# Value of g inside Earth

g is proportional to r, which can be dervied using newton's law of grav and substituting m = density * volume of a sphere,

so can g=GM/r^2 still be used inside the Earth?

I assume not but kinda confused as it was used to derive that g is proportional to r...

## Answers and Replies

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L-x
As you were asking this question I assume you are comfortable with integration.

g=GM/r^2 may be used on a "small piece" of the earth, you can then perform an integral over all the "small pieces" at a distance d from the centre to find that g is proportional to d inside the earth.

Spherical symmetry will make the integration easier. Remember that r in GM/r^2 will be different depending on which "piece" of earth you are considering, not just equal to d.

IIRC it turns out you need only consider contributions closer to the centre than the d you have chosen. The contributions from parts of the "shell" d<r<R where R is the radius of the earth and r is the distance from the center exactly cancel. For a full derivation you should prove that they cancel.

Edit: Obviously you don't need to integrate to find the contribution from the sphere 0<r<d, see below. Integration is actually only required to prove that the contributions from the shell sum to 0, and can therefore be ommitted.

Last edited:
gneill
Mentor
g is proportional to r, which can be dervied using newton's law of grav and substituting m = density * volume of a sphere,

so can g=GM/r^2 still be used inside the Earth?
Sure, but substitute for M the mass of the sphere of the Earth below your feet at your radial distance r. Ignore the spherical shell from r through R, where R is the Earth's surface radius.

Homework question: Supposing that the Earth had a uniform density $$\rho$$, find an expression for the local value of g, in terms of density and radius, at radius r, where 0 < r < R.