- #36
Hak
- 709
- 56
Thank you, @haruspex! I'll see if I can achieve the same, although it seems difficult. I will post my results even if they are wrong. However, I didn't understand one thing: from these expressions you got, how should you arrive at the answers to 1. and 2.? I didn't understand...haruspex said:I set ##R_1, R_2## to be the distances to the common mass centre. I found the sum of the centrifugal and M2-based gravitational potentials for two points, one at ##y## from the centre of M1, towards M2, and one at 'moonrise', i.e such that the point and M2 subtend an angle of 90° at the centre of M1.
I then took the difference between these two sums and discarded terms order ##y^3## and higher. (Note that a trap with this sort of calculation is taking approximations too soon and ending up with 0.)
I ended up with ##\frac{GM_2y^2(R_2+3R_1)}{2(R_1+R_2)^3R_1}##. See if you get the same.
If this results in a tide height difference of h then that potential difference should equal ##gh##.
I haven't validated it numerically by plugging in the numbers for Earth and Moon.