Question

Kepler's third law states that square of period revolution (T) of a planet around the sun is proportional to third power of average distance between sun and planet
i.e. ${{T}^{2}}=K{{r}^{3}}$ , here K is a constant.
if the mass of sun and planet are M and m respectively then as per Newton's law of gravitation the force of attraction between them is
$F=\dfrac{GMm}{{{r}^{2}}}$ , here GG is gravitational constant. The relation between G and K is described as
A. $K=G$
B. $K=\dfrac{1}{G}$
C. $GK=4{{\pi }^{2}}$
D. $GMK=4{{\pi }^{2}}$

Answer 1

Hint: Kepler’s third law tells us the relation between the time period of revolution and the distance between the sun. We can get this equation by using the gravitational and centripetal force and using the concepts of circular motion. Find the relation and compare it to the given equation in question. We will get our answer.

Complete step by step answer:
Let the mass of the planet is m and the mass of the sun is M. T is the time period of evolution for the planet around the sun.
From Kepler’s law we have that,
${{T}^{2}}=K{{r}^{3}}\text{ }\to \text{ 1}$
Now, let the velocity of the planet around the sun is v.
So, we can write that, $\begin{aligned} & v=\dfrac{2\pi r}{T} \\\ & T=\dfrac{2\pi r}{v} \\\ \end{aligned}$
Now, to find the value of v we use the concepts of circular motion.
The centripetal force acting on the planet which is moving in a circular orbit around the sun at a distance r is given by,
${{F}_{c}}=\dfrac{m{{v}^{2}}}{r}$
Law of universal gravitation gives the gravitational force between two massive objects with a distance r apart.
Again, the gravitational force on the planet due to sun is given by,
${{F}_{G}}=\dfrac{GMm}{{{r}^{2}}}$
equating the above force, we will get that,
$\begin{aligned} & {{F}_{c}}={{F}_{G}} \\\ & \dfrac{m{{v}^{2}}}{r}=\dfrac{GMm}{{{r}^{2}}} \\\ & {{v}^{2}}=\dfrac{GM}{r} \\\ \end{aligned}$
Now putting this value on the time period,
$\begin{aligned} & T=\dfrac{2\pi r}{v} \\\ & {{T}^{2}}=\dfrac{4{{\pi }^{2}}{{r}^{2}}}{{{v}^{2}}} \\\ & {{T}^{2}}=\dfrac{4{{\pi }^{2}}{{r}^{2}}}{\dfrac{GM}{r}} \\\ & {{T}^{2}}=\dfrac{4{{\pi }^{2}}}{GM}{{r}^{3}}\text{ }\to \text{ 2} \\\ \end{aligned}$
Comparing equation 1 and equation 2, we get that,
$\begin{aligned} & K=\dfrac{4{{\pi }^{2}}}{GM} \\\ & GMK=4{{\pi }^{2}} \\\ \end{aligned}$
So, the correct option is (D)
Note: Kepler’s laws give us the motions of the planets around the sun. The third law gives us the period of orbit of a planet around the sun. With these three laws we can find most of the physical quantities related to the motion of planets around the sun.