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Q.

A statellite is orbiting just above the surface of the rearth with period T. If d is the density of the earth and G is the universal contant of gravitation, the quantity $\frac{3 π}{Gd}$ represents:

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a

$\sqrt{T}$

b

T

c

T²

d

T³

answer is C.

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Detailed Solution

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The centripetal force of the satellite is given by gravitational force.
$\Rightarrow \frac{{GM}_{e} m}{R^{2}} = mR ω^{2}$
(here $ω = \frac{2 π}{T}, M_{e} = \frac{4 π}{3} R^{3} d)$
$\begin{array}{l} \Rightarrow \frac{G}{R^{2}} \cdot \frac{4 π}{3} R^{3} \cdot d \cdot m = m \cdot R {(\frac{2 π}{T})}^{2} \\ \Rightarrow T^{2} = \frac{3 π}{G d} \end{array}$

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