Time Period of Satellite at a Given Height – Rankers Physics

Planet and Satellite: Practice Problem & Solution

A geostationary satellite is orbiting the earth at a height of $5R$ above that surface of the earth, $R$ being the radius of the earth. The time period of another satellite in hours at a height of $2R$ from the surface of the earth is: (2012 Pre)
$5$
$10$
$6\sqrt{2}$
$\sqrt{2}$

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Planet and Satellite. Understanding the underlying formula is key to arriving at the correct answer below:

For the geostationary satellite, $T_1 = 24\text{ hours}$, $r_1 = R + 5R = 6R$. For the second satellite, $r_2 = R + 2R = 3R$. Using Kepler's third law $T^2 \propto r^3$, we have $T_2 = T_1 \left(\frac{r_2}{r_1}\right)^{3/2} = 24 \left(\frac{3R}{6R}\right)^{3/2} = 24 \left(\frac{1}{2}\right)^{3/2} = 6\sqrt{2}\text{ hours}$.

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