The Roche Limit: Tidal Forces vs. Gravitational Binding Force

By Joshua Winn, Princeton University

All the planets have a Roche limit, including the Earth. This was first worked out by Éduoard Roche, and it is the minimum distance within which tidal forces overcome the gravitational binding force of a moon, modeled as an idealized fluid body. The force of gravity depends on distance; it gets weaker as you get farther away from the attracting mass.

Destroyer of the Moons

The side of the Earth facing the Sun—the dayside—is pulled harder than the nightside. These differences in gravitational forces from one part of a body to another are called tidal forces. 

Suppose we have a planet of mass M. And let’s say it has a moon with a small mass, m, whirling around in a circular orbit of radius r. Actually, let’s start with an even simpler case. Let’s just drop the moon, starting from rest. What will happen? The moon will accelerate downward, and crash into the planet. It doesn’t orbit because we didn’t give it any angular momentum.

Now let’s give the moon a nonzero size. We could make it a sphere, but the math would get too hairy. In the spirit of keeping things as simple as possible, let’s just take one step beyond the point-mass approximation. We’ll model the moon not as a point mass, but as two rigid spherical rocks, each of mass m, with their centers separated by some small distance Delta-r. 

When we let go of those rocks, they both fall onto the planet, but will they stay together as they fall? No, the inner rock is closer to the planet, so it feels a stronger gravitational force than the outer one, leading to a larger acceleration, and causing the inner rock to pull away from the outer rock. So, the moon breaks apart as it falls.

Keeping It Intact

In general, the magnitude of the force from the planet is G × M × m over r². Let’s calculate the difference in force, Delta-F, experienced by the rocks. Since Delta-r is very small compared to r, Delta-F approximately equal to dF dr × Delta-r. Which gives Delta-F equals -2GMm over r³ times Delta-r. The minus sign means that the force weakens with distance.

We’ve learned that the part of the moon closest to the planet is pulled harder, by an amount proportional to Delta-r, the size of the moon, and M, the mass of the planet, and its inversely proportional to the cube of the moon’s orbital distance. Those are the hallmarks of tidal forces: they grow with the size of the body, the mass of the attractor, and fall off as the third power of distance.

Now, if we want to keep the moon intact, we’ll need to supply a force to counteract Delta-F. We could tie them together with some rope or, wait a minute, what about the gravitational force between the rocks themselves? They are attracted to each other, with a “self” gravitational force of Gm² over Delta-r². 

So, if we want them to stay together as they fall, we’d better hope the magnitude of Delta-F is smaller than F-self. That leads to an inequality in which the G’s cancel out, along with one of the m’s, and if we rearrange it to put all the moon properties on the same side, we get 2 × M over r³ is less than m over Delta-r³. So, if m is big enough and Delta-r is small enough, that is, if our rocks are massive and closely packed, we can satisfy this inequality, and they’ll hold together as they fall.

This article comes directly from content in the video series Introduction to Astrophysics. Watch it now, on Wondrium.

The Roche Limit and Orbiting Moons 

What if the moon is orbiting, not just falling in? Is there still a Roche limit? Yes. It’s the same, and in that case r-min refers to the minimum orbital distance. Suppose a moon made from a bundle of rocks is going in a circular orbit. 

The rocks that are closer to the star have slightly smaller orbital distances, so, by Kepler’s third law, they have shorter orbital periods. The outer rocks have longer periods. So, unless the moon’s self-gravity is strong enough, the rocks will drift apart over time, the inner ones moving ahead of the outer ones. The moon gets shorn into pieces and strung out into an arc around the star. Eventually, the arc reaches all the way around the planet, making a ring.

See where we’re going? This may be where planetary rings come from. Maybe a moon ventured inside the Roche limit, and the tidal forces from the planet sheared it out into a ring. Or maybe two moons collided, smashing into pieces, which ordinarily, given millions of years, would gravitationally reassemble into a moon, but inside the Roche limit, that’s impossible. The tidal forces prevent the moon from reforming.

And the number, 2.44, is a pretty good match to the observed sizes of the rings of the giant planets, which we saw range up to 2 or 2.5 times the radius of the planet. The density ratio is always of order unity, because the densities of the moons and planets are of the same order of magnitude, a few grams per cubic centimeter. So, the numbers fit the story.

The Moon’s mean density is around 3 grams per cubic centimeter, typical of rocks. The Earth’s is higher, it’s about 5.5, because the Earth’s stronger gravity compresses its interior and because the Earth has more iron in its core. Given those numbers, the Roche limit comes out to about 3 Earth radii. Whereas the Moon is orbiting at a distance of 60 Earth radii, so it’s not in any danger of tidal destruction.

A good example of what does happen when a body violates the Roche limit came in the summer of 1994, when a comet named Shoemaker-Levy-9 crashed into Jupiter. By the time it hit Jupiter, tidal forces had broken it into fragments, each of which punctured Jupiter’s clouds in a different place, making a series of brown spots.

Common Questions about the Roche Limit

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Kepler’s Second and Third Laws
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