[time-nuts] advice

Bruce Griffiths bruce.griffiths at xtra.co.nz
Wed Feb 22 19:09:52 EST 2017

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.

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

 > Δf/fo = g Δh/c2

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.

time-nuts mailing list -- time-nuts at febo.com
To unsubscribe, go to https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
and follow the instructions there.


More information about the time-nuts mailing list