Variation in measurements of $g$ on the Earth's surface due to the Moon's gravity A book, Exercises for the Feynman Lectures on Physics, has recently been published. It consists of homework problems to provide practice with the techniques and concepts used in the famous Feynman lectures, which had not had published problem sets to accompany them since they came out in 1964. I bought a copy, found what I considered to be an error in the answer given for one of the problems, and, as a matter of professional courtesy, emailed one of the book's editors, M. Gottlieb, to report the erratum. Dr. Gottlieb insists that I'm wrong. I'll quote the question here and give my own answer below.

3.12 In making laboratory measurements of $g$, how precise does one have to be to detect diurnal variations $\Delta g$ due to the Moon's gravitation? For simplicity, assume that your laboratory is so located that the Moon passes through the zenith and nadir. Also, neglect earth-tide effects.

The answer given in the back of the book is:

$\Delta g/g=7\times10^{-6}$

 A: The question refers to "laboratory" measurements, i.e., local ones. Such a measurement can only be sensitive to the gravitational acceleration of a test mass relative to the laboratory. For example, if one drops a mass in a vacuum column and measures the time it takes to hit the floor, the acceleration inferred is the acceleration of the mass relative to the floor. The same applies to any other local measurement, e.g., a measurement using a pendulum.
The question asks the student to calculate the "diurnal" variation in $g$, and this word means a variation with a period of 24 hours. As I'll show below, one expects nearly zero variation due to the moon with a period of 24 hours, under the simplifying assumptions of the problem as stated. The actual leading-order effect, under these assumptions, has a period of 12 hours, and is about two orders of magnitude smaller than the one claimed in the book.
Let $M$ be the mass of the earth, $m$ the mass of the moon, $R$ the center-to-center distance between earth and moon, and $r$ the radius of the earth. Let the subscript N refer to the condition in which the moon is at the nadir, and Z the one in which the moon is at the zenith. Let the moon and earth be on the $x$ axis, with the moon lying to the right of the earth.
The acceleration of the earth due to the moon's gravitational attraction is $a_E=Gm/R^2$. The acceleration of a test mass in the laboratory in the two conditions is $a=\pm GM/r^2+ Gm/(R\pm r)^2$, where $+$ is for N, $-$ for Z. Subtraction gives the acceleration of the test mass relative to the laboratory,
$$ a_R=a-a_E=\frac{\pm GM}{r^2}+\frac{Gm}{(R\pm r)^2} - \frac{Gm}{R^2}$$
which becomes, with the approximation $1/(1+\epsilon)^2-1\approx -2\epsilon$,
$$ a_R \approx \pm \left[g_0 -\frac{ 2Gmr}{R^3}\right] $$
In condition N this quantity is positive, while in condition Z it is negative. Since the apparatus rotates 180 degrees in 12 hours due to the rotation of the earth, what we actually measure is $|a_R|$, which is the same in both cases. (To see any variation over 12 hours, we'd have to go to the next order.)
At the times 6 hours before and after N and Z, when the moon is on the horizon, the moon's gravity acts at an angle $\theta\approx r/R$ below the horizon, increasing the $y$ acceleration of the test mass by $(Gm/R^2)\sin\theta\approx Gmr/R^3$. The earth's own acceleration has no component along the $y$ axis, so this increase is also observed by laboratory measurements.
In summary, there is a twice-diurnal (not diurnal) variation with a peak-to-peak amplitude, relative to $g$, of
$$ \frac{3Gmr/R^2}{g} = 1.7\times 10^{-7}.$$
This seems to be in agreement with the figure calculated at the bottom of this answer by David Hammen. (David starts by calculating the solar effect, which turns out to be much smaller, but then gives a figure at the end for the combined lunar and solar effect. Dividing his figure by 9.8 m/s2 seems to give the same thing I got.)
The incorrect answer given by the book appears to have been obtained by ignoring the fact that the earth accelerates in response to the moon's gravity. Under that assumption, one obtains a peak-to-peak variation
$$ \frac{\Delta g}{g} = 2\frac{m}{M}\left(\frac{r}{R}\right)^2 = 7\times10^{-6} $$
with the 24-hour period claimed by the book.
After I wrote up the calculations above, Floris pointed out this previous answer, which gives a graph showing experimental data. There appear to be two Fourier components of about equal amplitude, one with a period of 12 hours and one with a period of 24 hours. The amplitude of the 12-hour component appears to match my prediction. However, I'm having trouble understanding the 24-hour component, which has a peak-to-peak amplitude of about $10^{-7}g$. This is two orders of magnitude smaller than the book's result, but still larger than I can account for. If I continue the Taylor series that I truncated in my original answer, I get $1/(1+\epsilon)^2-1\approx -2\epsilon+3\epsilon^2$. The additional term gives a 24-hour oscillation with a peak-to-peak amplitude of $6Gmr^2/R^4$, which comes out to be about $6\times10^{-9}g$, which more than an order of magnitude too small to explain the observations. I wonder what the origin of this effect is? Unfortunately I don't have access to the Zumberge paper (is it this?). I've asked a separate question about this effect.
A: The answer depends on how you interpret the question. Choosing to define g as a difference in acceleration relative to the earth gives one answer, the answer is different if g is defined simply as 'the acceleration due to gravity' and no measurement method is assumed. The question does not say how the measurement is made. Setting aside GR, if it is made relative to an un-accelerated object well away from any masses (eg free body in interstellar space), then the answer is as in the book, to a reasonable degree of accuracy, as it is simply the change in acceleration due to the change in total force on a test mass divided by the mass of the object. The acceleration of other objects is irrelevant. This is a measurement-independent answer to the question, ie 7x10-6g. This can be measured in a lab, in a 'simple' way, if the motion of the earth around the earth-moon centre of gravity is accounted for in the analysis.
