Question

A test particle is considered to be in motion along a circular orbit in the gravitational field created by a mass density $\rho \left( r \right)=\dfrac{K}{{{r}^{2}}}$. Find out the exact relation between the radius $R$ of the orbit of the particle and the time period $T$.
A. $\dfrac{T}{{{R}^{2}}}$will be a constant
B. $TR$ will be a constant
C. $\dfrac{{{T}^{2}}}{{{R}^{3}}}$ will be a constant
D. $\dfrac{T}{R}$will be a constant

Answer 1

Hint: Mass of a body is given as the product of the mass density of the body and the area of the body. This should be used to find the mass of the test particle. After that take the ratio between the time period and the radius of the orbit which will lead us to the answer. This will be helpful when you answer this question.

Complete step by step answer:
Mass of a body is given as the product of the mass density of the body and the area of the body. The mass density of the particle is given as,
$\rho \left( r \right)=\dfrac{K}{{{r}^{2}}}$
And the radius of the orbit has been mentioned as $R$. The mass of the test particle can be found using the equation which can be written as,
$m=\int\limits_{0}^{R}{\rho \times 4\pi {{r}^{2}}}dr$
Let us substitute the density equation in this will give,
\[m=\int\limits_{0}^{R}{\dfrac{K}{{{r}^{2}}}\times 4\pi {{r}^{2}}}dr=\int\limits_{0}^{R}{K\times 4\pi }dr\]
Integrating the equation according to the limits will give,
\[m=\int\limits_{0}^{R}{K\times 4\pi }dr=4\pi K\left[ r \right]_{0}^{R}=4\pi KR\]
From this, we can write that,
\[v\propto \sqrt{4\pi K}\]
Now let us take the ratio of the time period and the radius of the orbit which can be written as \[\dfrac{T}{R}\].
As the one circular orbit is having the total circumference as,
\[T=2\pi \]
And the radius can be shown as,
\[R=\sqrt{4\pi K}\]
Substituting the values in the ratio will give,
\[\dfrac{T}{R}=\dfrac{2\pi }{\sqrt{4\pi K}}=\text{constant}\]
This value will be a constant. Therefore the answer will be given as option D.

Note:
In planetary motion, the relation between the time period and the radius of the orbit is being perfectly explained using a law known as Kepler’s law. It says that the square of the orbital time period will be equivalent to the cube of the radius of the orbit. This law is valid as all the planets are orbiting around the sun.