# When could the Roche limit be larger than the Hill sphere?

This answer to Are Saturn's rings stable? begins with:

Most of Saturn's rings are inside it's Roche limit, which means they will never clump together. Tidal forces prevent this from happening.

I always confuse Roche limit and Hill sphere, so I thought I'd ask a question about both.

Question: Under what conditions could a solar system body orbiting the Sun have a Roche limit larger than its Hill sphere?

• Another word for Hill sphere is Roche sphere, so . . . some confusion is understandable. en.wikipedia.org/wiki/Hill_sphere Commented Jan 28, 2020 at 8:49
• @userLTK that may be the original source of my trouble. It's possible that I learned it that way long ago.
– uhoh
Commented Jan 28, 2020 at 9:19
• @userLTK yes now I remember, "Roche lobe" (mentioned in this answer) is exactly the term I learned to use for what we also call the Hill sphere.
– uhoh
Commented Jan 29, 2020 at 0:22
• As a sidebar, Phobos is pretty close to qualifying as a real example of this. A calculation of Phobos' hill sphere puts it about 16 km from Phobos center, less than 1 radius away from it's surface. Of-course, an orbit in those conditions becomes close to impossible, so the Roche limit isn't all that relevant. Commented Jan 29, 2020 at 3:01

The Hill sphere is approximately given by

$$R_\mathrm{Hill} \approx a_\mathrm{p} \left(\frac{M_\mathrm{p}}{3M_\ast}\right)^\frac{1}{3}$$

Where $$a_\mathrm{p}$$ is the radius of the planet's orbit, and $$M_\mathrm{p}$$ and $$M_\ast$$ are the masses of the planet and the star respectively. This is an approximation to the size of the Roche lobe around the secondary.

The Roche limit (not to be confused with the Roche lobe) is the limit at which tidal forces will disrupt an object held together by its own gravity. Roche derived the following formula:

$$R_\mathrm{Roche} \approx 2.44r_\mathrm{p} \left(\frac{\rho_\mathrm{p}}{\rho_\mathrm{s}}\right)^\frac{1}{3}$$

Where $$r_\mathrm{p}$$ is the radius of the planet, and $$\rho_\mathrm{p}$$ and $$\rho_\mathrm{s}$$ are the densities of the planet and the satellite respectively. Note that there are various different formulae for the Roche limit depending on the different assumptions being made, see the Wikipedia Roche limit article for details. For a spherical planet, this can be rewritten in terms of the planet's mass:

$$R_\mathrm{Roche} \approx 2.44 \left(\frac{3M_\mathrm{p}}{4 \pi \rho_\mathrm{s}}\right)^\frac{1}{3} \approx 1.51 \left(\frac{M_\mathrm{p}}{\rho_\mathrm{s}}\right)^\frac{1}{3}$$

So the condition you're interested in becomes:

$$1.51 \left(\frac{M_\mathrm{p}}{\rho_\mathrm{s}}\right)^\frac{1}{3} \gtrsim a_\mathrm{p} \left(\frac{M_\mathrm{p}}{3M_\ast}\right)^\frac{1}{3}$$

Cancelling and rearranging gives:

$$a_\mathrm{p} \lesssim 2.18 \left(\frac{M_\ast}{\rho_\mathrm{s}}\right)^\frac{1}{3}$$

For a satellite density of 3300 kg/m3 (similar to the Moon) with a host star the mass of the Sun, this corresponds to a planetary orbit of about 2.6 solar radii.

Needless to say, this is an extremely close star–planet separation that invalidates a lot of the approximations used to compute that limit. For example, the planet will be close to or within its Roche limit with respect to the star. If it is not itself being disrupted, it will likely be significantly non-spherical.

• At $2.6$ solar radii distance, the planet might have crossed itself the Roche-radius of the star, that might be worth calculating... Commented Jan 28, 2020 at 22:40
• @AtmosphericPrisonEscape - there's a good chance that a planet at such a distance would already have substantial deviations from being spherical even if it avoided disruption, which would require some changes to the formulae.
– user24157
Commented Jan 28, 2020 at 22:51
• In fact, $R_{\rm Roche}$ is 2.44 $R_{\odot}$ if we assume the Moon is coupled to a Jupiter-like gas giant of the same average density as the star. For a rocky planet this can be farther in, if we take Earths density, we get 1.6 $R_{\odot}$, but of course then the Hill radius is smaller already. So I don't see a physical possibility for $R_{\rm Hill}<R_{\rm Roche}$. Commented Jan 28, 2020 at 23:04
• Great, thanks! I see that this is not going to happen very often. Also thank you for mentioning Roche lobe, it clears up why I have always gotten Roche limit and Hill sphere mixed up.
– uhoh
Commented Jan 29, 2020 at 0:30