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(C) The time period $T$ of a satellite orbiting just above the surface of a planet of radius $R$ is given by $T = 2 \pi \sqrt{\frac{R}{g}}$.Squaring b

Vedclass · Accepted Answer

$\frac{3 \pi}{G}$

Vedclass · Answer

(C) The time period $T$ of a satellite orbiting just above the surface of a planet of radius $R$ is given by $T = 2 \pi \sqrt{\frac{R}{g}}$.Squaring both sides,we get $T^2 = 4 \pi^2 \frac{R}{g}$.We know that the acceleration due to gravity $g$ at the surface of a planet is given by $g = \frac{GM}{R^2}$.Since the mass $M$ of the planet is $M = 	ext{Volume} 	imes 	ext{Density} = \frac{4}{3} \pi R^3 ho$,we substitute $M$ into the expression for $g$:$g = \frac{G}{R^2} \left( \frac{4}{3} \pi R^3 ho ight) = \frac{4}{3} \pi ho G R$.Now,substitute this expression for $g$ into the equation for $T^2$:$T^2 = 4 \pi^2 \frac{R}{\frac{4}{3} \pi ho G R} = 4 \pi^2 	imes \frac{3}{4 \pi ho G} = \frac{3 \pi}{ho G}$.Rearranging the terms,we get $T^2 ho = \frac{3 \pi}{G}$.

$A$ satellite is orbiting just above the surface of a planet of density $\rho$ with a periodic time $T$. The quantity $T^2 \rho$ is equal to ($G=$ universal gravitational constant).

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