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.

How do you go from this:

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

to this?

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

To leave RcW^2 the only term on one side you must add AW^2 to both sides; how does this change M/(Rc^2 - 2RcA + A^2) to M(Rc^2 - 2RcA + A^2) ? Looks like you lost a division sign along the way.

That's more math than I would want to wade into to analyze this question. But please tell me if I'm missing something, isn't the issue simply that gravity as a function of distance is concave downward? This much is intuitively obvious, and it would seem to follow at once that it diminishes more from Earth's moonside to Earth's center than it does from Earth's center to the opposite side.

Err, I meant upward. I was thinking of the "gravity well," which of course is an upside down picture.

Yes, I obviously lost that division sign.

Assuming your last 2 equations are correctly derived, one is larger than the other according as

4M*Rc is greater than or less than w^2, with equality when 4M*Rc=w^2

and A is then an amplifying factor on the difference, assuming I've done the math right ;-)

Can you write the equtions out again carefully with all parentheses accounted for?

That would certainly make intuitive sense, TnT. But bear in mind that W is a function of Rc. Specifically, W = Sqrt(M/Rc^3)

Perhaps this link on tidal forces and common misconceptions in textbooks about tides will help.

http://www.lhup.edu/~dsimanek/scenario/tides.htm

My only thought was, that with a satelite, even with minimal (which the moons effect is not) interaction would still shift the center of gravity of any body, around which it is revolving.

So if the center of gravity is always shifting, then obviously tidal forces are not uniformly distributed, and tides in terms of water cannot be.

Don't have the math, thats just obvious logic.

Don't those equations assume point masses for the sake of simplification? Since the Earth is not a point mass the gravitational attraction of the Moon for a point on the side of the earth facing the Moon is going to be greater than the gravitational attraction of a point on the surface of the Earth on the far side by the ratio of the gravitational attraction over the 8000 mile further distance. This difference is small, on the order of a few microgee but it is there.

Another FYI, the gravitational attraction of the Moon on the nearside of the Earth is enough to lift the north american continent by about 2 inches when the Moon is overhead.

The accelerometers that we built for microgravity measurement on the Shuttle could easily see the delta in the gravity potential.

The way I think of it, is that tidal force is just the difference in gravitational attraction between nearby points (eg, between two ends of a rod). Since gravity is proportional to R^-2 (R to the minus -2 power), this implies that the difference between the force of gravity acting on two points is approximately dR*R^-3 where dR is the difference in the distance between the two objects and the massive body like the Moon.

Note that since the tidal force is approximately proportional to R^-3, and since the nearer side of the Earth is about 3% closer than the far side, you get roughly that the tidal force will be around 10% stronger (inverse cube approximately decreases as triple the relative change when the relative change is small) on the near side than on the far side. Hence, tidal force is asymmetric from the point of view of Earth.