I have a question about tidal forces on the far side of a body experiencing gravitational attraction from another body.

Let's assume we have two spherical bodies $A$ and $B$ whose centers are $D$ apart, and who have radiuses $R_{A}$ and $R_{B}$, much smaller than $D$.

Gravitation force has a law in $1/distance^{2}$. On the line $AB$, a mass $m$ at the point point of $B$ closest to $A$ experiences a pull towards the center of $A$ of magnitude $Km/(R_{B}-D)^{2}$, and the point farthest from $A$ experiences a pull of magnitude $Km/(R_{B}+D)^{2}$.

Note that I did not mention the direction of the force at that second point. It seems that the centers of mass being all on one side of that point, the force should point towards the center of $A$ on the line $AB$. However, the tidal bulge on this "far side" suggests that some force (?) is pulling on matter away from the center $A$.

How can we explain this tidal bulge on the far side? - I am specifically interested in a clear derivation using classical mechanics.


It has to do with the fact that the entire earth is in an accelerating reference from due to the attraction from the moon.

Imagine 3 points on a line out from the moon. $N$ is the oceans near the moon, $C$ is the center of the earth, and $F$ is the far oceans. The attraction falls off with distance so the forces $F_N > F_C > F_F$ as you know. Also, all the forces are directed towards the body $A$ as you know.

Since the oceans are not rigidly attached to the earth, these points all can move independently to some degree. Due to the high force, $N$ will get pulled towards $A$ by a large amount. $C$ will get pulled a little. While $F$ will barely get pulled at all. What's important is that due to the different amounts they move, the distances between the three points have increased. $N$ and $C$ are farther apart than they were before, and $C$ and $F$ are farther apart too.

Now imagine your standing on the earth at $C$. The oceans at $N$ are farther away from you now, so the ocean appears to bulge out at that point. This seems to make sense since that side is closer to the moon. But also, the oceans at $F$ are farther away from you too, so they also appear to be bulging.

In essence, it is not that those far oceans are being pulled away from the earth, it's that the earth is getting pulled away from those oceans.

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    $\begingroup$ Todd - thanks for this explanation. Let me ask some further questions. Suppose that somehow A and B are perfectly still. There is no movement (I know it's not possible, but let's make this imaginary experiment). Suppose further that we can "turn on" gravitation at some point. Before we "turn on" gravitation, the 2 bodies occupy perfect spheres, and they are elastic. Now we turn on gravitation. We should see 4 bulges appear along AB, on the near and far sides of A and B. Are the bulges on the far sides inside the initial spheres that A and B occupied, or outside? I think they are outside. $\endgroup$ – Frank Dec 14 '12 at 2:08
  • $\begingroup$ So if they were perfectly still and we turned on gravity they would just go right towards each other. I understand what you're saying though. Let's have them orbiting each other but at first their attractive force is from a "rigid rod" connecting them so that there is no distance dependence on the force of attraction. Now remove the rod and turn on gravity in such a way that the orbits remain unchanged, but tidal forces can now be observed. So that's the set up. $\endgroup$ – Todd R Dec 14 '12 at 2:34
  • $\begingroup$ My first thought is that the bulges would be inside the initial spheres of occupation. But we also have to ask the question, are these bodies rotating about their own axis? If that's the case then the near sides and far sides are constantly switching. It's like a rotisserie for tidal forces. In this way, perhaps the stronger pull on the front could cause a bulge so large that when it turns to the back it sticks out past the initial sphere. I really don't know what would happen, but another great question. $\endgroup$ – Todd R Dec 14 '12 at 2:39
  • $\begingroup$ I think gravity is all we need and we don't need to add rotations. The bulge on the near side has to be outside the initial sphere of occupation, IMHO. From everything I could see so far on the internet, it looks like the situation is truly symmetric and the bulge on the far side is also out. It's not intuitive at all, but I think it has something to do with the Earth's reference frame not being inertial. $\endgroup$ – Frank Dec 14 '12 at 3:20
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    $\begingroup$ The two planets will start to fall towards each other. The ocean close to the other planet will have greater acceleration because 1/r^2 and the ocean away from the other planet will have slower acceleration. So the ocean on the far side will be left behind as the planet falls in towards the other planet. That will look like a bulge to an observer on the planet. $\endgroup$ – jcohen79 Dec 16 '12 at 21:29

The easiest way to understand this is to represent the moon's gravitational field in the vicinity of the earth as a "multipole expansion". The field lines of the moon converge radially towards the moon. You can represent this as the superposition of a constant field (parallel field lines) and a distortional component called the "quadropole field". The constant field has no effect on the tides, which are completely caused by the quadrupole field. I explain it in this blogpost: enter image description here

  • $\begingroup$ Can you post equations here? Thanks! $\endgroup$ – Frank Dec 16 '12 at 22:07
  • $\begingroup$ electrostatic force is also in 1/r^2. Suppose I have two immobile spheres somewhat elastic that I can charge with opposite charges so they attract like gravity. Initially, they do not have a charge. I flip a switch, and they get charged. If the charges are strong enough, they could deform the elastic material. I would see 4 bulges, 2 on each sphere, somewhat asymmetric as per Nick, on the line of the centers, without anything moving at all. Is that correct? $\endgroup$ – Frank Dec 16 '12 at 22:16
  • $\begingroup$ Yes, the electrostatic case is the same as gravity. The immobility condition however is problematic. An important aspect of the tidal system is that the earth and moon are in free-fall about each other. That's why the average field has no visible effect, and only the distortional component is at work. Also, I question Nick's asymmetry. In my analysis, the quadrupole force is symmetric. $\endgroup$ – Marty Green Dec 16 '12 at 23:55
  • $\begingroup$ Marty - how does the free-fall show up in the equations? I was somehow under the impression that for the case of gravity, being accelerated (free-fall) or simply being in the gravity field were the same thing (I am probably confusing several things here). $\endgroup$ – Frank Dec 17 '12 at 0:18

The point on the surface of 'b' that is farthest from 'a' is being pulled towards both a & b, but 'b' is being pulled towards 'a' even harder. The net result is a tidal force pulling the point on the surface away from the center of the body.

Tidal distortion
The attracting body is on the right (not shown). The arrows represent the acceleration vectors at each point on a spherical surface. These tidal accelerations cause two bulges with a ring of low tide separating them. The tidal bulges are not symmetric. It is assumed that the body and the ocean are of the same density. Due to the difficulty of calculating the effects of self gravitation this diagram is only an approximation, it is also exaggerated for effect.

The case of a spherical rigid body covered with an ocean of negligible density is more amenable to calculation. The surface of the ocean is simply the equipotential surface of two point masses. In the limit of a zero density ocean, these diagrams are exact.

When the two masses are being held apart at fixed difference:
Equipotential surface when held apart
This dashed circle is the undisturbed spherical ocean level. The solid line is the ocean level (equipotential surface) in the proximity of the attracting mass.

When the two bodies are in free fall:
Equipotential surface in free fall

To reproduce this graph:

Start with the equation: $ U = \frac{-G n}{ \sqrt{x^2+y^2} } - \frac{G m}{ \sqrt{ x^2 -2 d x+y^2+d^2 }} +\frac{G m}{d^2} $
where U is potential at the point [x,y], m is the mass of the perturbing body a, n is the mass of b, and d is the distance between them.

Guess an initial value for the potential of the distorted surface $U$. Use a solver to calculate the surface height $y$ in terms of $x$ $(y=f(x,U))$. Then integrate this equation to find the volume enclosed by the equipotential surface ($ \pi \int_{x_\mathrm{min}}^{x_\mathrm{max}} f(x,U)^2 dx $. Adjust value of the potential and iterate until the distorted volume is equal to the original volume. Finally, convert the function $y=f(x,U)$ to polar coordinates ($ r = g(\theta,U) )$ and graph.

  • $\begingroup$ Nick - Can you explain how you made this graphic? What software did you use? What is the model? What are the equations? - I want to try and reproduce that myself. Thanks! $\endgroup$ – Frank Dec 16 '12 at 22:05
  • $\begingroup$ @Frank I used a Texas Instruments TI-Nspire CAS, which I highly recommend. Mathematica is more suited for this type of thing, if you have the means. $\endgroup$ – Nick Dec 20 '12 at 21:58
  • $\begingroup$ I've read somewhere that the asymmetry between the 2 bulges is something like 5% in the case of Earth/Moon. Also, your coordinate system seems to indicate that you are working in "Earth's frame of reference". In that frame of reference, don't we observe 2 bulges, without having to talk about the "center of figure"? $\endgroup$ – Frank Dec 21 '12 at 2:51
  • $\begingroup$ Maybe another way of seeing this: suppose Earth deforms much less than the oceans, so that the initial dark disk stays where it is. Shouldn't we see 2 bulges on both sides of that initial dark disk? $\endgroup$ – Frank Dec 21 '12 at 3:51
  • $\begingroup$ Duh, you are absolutely right. The 2nd diagram represents two masses held apart at a fixed distance. I have added the relevant diagram for two masses in free fall. Regarding "bulges", I just wanted to point out that number of lobes on a cam depends the location of the spin axis, and not just its shape. $\endgroup$ – Nick Dec 22 '12 at 5:36

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