Question 340: 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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