This post set off a discussion in which I pointed out that tidal forces are asymmetric. Carl Pham expressed skepticism at this, asking if I was saying that the tide rose higher on the side of the earth closer to the moon. I hadn’t previously thought about this before, but since I do believe that tidal forces are asymmetric, this probably followed. Or at least it followed that they were different.

One attempt was made by Ilya to prove it, but I thought it flawed and oversimplified for reasons I pointed out in comments there, because one has to consider both centripetal effects and gravitational effects when analyzing tides.

Here’s my attempt. Caution, math to follow:

Think about it this way. Let’s take a one and a half body problem (a small satellite, which, unlike the moon, allows us to ignore the complication of its mass and assume that the center of mass of the system is at the earth’s center). At orbital altitude in a circular orbit, gravity exactly offsets the centripetal acceleration, so there’s no local vertical velocity, and the orbit stays circular, by definition. The tidal force is the difference between the two accelerations, which at orbital altitude is zero (that is, they’re equal).

That is, M/Rc^2 = RcW^2, where M is the planetary mass times the universal constant, Rc is the radius of the circular orbit, and W is omega, the angular velocity. Now since the body is connected, W is a constant. If we vary Rc by distance A, above and below, we get for the case below:

M/(Rc – A)^2 = (Rc – A)W^2

or M/(Rc^2 – 2RcA + A^2) = RcW^2 – AW^2

or RcW^2 = M/(Rc^2 – 2RcA + A^2) + AW^2

Since RcW^2 is the centripetal acceleration at the orbital altitude, we can substitute the gravitational term at that altitude for it, so

M/Rc^2 = M/(Rc^2 – 2RcA + A^2) + AW^2

and the tidal force is the difference between them. Designating this Tb (for tidal force below) yields:

Tb = M/Rc^2 – M/(Rc^2 – 2RcA + A^2) – AW^2

Similarly, for the tidal force at distance A above:

Ta = M/Rc^2 – M/(Rc^2 + 2RcA + A^2) + AW^2

OK, it’s not obvious on inspection (at least to me) which number is larger and which is smaller, but I hope that it’s obvious that in general, for any value of A other than zero, they’re two different numbers in magnitude (there may be some non-zero value of A for which they’re equal, but as I said, in general they are not). Therefore, the gravity-gradient acceleration field is asymmetric.

[Update a couple minutes later]

In proofreading it, I realize that there’s no purpose in substituting the gravity term in. But it doesn’t hurt anything, and trying to fix it at this point would probably introduce an error.