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Question

(B) The time period $T$ of a satellite orbiting just above the surface of the Earth is given by $T = 2 \pi \sqrt{\frac{R}{g}}$,where $R$ is the radius

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$T^2$

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(B) The time period $T$ of a satellite orbiting just above the surface of the Earth is given by $T = 2 \pi \sqrt{\frac{R}{g}}$,where $R$ is the radius of the Earth and $g$ is the acceleration due to gravity.We know that $g = \frac{GM}{R^2}$,where $M$ is the mass of the Earth.The mass of the Earth can be expressed in terms of its density $d$ as $M = \frac{4}{3} \pi R^3 d$.Substituting $M$ into the expression for $g$,we get $g = \frac{G (\frac{4}{3} \pi R^3 d)}{R^2} = \frac{4}{3} \pi G R d$.Now,substitute $g$ back into the time period formula: $T = 2 \pi \sqrt{\frac{R}{\frac{4}{3} \pi G R d}} = 2 \pi \sqrt{\frac{3}{4 \pi G d}} = \sqrt{\frac{4 \pi^2 \cdot 3}{4 \pi G d}} = \sqrt{\frac{3 \pi}{G d}}$.Squaring both sides,we get $T^2 = \frac{3 \pi}{G d}$.

$A$ satellite is orbiting just above the surface of the earth with period $T$. If $d$ is the density of the earth and $G$ is the universal constant of gravitation,the quantity $\frac{3 \pi}{G d}$ represents :

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