?Earth's mass below a given elevation level? [Archive]

Gene Partlow

Aug4-04, 01:20 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>This is partly a question in geophysics, but I have to begin with a\nbrief excursion in gravitation physics...\n\nThe gravitational redshift predicted by general relativity was veri-\nfied by the Pound-Rebka experiment, where they measured the\nredshift in wavelengths of photons traveling from the ground at\nHarvard Univ. straight up to a device 22.6 meters higher in the\ngravitational field. They measured the gravitational acceleration\ng at both levels and used that to predict the redshift theoretically,\nwhich indeed closely matched what they observed experimen-\ntally.\n\nWhat I would like is this: without \'measuring\' g at either level,\nit would be interesting to predict what g should be at both\nlevels, since\n\ng = (G M) / r^2.\n\nG is the Newtonian grav. constant, M the mass of the earth below\nthe surface of the geocentric sphere whose radius is r.\n\nLet\'s say for sake of discussion that M_o is earth\'s mass below that\n\'surface\' at r_o (ground level) and that it\'s 5.98*10^24 kg. exactly.\nBut this means that M_1 is the earth\'s (greater) mass below the\nlarger surface at r_1 (22.6 meters higher). That is, the spherical\nshell between r_o and r_1 will contain all sorts of mass from air,\nwater, rock, lawyers etc. from masses whose elevation, all around\nthe earth, is somewhere between r_o and r_1.\n\nMy problem in a nutshell is to estimate the small but non-zero mass\ndifference between M_o and M_1. While such a delta M is not\ngoing to materially influence either the classical P-R experiment\nor the calculation of g from \'scratch\', I still would like to know its\neffect *in principle*. Sooo...\n\nIs there any data set, perhaps at USGS?, which allows such an\nestimation to be made, or at least interpolated?\n\ncheers,\n\'Gene\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>This is partly a question in geophysics, but I have to begin with a
brief excursion in gravitation physics...

The gravitational redshift predicted by general relativity was veri-
fied by the Pound-Rebka experiment, where they measured the
redshift in wavelengths of photons traveling from the ground at
Harvard Univ. straight up to a device 22.6 meters higher in the
gravitational field. They measured the gravitational acceleration
g at both levels and used that to predict the redshift theoretically,
which indeed closely matched what they observed experimen-
tally.

What I would like is this: without 'measuring' g at either level,
it would be interesting to predict what g should be at both
levels, since

g = (G M) / r^2.

G is the Newtonian grav. constant, M the mass of the earth below
the surface of the geocentric sphere whose radius is r.

Let's say for sake of discussion that M_o is earth's mass below that
'surface' at r_o (ground level) and that it's 5.98*10^24 kg. exactly.
But this means that M_1 is the earth's (greater) mass below the
larger surface at r_1 (22.6 meters higher). That is, the spherical
shell between r_o and r_1 will contain all sorts of mass from air,
water, rock, lawyers etc. from masses whose elevation, all around
the earth, is somewhere between r_o and r_1.

My problem in a nutshell is to estimate the small but non-zero mass
difference between M_o and M_1. While such a \delta M is not
going to materially influence either the classical P-R experiment
or the calculation of g from 'scratch', I still would like to know its
effect *in principle*. Sooo...

Is there any data set, perhaps at USGS?, which allows such an
estimation to be made, or at least interpolated?

cheers,
'Gene

grelbr

Aug5-04, 03:24 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>stargene@earthlink.net (Gene Partlow) wrote in message news:<c504f3da.0408022351.1d8d553@posting.google.c om>...\n> What I would like is this: without \'measuring\' g at either level,\n> it would be interesting to predict what g should be at both\n> levels, since\n>\n> g = (G M) / r^2.\n>\n> G is the Newtonian grav. constant, M the mass of the earth below\n> the surface of the geocentric sphere whose radius is r.\n\nThere are lots of complications here. This applies to mass in\nspherical symmetry only. If you are going to take into account\nsuch variations as mountains, tidal bulge, possibly differences\nin density of various parts, and whatnot, then you need to be\nmore careful with your form of g also.\n\n[snip]\n> Is there any data set, perhaps at USGS?, which allows such an\n> estimation to be made, or at least interpolated?\n\nMy expectation is that the leading non-spherical effect will\nlikely be due to the Earth not being spherical but slightly\n"squashed" due to rotation. After that, you\'ve got tidal bulge,\nmountains, ocean basins, and maybe density fluctuations.\n\nDoing such things at any more accuracy than you could measure g\nis likely to be quite a chore.\n\nYou could start with the folks doing gravity measurements down\nmines a few years ago. They thought they had a fifth force.\nThat seems to have gone away after more careful measurements.\ngrelbr\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>stargene@earthlink.net (Gene Partlow) wrote in message news:<c504f3da.0408022351.1d8d553@posting.google.com>...
> What I would like is this: without 'measuring' g at either level,
> it would be interesting to predict what g should be at both
> levels, since
>
> g = (G M) / r^2.
>
> G is the Newtonian grav. constant, M the mass of the earth below
> the surface of the geocentric sphere whose radius is r.

There are lots of complications here. This applies to mass in
spherical symmetry only. If you are going to take into account
such variations as mountains, tidal bulge, possibly differences
in density of various parts, and whatnot, then you need to be
more careful with your form of g also.

[snip]
> Is there any data set, perhaps at USGS?, which allows such an
> estimation to be made, or at least interpolated?

My expectation is that the leading non-spherical effect will
likely be due to the Earth not being spherical but slightly
"squashed" due to rotation. After that, you've got tidal bulge,
mountains, ocean basins, and maybe density fluctuations.

Doing such things at any more accuracy than you could measure g
is likely to be quite a chore.

You could start with the folks doing gravity measurements down
mines a few years ago. They thought they had a fifth force.
That seems to have gone away after more careful measurements.
grelbr

Gene Partlow

Aug6-04, 03:06 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>grelbr@hotmail.com (grelbr) wrote in message news:<\n> > Is there any data set, perhaps at USGS?, which allows such an\n> > estimation to be made, or at least interpolated?\n>\n> My expectation is that the leading non-spherical effect will\n> likely be due to the Earth not being spherical but slightly\n> "squashed" due to rotation. After that, you\'ve got tidal bulge,\n> mountains, ocean basins, and maybe density fluctuations.\n>\n> Doing such things at any more accuracy than you could measure g\n> is likely to be quite a chore.\n\nRight...I agree it\'s messy, but I\'ve just found out belatedly that modern\ngravimeters (eg: an absolute gravimeter called the \'FG5\') can actually\nmeasure differences in g over elevation differences of 3 mm (!). In\nprinciple such a device would automatically take all that mess into\naccount and yield an appropriate delta g between the top and\nbottom of the tower that Pound et al used at Harvard.\n\nAdditionally, I just found that a formula exists for calculating g to good\napproximation at any latitude and at any elevation. It\'s called Helmert\'s\nequation where\n\ng=980.616 - 2.5928(cos2phi) + 0.0069(cos2phi)^2 - 3.086*10^-6 H ,\n\nin cgs units and phi is the latitude and H is the elevation in cm. I need\nto find out if, to good approximation, this equation takes all the above\nmess into account, if the point being considered is not too far above the\ngeoid. Then I won\'t need to send a team of ninja gravimetrists armed\nwith an FG5 into that Harvard tower under cover of dark.\n\n;-]\nGene\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>grelbr@hotmail.com (grelbr) wrote in message news:<
> > Is there any data set, perhaps at USGS?, which allows such an
> > estimation to be made, or at least interpolated?
>
> My expectation is that the leading non-spherical effect will
> likely be due to the Earth not being spherical but slightly
> "squashed" due to rotation. After that, you've got tidal bulge,
> mountains, ocean basins, and maybe density fluctuations.
>
> Doing such things at any more accuracy than you could measure g
> is likely to be quite a chore.

Right...I agree it's messy, but I've just found out belatedly that modern
gravimeters (eg: an absolute gravimeter called the 'FG5') can actually
measure differences in g over elevation differences of 3 mm (!). In
principle such a device would automatically take all that mess into
account and yield an appropriate \delta g between the top and
bottom of the tower that Pound et al used at Harvard.

Additionally, I just found that a formula exists for calculating g to good
approximation at any latitude and at any elevation. It's called Helmert's
equation where

g=980[/itex].616 - 2.5928(cos2phi) + .0069(cos2phi)^2 - 3.[itex]086*10^-6 H ,

in cgs units and \phi is the latitude and H is the elevation in cm. I need
to find out if, to good approximation, this equation takes all the above
mess into account, if the point being considered is not too far above the
geoid. Then I won't need to send a team of ninja gravimetrists armed
with an FG5 into that Harvard tower under cover of dark.

;-]
Gene

Uncle Al

Aug7-04, 05:04 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Gene Partlow wrote:\n>\n> grelbr@hotmail.com (grelbr) wrote in message news:<\n> > > Is there any data set, perhaps at USGS?, which allows such an\n> > > estimation to be made, or at least interpolated?\n> >\n> > My expectation is that the leading non-spherical effect will\n> > likely be due to the Earth not being spherical but slightly\n> > "squashed" due to rotation. After that, you\'ve got tidal bulge,\n> > mountains, ocean basins, and maybe density fluctuations.\n> >\n> > Doing such things at any more accuracy than you could measure g\n> > is likely to be quite a chore.\n>\n> Right...I agree it\'s messy, but I\'ve just found out belatedly that modern\n> gravimeters (eg: an absolute gravimeter called the \'FG5\') can actually\n> measure differences in g over elevation differences of 3 mm (!). In\n> principle such a device would automatically take all that mess into\n> account and yield an appropriate delta g between the top and\n> bottom of the tower that Pound et al used at Harvard.\n>\n> Additionally, I just found that a formula exists for calculating g to good\n> approximation at any latitude and at any elevation. It\'s called Helmert\'s\n> equation where\n>\n> g=980.616 - 2.5928(cos2phi) + 0.0069(cos2phi)^2 - 3.086*10^-6 H ,\n>\n> in cgs units and phi is the latitude and H is the elevation in cm. I need\n> to find out if, to good approximation, this equation takes all the above\n> mess into account, if the point being considered is not too far above the\n> geoid. Then I won\'t need to send a team of ninja gravimetrists armed\n> with an FG5 into that Harvard tower under cover of dark.\n\nhttp://www.mazepath.com/uncleal/eotvos.htm#b32\nFormulae after Table VII, WGS84 fits.\n\nHigher order corrections have not been included in the above. Go\nto the original text if you want them.\n\n--\nUncle Al\nhttp://www.mazepath.com/uncleal/\n(Toxic URL! Unsafe for children and most mammals)\nhttp://www.mazepath.com/uncleal/qz.pdf\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Gene Partlow wrote:
>
> grelbr@hotmail.com (grelbr) wrote in message news:<
> > > Is there any data set, perhaps at USGS?, which allows such an
> > > estimation to be made, or at least interpolated?
> >
> > My expectation is that the leading non-spherical effect will
> > likely be due to the Earth not being spherical but slightly
> > "squashed" due to rotation. After that, you've got tidal bulge,
> > mountains, ocean basins, and maybe density fluctuations.
> >
> > Doing such things at any more accuracy than you could measure g
> > is likely to be quite a chore.
>
> Right...I agree it's messy, but I've just found out belatedly that modern
> gravimeters (eg: an absolute gravimeter called the 'FG5') can actually
> measure differences in g over elevation differences of 3 mm (!). In
> principle such a device would automatically take all that mess into
> account and yield an appropriate \delta g between the top and
> bottom of the tower that Pound et al used at Harvard.
>
> Additionally, I just found that a formula exists for calculating g to good
> approximation at any latitude and at any elevation. It's called Helmert's
> equation where
>
> g=980.616 - 2.5928(cos2phi) + .0069(cos2phi)^2 - 3.086*10^-6 H ,
>
> in cgs units and \phi is the latitude and H is the elevation in cm. I need
> to find out if, to good approximation, this equation takes all the above
> mess into account, if the point being considered is not too far above the
> geoid. Then I won't need to send a team of ninja gravimetrists armed
> with an FG5 into that Harvard tower under cover of dark.

http://www.mazepath.com/uncleal/eotvos.htm#b32
Formulae after Table VII, WGS84 fits.

Higher order corrections have not been included in the above. Go
to the original text if you want them.

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf

Igor Khavkine

Aug7-04, 05:07 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>stargene@earthlink.net (Gene Partlow) wrote in message news:<c504f3da.0408052309.4f59d87a@posting.google. com>...\n\n> Additionally, I just found that a formula exists for calculating g to good\n> approximation at any latitude and at any elevation. It\'s called Helmert\'s\n> equation where\n>\n> g=980.616 - 2.5928(cos2phi) + 0.0069(cos2phi)^2 - 3.086*10^-6 H ,\n>\n> in cgs units and phi is the latitude and H is the elevation in cm. I need\n> to find out if, to good approximation, this equation takes all the above\n> mess into account, if the point being considered is not too far above the\n> geoid. Then I won\'t need to send a team of ninja gravimetrists armed\n> with an FG5 into that Harvard tower under cover of dark.\n\nThis Wikipedia article might be of help (http://en.wikipedia.org/wiki/Gee).\nIt gives a similar formula and talks about what is taken into account.\n\nHope this helps.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>stargene@earthlink.net (Gene Partlow) wrote in message news:<c504f3da.0408052309.4f59d87a@posting.google.com>...

> Additionally, I just found that a formula exists for calculating g to good
> approximation at any latitude and at any elevation. It's called Helmert's
> equation where
>
> g=980.616 - 2.5928(cos2phi) + .0069(cos2phi)^2 - 3.086*10^-6 H ,
>
> in cgs units and \phi is the latitude and H is the elevation in cm. I need
> to find out if, to good approximation, this equation takes all the above
> mess into account, if the point being considered is not too far above the
> geoid. Then I won't need to send a team of ninja gravimetrists armed
> with an FG5 into that Harvard tower under cover of dark.

This Wikipedia article might be of help (http://en.wikipedia.org/wiki/Gee).
It gives a similar formula and talks about what is taken into account.

Hope this helps.

Igor

J. J. Lodder

Aug7-04, 05:07 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Gene Partlow <stargene@earthlink.net> wrote:\n\n> Right...I agree it\'s messy, but I\'ve just found out belatedly that modern\n> gravimeters (eg: an absolute gravimeter called the \'FG5\') can actually\n> measure differences in g over elevation differences of 3 mm (!). In\n> principle such a device would automatically take all that mess into\n> account and yield an appropriate delta g between the top and\n> bottom of the tower that Pound et al used at Harvard.\n>\n> Additionally, I just found that a formula exists for calculating g to good\n> approximation at any latitude and at any elevation. It\'s called Helmert\'s\n> equation where\n>\n> g=980.616 - 2.5928(cos2phi) + 0.0069(cos2phi)^2 - 3.086*10^-6 H ,\n>\n> in cgs units and phi is the latitude and H is the elevation in cm. I need\n> to find out if, to good approximation, this equation takes all the above\n> mess into account, if the point being considered is not too far above the\n> geoid. Then I won\'t need to send a team of ninja gravimetrists armed\n> with an FG5 into that Harvard tower under cover of dark.\n\nYou might want to search under \'fifth force\'.\nWhile that idea was in fashion measurements of g(h) were made\ndown mineshafts, up TV towers, and on cliff faces.\n\nInitially claims of non-Newtonian behaviour were made,\nbut people soon realised that predicting the detailed g(h) behaviour\nis far more complicated than you would naively think.\nGravity -really- is a long range force.\n\nIn the end Newton prevailed,\n\nJan\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Gene Partlow <stargene@earthlink.net> wrote:

> Right...I agree it's messy, but I've just found out belatedly that modern
> gravimeters (eg: an absolute gravimeter called the 'FG5') can actually
> measure differences in g over elevation differences of 3 mm (!). In
> principle such a device would automatically take all that mess into
> account and yield an appropriate \delta g between the top and
> bottom of the tower that Pound et al used at Harvard.
>
> Additionally, I just found that a formula exists for calculating g to good
> approximation at any latitude and at any elevation. It's called Helmert's
> equation where
>
> g=980.616 - 2.5928(cos2phi) + .0069(cos2phi)^2 - 3.086*10^-6 H ,
>
> in cgs units and \phi is the latitude and H is the elevation in cm. I need
> to find out if, to good approximation, this equation takes all the above
> mess into account, if the point being considered is not too far above the
> geoid. Then I won't need to send a team of ninja gravimetrists armed
> with an FG5 into that Harvard tower under cover of dark.

You might want to search under 'fifth force'.
While that idea was in fashion measurements of g(h) were made
down mineshafts, up TV towers, and on cliff faces.

Initially claims of non-Newtonian behaviour were made,
but people soon realised that predicting the detailed g(h) behaviour
is far more complicated than you would naively think.
Gravity -really- is a long range force.

In the end Newton prevailed,

Jan

J. J. Lodder

Aug7-04, 05:08 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Gene Partlow <stargene@earthlink.net> wrote:\n\n> This is partly a question in geophysics, but I have to begin with a\n> brief excursion in gravitation physics...\n>\n> The gravitational redshift predicted by general relativity was veri-\n> fied by the Pound-Rebka experiment, where they measured the\n> redshift in wavelengths of photons traveling from the ground at\n> Harvard Univ. straight up to a device 22.6 meters higher in the\n> gravitational field. They measured the gravitational acceleration\n> g at both levels and used that to predict the redshift theoretically,\n> which indeed closely matched what they observed experimen-\n> tally.\n>\n> What I would like is this: without \'measuring\' g at either level,\n> it would be interesting to predict what g should be at both\n> levels, since\n>\n> g = (G M) / r^2.\n>\n> G is the Newtonian grav. constant, M the mass of the earth below\n> the surface of the geocentric sphere whose radius is r.\n\nGiven the attainable accuracy the g-variation is completely irrelevant.\nAll the experiment is sensitive to\nis the Newtonian potential difference between the two places,\nwhich is just g dh, or if you insist <g(h)> dh.\n\nJan\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Gene Partlow <stargene@earthlink.net> wrote:

> This is partly a question in geophysics, but I have to begin with a
> brief excursion in gravitation physics...
>
> The gravitational redshift predicted by general relativity was veri-
> fied by the Pound-Rebka experiment, where they measured the
> redshift in wavelengths of photons traveling from the ground at
> Harvard Univ. straight up to a device 22.6 meters higher in the
> gravitational field. They measured the gravitational acceleration
> g at both levels and used that to predict the redshift theoretically,
> which indeed closely matched what they observed experimen-
> tally.
>
> What I would like is this: without 'measuring' g at either level,
> it would be interesting to predict what g should be at both
> levels, since
>
> g = (G M) / r^2.
>
> G is the Newtonian grav. constant, M the mass of the earth below
> the surface of the geocentric sphere whose radius is r.

Given the attainable accuracy the g-variation is completely irrelevant.
All the experiment is sensitive to
is the Newtonian potential difference between the two places,
which is just g dh, or if you insist <g(h)> dh.

Jan