Question 324: Gravitation

Question

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]

Accepted Answer

$\begin{aligned}
\mathrm{T}=2 \pi \sqrt{\frac{(\mathrm{R}+\mathrm{h})^3}{\mathrm{gR}^2}} & =2 \pi \sqrt{\frac{(2 \mathrm{R})^3}{\mathrm{gR}^2}} \quad \cdots .(\because \mathrm{h}=\mathrm{R}) \
& =4 \pi \sqrt{\frac{2 \mathrm{R}}{\mathrm{g}}}
\end{aligned}$

Answer

$2 \pi \sqrt{\frac{2 R}{g}}$

Answer

$4 \pi \sqrt{\frac{2 R}{g}}$

Answer

$2 \pi \sqrt{\frac{R}{g}}$

Answer

$8 \pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$

Gravitation

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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