Question

A moon of Saturn has a nearly circular orbit of radius R and an orbit period of T. Which of the following expressions gives the mass of Saturn?
$\text{A}\text{. }\dfrac{2\pi R}{T}$
$\text{B}\text{. }\dfrac{4{{\pi }^{2}}R}{T}$
$\text{C}\text{. }\dfrac{\text{2}\pi {{R}^{3}}}{G{{T}^{2}}}$
$\text{D}\text{. }\dfrac{4{{\pi }^{2}}{{R}^{3}}}{G{{T}^{2}}}$

Answer 1

The gravitational pull by Saturn provides the centripetal force for the circular motion of its moon. Compare gravitational force and centripetal force to determine the expression for mass of Saturn. Use the orbit period to obtain an expression for velocity of moon.

Formula used: ${{F}_{g}}=\dfrac{GMm}{{{R}^{2}}}$, ${{F}_{cp}}=\dfrac{m{{v}^{2}}}{R}$ and $T=\dfrac{2\pi R}{V}$

Complete step by step answer:
The moon is moving in a circular orbit around Saturn. The net centripetal force acting upon this moon is given by the relation
${{F}_{cp}}=\dfrac{m{{v}^{2}}}{R}$
Where
${{F}_{cp}}=$Centripetal force acting upon the moon
$m=$Mass of the moon
$v=$Velocity of the moon
$R=$Radius of the circular orbit
This net centripetal force is the consequence of the gravitational force that attracts the moon towards the Saturn and can be represented as
${{F}_{g}}=\dfrac{GMm}{{{R}^{2}}}$
Where
$G=$Gravitational constant=$6.673\times {{10}^{-11}}\text{ N}{{\text{m}}^{\text{2}}}\text{k}{{\text{g}}^{\text{-2}}}$
$M=$Mass of Saturn
Since ${{F}_{g}}={{F}_{cp}}$, the above expressions for gravitational force and centripetal force can be equated. Therefore,
$\dfrac{GMm}{{{R}^{2}}}=\dfrac{m{{v}^{2}}}{R}$
We observe that the mass of the moon is present on both sides of the equation; which can be cancelled by dividing through by mass of the moon. Then both sides of the equation can be multiplied with $R$. Then we rearrange the equation and obtain
${{v}^{2}}=\dfrac{GM}{R}$
Now we know that velocity of a body moving in circular orbit of radius $R$ with time period $T$, the speed $v$ is given by
$v=\dfrac{2\pi R}{T}$
We now square this equation and then equate with its former equation.
${{\left( \dfrac{2\pi R}{T} \right)}^{2}}=\left( \dfrac{GM}{R} \right)$
On rearranging above equation, we get
$M=\dfrac{4{{\pi }^{2}}{{R}^{3}}}{G{{T}^{2}}}$

So, the correct answer is “Option D”.

Note: Generally, mass of an object is a constant quantity. Here, we can observe from the result that ${{R}^{3}}\propto {{T}^{2}}$. This is Kepler’s third law which states that, “the square of the period is proportional to the cube of the radius of the circular orbit.” Kepler’s laws are also valid for elliptical paths. In that case, the radius of the circular orbit in third law gets replaced with the semi-major axis of the ellipse.