# Change in gravitational potential below the surface of the Earth

Gold Member

## Homework Statement

Hi,
Infinitely far away from a mass-->gravitational potential is zero.
As get closer-->becomes negative.
At surface-->it is the smallest value of r, i.e. the radius of the mass, hence the most negative value for gravitational potential.
But as you go below surface of Earth (say Earth is the mass, and the only mass), what happens to the value of gravitational potential as you go below surface of the Earth?

V=-GM/r

## The Attempt at a Solution

I thought that, as the mass decreases by a cubic factor (volume), but the radius decreases by a linear factor, then the gravitational potential would increase (become more positive) by a factor to the power of 2, reaching zero at centre of earth?

Is this right?
Thanks!

BvU
Homework Helper
Afraid not. Why would the potential be zero at the center ? Gravitation force IS zero there, though.

Compare (897) here with (4.10.3) here

Gold Member
Afraid not. Why would the potential be zero at the center ? Gravitation force IS zero there, though.

Compare (897) here with (4.10.3) here
Thanks for your reply. Could you explain why it is not zero? As in, what would it be? A non-zero finite negative value? Infinite? What would it be?

haruspex
Homework Helper
Gold Member
2020 Award
what would it be?

Once you move inside the radius of the Earth, the field gets weaker, but it is still directed towards the Earth's centre. So the potential must be still decreasing.

## Homework Statement

Hi,
Infinitely far away from a mass-->gravitational potential is zero.
As get closer-->becomes negative.
At surface-->it is the smallest value of r, i.e. the radius of the mass, hence the most negative value for gravitational potential.
But as you go below surface of Earth (say Earth is the mass, and the only mass), what happens to the value of gravitational potential as you go below surface of the Earth?

V=-GM/r

## The Attempt at a Solution

I thought that, as the mass decreases by a cubic factor (volume), but the radius decreases by a linear factor, then the gravitational potential would increase (become more positive) by a factor to the power of 2, reaching zero at centre of earth?

Is this right?
Thanks!

If you know calculus already, the proper way to solve this would be to express the potential as an integral of the force divided by the test mass:

$$V(r) = -\int_r^\infty\frac{GM(r)}{r^2}dr$$

While you're above the Earth's surface, M is constant and the integral is simply the integral of ##\frac{1}{r^2}##. When you're below you replace M by ##\frac{4}{3}\pi\rho r^3## (with ##\rho## density of the Earth, if we consider it homogeneous, which it isn't but let's not go there). So the integral does not explode at ##r=0## and has instead a finite value, but still, it's not zero.

Without calculus I'm not sure how you can derive the correct formula, but you can realise at least that the rate of change of the potential never changes sign. If the potential is zero at infinity and negative at the surface, it would take a repulsive force to bring it back to zero, and that never happens. Even when you're below the surface all the layers above you merely cancel out their attraction to zero, they don't pull you upwards. So the potential keeps becoming more negative, it simply does so at a slower rate and does not become infinitely negative.