# Can you tell just from its gravity whether the Moon is above or below you?

If you are on a place of Earth where the Moon is currently directly above or directly below you, you experience a slightly reduced gravitational acceleration because of Moon's gravity. This is what causes tides. My question here asks how difficult it is to decide with a measurement whether the Moon is above or below you.

You must do the measurement using only the gravitational interaction of the Moon, not by optical or radio observation which is much easier. Suppose if you wish that an evil wizard has turned the Moon to completely transparent dark matter that does not interact electromagnetically in any way, without changing its mass and orbit significantly.

Let's do some calculations. The Moon changes the gravitational acceleration by approximately $2 G m r d^{-3} \approx 9\cdot 10^{-7} \mathrm{m}/\mathrm{s}^{2}$, where $G$ is the gravity constant, $m$ is the mass of the Moon, $r$ is the radius of Earth, and $d$ is the radius of orbit of the Moon. (This is a greatly simplified computation that assumes that Earth is a rigid perfect sphere.) The difference of this extra accleration between the side of Earth closest and farthest from the Moon should be approximately $12 G m r^2 d^{-4} \approx 9\cdot 10^{-8} \mathrm{m}/\mathrm{s}^2$.

Compared to this, Earth is said to have local gravitational variation on the order of magnitude of $10^{-3} \mathrm{m}/\mathrm{s}^2$ because of uneven surface, and that's not even counting the even greater effects of varying latitude and altitude. As you know, because of Earth's rotation and Earth's elliptical shape where the equatorial radius is greater than the polar radius, the acceleration depends very much on the latitutde.

Now I could imagine a measurement that's done carefully on constant latitude (say on the Equator only) and on constant altitude somehow, but it seems difficult to account for the local variations, so you'd think that the local variations always dwarf the effect of the Moon. However, the Moon still causes easily observable tides of the oceans, despite that the above calculation shows that the tidal effects are three order of magnitudes weaker than the local variations. This shows that the local variations are likely constant in time, so if you measure the gravitational acceleration in a fixed location, these local variations likely won't affect you much. Even without the ocean, I imagine the magnitude of the gravitational acceleration should be possible to measure very accurately. From the above calculation it appears that the Moon's tidal effect is as much as 9 percent stronger on the side closest to Moon than on the side farthest. This suggests that such a measurement should be possible, but then there might be other distracting effects I did't think of.

In any case, it's clear that some kind of measurement must be possible: in the worst case you can send an unmanned rocket probe to the presumed location of the Moon where its gravitational effect should be obvious. Such a space probe is very expensive though. This is why I'm asking not whether such a measurement is possible, but how easy it is.

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Oddly enough, the tidal effect is symmetric. Tides are high near and far from the moon and low at places where the moon lies on the horizon. This is because the moon pulls on the bulk of the Earth too. When the moon is above you it pulls you away from the Earth, but when the moon is under your feet it pulls the Earth away from you. The residual is the same either way. – dmckee May 22 '14 at 16:06

In the summer of 2010, I had the opportunity to attend a presentation by Reiner Rummel, involved in the GOCE satellite containing a very precise gradiometer. The presentation can be found on the ESA website. It contains a table with orders of magnitude for the accelerations they could measure when still in the lab.

The gravitational acceleration in the lab in Munich they measured is $g=9.80724672 m s^{-2}$. Components (quoted literally from the aforelinked presentation), all units in $m s^{-2}$:

• $10^{0}$ spherical Earth
• $10^{-3}$ flattening & centrifugal acceleration
• $10^{-4}$ mountains, valleys, ocean ridges, subduction
• $10^{-5}$ density variations in crust and mantle
• $10^{-6}$ salt domes, sediment basins, ores
• $10^{-7}$ tides, atmospheric pressure
• $10^{-8}$ temporal variations: oceans, hydrology
• $10^{-9}$ ocean topography, polar motion
• $10^{-10}$ general relativity

From the number of significant digits in their measurement, it can be seen that they are able to measure down to a $10^{-8} m s^{-2}$ precision. So it might just be possible. However, I remember an anecdote that in the lab, they could measure the metro passing by several hundred meters away. So... good luck.

Note: there are surely more authoritative sources than the one I linked, but as it's not my field of expertise, I don't know them.

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The presentation seems to have disappeared from the website, but a copy is available from the Wayback Machine at web.archive.org/web/20130430131234/http://earth.eo.esa.int/… – b_jonas Aug 31 '15 at 11:58