GATE-2018 (24) :Wyllie – Time average equation

 

Compared to the earth Neptune is approximately 30 times farther away from the sun, its period of revolution around the sun ( in Neptune years) would be___

(Thanks to Chandrasekhar, ANU)

Solution:

Kepler’s Third law states that “the square of the period of revolution of the planet (T) is directly proportional to the cube of the distance of the planet from the sun (a)”. 

$T^{2}\propto a^{3}$

$ T^{2}=\frac{4\pi^{2}}{GM}a^{3}$--------(1)

 For the above formula

 For the earth $ T_{E}^{2}=\frac{4\pi^{2}}{GM}a_{E}^{3}$

 For the Neptune $ T_{N}^{2}=\frac{4\pi^{2}}{GM}a_{N}^{3}$

By dividing the above two equations

 

$\frac{T_{N}^{2}}{T_{E}^{2}}=\frac{a_{N}^{3}}{a_{E}^{3}}$

 

$\frac{T_{N}^{2}}{T_{E}^{2}}=\frac{(30\times a_{E})^{3}}{a_{E}^{3}}$

 

$\frac{T_{N}^{2}}{T_{E}^{2}}=\frac{(27000)a_{E}^{3}}{a_{E}^{3}}$

 

$\frac{T_{N}^{2}}{T_{E}^{2}}=27000$

 

$(\frac{T_{N}}{T_{E}})^{2}=27000$

 

$(\frac{T_{N}}{T_{E}})=(27000)^{1/2}$

 

$(\frac{T_{N}}{T_{E}})=165.31$

 

$T_{N}=165.31 T_{E}$

 The period of revolution of Neptune is 165 times of Earth.

 We know that the one Neptune year is equal to 165 years.

 

So the Period of revolution of Neptune around the sun is 1 Neptune year