The time period of a satellite, revolving above earth’s surface at a height equal to will be (Given , R = radius of earth) [2023]

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Solution

Q.
The time period of a satellite, revolving above earth’s surface at a height equal to $R$ will be (Given $g = π^{2} {m/s}^{2}$ , R = radius of earth) [2023]

1 $\sqrt{4 R}$

2 $\sqrt{8 R}$

3 $\sqrt{32 R}$

4 $\sqrt{2 R}$

Ans.
(3)

$\frac{m v^{2}}{2 R} = \frac{G M m}{{(2 R)}^{2}}$

$\Rightarrow v = \sqrt{\frac{G M}{2 R}} = \sqrt{\frac{R g}{2}}$

$T = \frac{2 π (2 R)}{v} = \frac{4 π R \sqrt{2}}{\sqrt{R g}} = \sqrt{32 R}$

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