Periodic time of a satellite revolving above the earth's surface at a height equal to radius of the earth '…

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]
  1. $2 \pi \sqrt{\frac{2 R}{g}}$
  2. $4 \pi \sqrt{\frac{2 R}{g}}$
  3. $2 \pi \sqrt{\frac{R}{g}}$
  4. $8 \pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$

Solution

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

Asked in: MHT CET 2023 (09 May Shift 2)

Practice more Gravitation questions on Aicharya