# 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) \[\sqrt {\dfrac{{3\pi G}}{\rho }} \]

Answer

Verified

290.1k+ views

**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.

Last updated date: 03rd Jun 2023

•

Total views: 290.1k

•

Views today: 4.46k

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Name the Largest and the Smallest Cell in the Human Body ?

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

A ball impinges directly on a similar ball at rest class 11 physics CBSE

Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

How do you define least count for Vernier Calipers class 12 physics CBSE

Two balls are dropped from different heights at different class 11 physics CBSE

A 30 solution of H2O2 is marketed as 100 volume hydrogen class 11 chemistry JEE_Main