Difference in gratitational field due to lunar apogee/perigee

Question

I am trying to answer this Phys.SE question but got an unreasonable answer:

In making laboratory measurements of $g$, how precise does one have to be to detect diurnal variations due to the moon's gravity? [Clarifications omitted.]

I figured I can calculate the change in GPE for a test mass

$$GPE_{earth} - GPE_{moon}$$ at apogee and perigee, and take a ratio to see how many significant figures are affected.

Where $d$ is lunar distance,

$$\frac{m(g_{earth}R_{earth}-g_{moon}d_{min})}{m(g_{earth}R_{earth}-g_{moon}d_{max})} = \frac{m_{earth}R_{earth}-m_{moon}d_{min}}{m_{earth}R_{earth}-m_{moon}d_{max}}$$

I asked Wolfram Alpha to calculate this and got a ratio of about 1.4, which seems unreasonable.

What has gone wrong?

Answer 1

The main effect the moon has on the measured strength of gravity on earth has not to do with changes in its distance from earth, but with its direction. It is a tidal effect, having to do not with the strength of the moon's gravitational field, but with its gradient: basically the difference in the moon's gravitational field between the center of the earth and the spot where you're measuring. That's why the moon has a stronger tidal effect on the earth than the sun even though the strength of the sun's gravitational field is much higher.

If the moon is directly overhead, it pulls on you a bit harder than it would if you were at the center of the earth, so you feel a touch lighter. Same thing if the moon is directly underfoot: its pull at the center of the earth is stronger than its pull on you, so again, you feel a bit lighter than you would without the moon. But if the moon is at the horizon, pulling sideways on you, it actually makes you feel a bit heavier than you would otherwise.

The math is gone through in Ben Crowell's answer at the question you linked to.