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A satellite is revolving very close to a planet of density ρ. The period of revolving of satellite is:
A) \[\sqrt {\dfrac{{3\pi \rho }}{G}} \]
B) \[\sqrt {\dfrac{{3\pi }}{{2\rho G}}} \]
C) \[\sqrt {\dfrac{{3\pi }}{{\rho G}}} \]
D) \[\sqrt {\dfrac{{3\pi G}}{\rho }} \]

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Last updated date: 30th May 2024
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Answer
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Hint: Try to recall the formula for the time period of a body revolving around the planet of mass M having at height equal to radius of the planet. With the given density in the question we can calculate the mass of the planet. In extras you can eliminate the options by checking the dimensions of each option if you don’t know much about the concept and proceed with any one option which has the dimensions of time.

Complete step by step answer:
First if you don't know the formula of Time period you do not have to worry It is very easy to calculate.
Just equate centripetal force on satellite to the gravitational force between satellite and planet.
$\implies$ \[\dfrac{{m{v^2}}}{r} = \dfrac{{GMm}}{{{r^2}}}\]
$\implies$ \[v = \sqrt {\dfrac{{GM}}{r}} \] ------(1)
Where \[M\] = mass of the planet , \[m\]= mass of the satellite , \[r\]= radius of the satellite , \[v\]= velocity with which the satellite is revolving around the planet
$\implies$ \[T = \dfrac{{2\pi }}{\omega }\] and \[\omega = \dfrac{v}{r}\]
Putting value of \[v\] from eq. 1
$\implies$ \[\omega = \sqrt {\dfrac{{GM}}{{{r^3}}}} \]
Putting \[\omega \] in Time Period formula we get
$\implies$ \[T = 2\pi \sqrt {\dfrac{{{r^3}}}{{GM}}} \]-----(2)
Now here M = mass of the planet , we know mass = Volume × density
$\implies$ \[M = \rho \times V = \rho \times \dfrac{4}{3}\pi {r^3}\]
Putting value of M in (2)
$\implies$ \[T = 2\pi \sqrt {\dfrac{{{r^3}}}{{G \times \rho \times \dfrac{4}{3}\pi {r^3}}}} \]
$\implies$ \[T = \sqrt {\dfrac{{4{\pi ^2}{r^3}}}{{G \times \rho \times \dfrac{4}{3}\pi {r^3}}}} \]
$\implies$ \[T = \sqrt {\dfrac{{3\pi }}{{G\rho }}} \]
Thus option C is correct.

Note: In the given question we have found the time period of the revolution by equating the centripetal force equal to the gravitational force. You can remember this formula for faster calculation. In advanced we can also calculate the time period of revolution at height H above the Earth. Notice here that time period is independent of the mass of the satellite.