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If the density of planet is $\rho$ and a satellite is revolving in circular orbit of radius r then time of revolution vareies with density $\rho$ as :
Option: 1
$T\propto \sqrt{\rho }$

Option: 2
$T\propto \frac{1}{\rho}$

Option: 3
$T\propto \rho$

Option: 4
$T\propto \frac{1}{\sqrt{\rho}}$

Answers (1)

best_answer

As we learn

Time period of satellite in terms of density -

$T=\sqrt{\frac{3\pi}{G\rho }}$

$\rho \rightarrow$ Density of planet

$T\rightarrow$ Time period

$G\rightarrow$ Gravitational constant

- wherein

$\rho= 5478.4Kg/m^{3}$ for earth

As we known

$T = \frac{2\pi r^{\frac{3}{2}}}{\sqrt{GM}}$

$M = \rho *\frac{4}{3}\pi r^{3}$

so, $T = \frac{2\pi r^{\frac{3}{2}}}{\sqrt{G\rho *\frac{4}{3}\pi R^{3}}}$

$T =\sqrt{\frac{3\pi }{G\rho }}$

$T \propto \frac{1}{\sqrt{\rho }}$

Posted by

manish

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