Does that work when going down below the surface as well as when up above it?

(My last physics class was a long time ago.  I remember doing the integrals
for computing the gravity inside a sphere, but don't remember the answer.  I
wouldn't be surprised if a factor of 2 or pi was involved below the surface.)

--
These are my opinions.  I hate spam.

For a spherical body of uniform density the value of g below the surface is proportional to  the radial distance of the location from the centre of the sphere.  For a spherically symmetric body only the mass contained within the sphere  below the point has any effect on the measured value of g.
The record for optical clocks is detecting the effect on frequency of a change in elevation of 1cm or so.
Bruce

On Thursday, 23 February 2017 1:02 PM, Hal Murray <hmurray@megapathdsl.net> wrote:

Does that work when going down below the surface as well as when up above it?

(My last physics class was a long time ago.  I remember doing the integrals
for computing the gravity inside a sphere, but don't remember the answer.  I
wouldn't be surprised if a factor of 2 or pi was involved below the surface..)

--
These are my opinions.  I hate spam.

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