A body weighs 64 N on the surface of the Earth. What is the gravitational force on it (in N) due to the Earth at a height equal to one-third of the radius of the Earth?

Answer

`1.85xx10^8` year

Explanation

Total number of helium atoms =`(6.023xx10^23)/(4xx10^(-3))xx5.0xx10^32` atoms
=`7.52875 xx10^58` atoms
3 helium atoms produce 7.27 MeV of energy =`7.27xx1.6xx10^(-13)`J
`"Total energy produced by all the atoms"/"Power"`=Time taken