MHT CET · Physics · 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]
- A \(2 \pi \sqrt{\frac{2 R}{g}}\)
- B \(4 \pi \sqrt{\frac{2 R}{g}}\)
- C \(2 \pi \sqrt{\frac{R}{g}}\)
- D \(8 \pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}}}\)
Answer & Solution
Correct Answer
(B) \(4 \pi \sqrt{\frac{2 R}{g}}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{T}=2 \pi \sqrt{\frac{(\mathrm{R}+\mathrm{h})^3}{\mathrm{gR}^2}}=2 \pi \sqrt{\frac{(2 \mathrm{R})^3}{\mathrm{gR}^2}} \quad \ldots .(\because \mathrm{h}=\mathrm{R})\)
\(=4 \pi \sqrt{\frac{2 \mathrm{R}}{\mathrm{g}}}\)
\(=4 \pi \sqrt{\frac{2 \mathrm{R}}{\mathrm{g}}}\)
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