time period of satellite

Question

A satellite revolves in a circular orbit where the radius is 10⁷ m and the acceleration due to gravity is 10 m/s². What is its time period?

RankGuru · Accepted Answer

1. **Centripetal acceleration**: The local value of gravity $g'$ at that path provides the needed radial acceleration.
   - $g' = \frac{v^2}{r} \implies v = \sqrt{g' r}$.
2. **Velocity Formula**: $v$ is the speed of traversal.
3. **Period Definition**: $T = \frac{2\pi r}{v} = \frac{2\pi r}{\sqrt{g' r}} = 2\pi \sqrt{\frac{r}{g'}}$.
4. **Apply Numbers**:
   - $T = 2\pi \sqrt{\frac{10^7}{10}} = 2\pi \sqrt{10^6}$.
5. **Final Value**: $T = 2\pi 	imes 10^3$ seconds.

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time period of satellite

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