Processing math: 100%

Introduction to gpsurya blog

Dear Friends, In this blog you will find around 320 solved Geophysics GATE and CSIR-NET and other competitive exams solutions with a better explanation. Please follow and share if you like it. Thanks, gpsurya and group

CSIR-NET 2020 JUNE

Post a Comment

 

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

Related Posts

There is no other posts in this category.

Post a Comment

Subscribe Our Newsletter