### Neet Solved Paper 2015 Question 13

##### Question: Kepler’s third law states that square of period of revolution (T) of a planet around the sun, is proportional to third power of average distance r between the sun and planet i.e. $ T^{2}=Kr^{3}, $ here K is constant. If the masses of the sun and planet are M and m respectively, then as per Newton’s law of gravitation force of attraction between them is $ F=\frac{GMm}{r^{2}} $ , here G is gravitational constant. The relation between G and K is described as

#### Options:

A) $ GK=4{{\pi }^{2}} $

B) $ GMK=4{{\pi }^{2}} $

C) $ K=G $

D) $ K=\frac{l}{G} $

## Show Answer

#### Answer:

Correct Answer: B

#### Solution:

The gravitational force of attraction between the planet and sun provide the centripetal force

i.e. $ \frac{GMm}{r^{2}}=\frac{mv^{2}}{r}\Rightarrow v=\sqrt{\frac{GM}{r}} $

The time period of planet will be $ T=\frac{2\pi r}{v} $

$ \Rightarrow T^{2}=\frac{4{{\pi }^{2}}r^{2}}{\frac{Gm}{r}}=\frac{4{{\pi }^{2}}r^{3}}{GM} $ ..(i)

Also from Kepler’s third law $ T^{2}=Kr^{3} $ ..(ii)

From Eqs. (i) and (ii), we get $ \frac{4{{\pi }^{2}}r^{3}}{GM}=Kr^{3}\Rightarrow GMK=4{{\pi }^{2}} $