# In the book of physics it is Given that the Gravitational Potential Energy at the center of earth(Vc) = 3/2(Vs?

Where (Vs) is the G.P.E. at the surface of earth.

=>Vc = -3/2 [GM/r], Can any one tell me How & from Where This (Vc) = 3/2 (Vs) is derived?

Relevance

The gravitational potential energy of an object on the surface of the Earth is the integral of the acceleration of gravity from r = infinity to R:

∫ GM(earth)/r² dr

This quantity equals -GM(earth)/R.

If Earth is assumed to be a sphere of uniform density (it isn't, but for the sake of argument) the difference in energy between the surface and the center is also the integral of the acceleration of gravity from R to 0, but the acceleration of gravity follows a different curve.  Inside a sphere of uniform density the mass inside the radius decreases as the cube of the radius:  m = 4/3πρr³.  Since the acceleration is Gm/r², the acceleration is 4/3Gπρr.  If you work this out, the ΔE between r=0 and r=R is -½GM(earth)/R.

Add the two together and you get -3/2GM/R.

Source(s): Worked this out for my own amusement, or maybe it was physics homework one day