi \sqrt{\frac{2 R}{g}}$
$\mathrm{T}=2 \pi \sqrt{\frac{(\mathrm{R}+\mathrm{h})^3}{\mathrm{gR}^2}}=2 \pi \sqrt{\frac{(2 \mathrm{R})^3}{\mathrm{gR}^2}} \quad \ldots .(\because \mathrm{h}=\mathrm{R})$
$=4 \pi \sqrt{\frac{2 \mathrm{R}}{\mathrm{g}}}$
$=4 \pi \sqrt{\frac{2 \mathrm{R}}{\mathrm{g}}}$
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