Earth & Moon Gravity Gradient: Investigating Its Impact

[Charles Link]

If you look at the red dots and continue them through the whole circle, you will see they are doing circular motion about the center of the earth. Yes, the moon has a rotation whose period is the same as its period of revolution, but that rotation simplifies this problem, by making the motion of the object simply as a circle around the earth.

The object is lighter than the gravity by the moon, because the period of its orbit is determined by the time $T$ that the center of the moon orbits. $F_{gravity \, earth}=GMm/(r-\Delta r)^2$ , while centripetal force needed is $F_c=[(2 \pi (r- \Delta r)/T]^2/(r- \Delta r)$, with center of moon in orbit obeys $GMm/r^2=(2 \pi r/T)^2/r$. The difference between the Earth's gravity and the centripetal force needed is how much lighter the scale reads from the gravity of the moon pulling on the object.

 

[Charles Link]

If you solve for $F_c$ eliminating $T$, you get $F_c=GMm(r-\Delta r)/r^3$. This gives
$F_g-F_c=GMm \frac{r^3-(r-\Delta r)^3}{r^3 (r-\Delta r)^2}$.
With a little algebra you get $F_g-F_c \approx 3GMm \Delta r/r^3=3 (GMm/r_e^2) r_e^2 \Delta r /r^3$.
In more detail, the above expression gives $F_g-F_c=(GMm/r^2) (3(\frac{\Delta r}{r})+3 (\frac{\Delta r}{r})^2+...)$
$GMm/r_e^2$ is the weight of the object on earth.

You can do a similar thing when the object is on the far side of the moon. Then the reduction in weight is $F_c-F_g$. To first order this gives an identical result to the above. The second order result for the difference $|F_{c1}-F_{g1}|-|F_{c2}-F_{g2}| =6GMm (\Delta r)^2 /r^4$. ( Note this difference is zero in first order). This result was first obtained by @PeroK post 22. (The second order term for the far side expression has the opposite sign, thereby the factor of 6).

 

[Richard R Richard]

[QUOTE = "PeroK, post: 6435672, member: 493650"]

$$ N_f = F_M + F_ {cMc} $$
And the effective gravity on the Moon depends on which side you are standing on?
[/QUOTE]
You're right, maybe I mix ideas, actually on the opposite side in the same frame of reference I must have written something like

$$-N_f = -F_M + F_ {cMc}$$
That is, when we include the Earth
$$N_c = F_M-F_ {Ec} F_ {cEc} -F_ {cMc}$$$$-N_f = -F_M - F_ {Ef} + F_ {cEf} + F_ {cMc}$$

Which are the same equations, but the sign only indicates the vector sense in that frame of reference.
Actually, the difference between the measured values between one side and the other is due to the difference between

$$ F_ {Ec} - F_ {cEc} \neq F_ {Ef} - F_ {cEf} $$
Edit I still haven't read the additional messages I've been replying to

 

[Charles Link]

It may be worth mentioning for the arithmetic of the results of post 32, etc., it is kind of simple, if you use for the radius of the Earth $r_e=4000$ miles, the Earth to moon distance $r=240,000$ miles, and the radius of the moon $\Delta r=1000$ miles. Then $r_e/r=1/60$, and $\Delta r/r=1/240$.