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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:

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

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Please see this earlier answer in which I reproduce a graph showing the diurnal variation in $g$ from the excellent book by Gordon Squires, "Practical Physics". Note that this shows both the effect of the sun and the moon - but it should help convince you of the magnitudes. –  Floris Jul 29 '14 at 22:09
You might also like the answer by John Rennie to the same question which computes the value of $2.6 \cdot 10^{-8}$ for the fractional change in gravity due to the sun. The moon has a much larger effect - which you can easily compute by substituting appropriate values in his equation. –  Floris Jul 29 '14 at 22:18
@Floris - John Rennie's calculation is incorrect. It's off by a factor of three. The correct value is $7.7\times10^{-8}$. The tides caused by the Sun are a bit less than half those caused by the Moon. –  David Hammen Jul 30 '14 at 0:13
related: physics.stackexchange.com/q/128816 –  Ben Crowell Jul 30 '14 at 17:17
Honestly, this comes across like one of those check-my-work homework questions we've been debating lately. My reading of it is that what you've written sets the stage for a potentially very interesting question, you just stopped writing before actually asking the question. Thoughts? (from anyone) –  David Z Jul 31 '14 at 19:37

1 Answer 1

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.

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I think "diurnal" is fine. It does not imply a 24 hour period but "happening every day", so something that happen roughly twice a day is fine. –  dmckee Jul 29 '14 at 21:24
@dmckee: You may be right. I was just trying to figure out how they got their answer, and it seemed like the erroneous calculation needed to get their numerical result would also result in a prediction of a 24-hour period, which would probably be the more common interpretation of "diurnal." –  Ben Crowell Jul 29 '14 at 21:30
+1. I agree that your $1.6\times10^{-7}$ is the correct result. The factor of 1.5 reflects the fact that g is greatest when the Moon is rising (or setting), but the difference between nominal in these circumstances is only half the reduction in g when the Moon is at nadir or zenith. –  David Hammen Jul 29 '14 at 22:10
The correct term is "semidiurnal", not "diurnal". –  David Hammen Jul 30 '14 at 0:24

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