First we'll start with the formula for the radius of the Hill sphere.

[math]R \approx a \sqrt[3]{\frac{m}{3M}}[/math]

Where a is the distance between our two main bodies.

m is the mass of the body that we are finding the Hill sphere for.

M is the mass of the body that the smaller body is orbiting.

Now we take the formula for the period of a satellite:

[math]T = 2 \pi \sqrt{\frac{r^3}{Gm}}[/math]

If we substitute

**R**from the first equation for

**r**in in the second, we get an equation that gives the period for a satellite orbiting at the edge of the Hill sphere. Since no satellite can exist outside the Hill sphere, this is also the longest period a satellite can have around the body:

[math]T = 2 \pi \sqrt{\frac{ \left(a \sqrt[3]{\frac{m}{3M}} \right)^3}{Gm}}[/math]

[math]T = 2 \pi \sqrt{\frac{ a^3\frac{m}{3M}}{Gm}}[/math]

[math]T = 2 \pi \sqrt{\frac{ a^3}{3GM}}[/math]

Now we compare this period with the rotation of our satellite, which is equal to the orbital period of that satellite in a tidal lock situation:

[math]T_2 = 2 \pi \sqrt{\frac{a^3}{GM}}[/math]

Comparing the two, we get:

[math]ratio = \frac{ 2 \pi \sqrt{\frac{ a^3}{3GM}}}{2 \pi \sqrt{\frac{a^3}{GM}}}[/math]

[math]ratio =\sqrt{ \frac{ \pi \frac{ a^3}{3GM}}{2 \pi \frac{a^3}{GM}}}[/math]

[math]ratio =\sqrt{ \frac{ \frac{ a^3}{3GM}}{ \frac{a^3}{GM}}}[/math]

[math]ratio = \sqrt{1/3} = 0.57735...[/math]

Which shows that the longest period our satellite can have will always be shorter than the orbital period of the body it is orbiting, and by extension, shorter than that of the rotational period of that body if it is tidally locked.

Now let's look at the other extreme for satellite orbits; the Roche limit. The Roche limit is the point where tidal forces become great enough to tear the body apart. (one small caveat: The Roche limit only effects satellites large enough that the tidal forces can overcome the tensile strength of the satellite.)

Often you will see the Roche Limit expressed as

[math]r_{Roche} \approx 2.423 R[/math]

Where

**R**is the radius of the body our satellite is orbiting.

This expression is somewhat vague as it makes two assumptions. One is that the satellite and main body are of equal density, and the other is that the bdy is "fluid" in nature. (not rigid, which is more descriptive of most moons)

Correcting for these assumptions we get the following relation:

[math]r_{Roche} \approx R \sqrt[3]{2 \frac{\rho M}{\rho m}}[/math]

Where [math]\rho M[/math] and [math]\rho m[/math] are the densities of the primary and satellite respectively.

Now we can determine the period of a satellite orbiting at the Roche limit by inserting this equation into the orbital period equation as we did above:

[math]T = 2 \pi \sqrt{\frac{\left( R \sqrt[3]{2 \frac{\rho M}{\rho m}}\right)^3}{GM}}[/math]

[math]T = 2 \pi \sqrt{\frac{ R^3{2 \frac{\rho M}{\rho m}}}{GM}[/math]

[math]T = 2 \pi \sqrt{\frac{2R^3 \rho M}{GM \rho m}}[/math]

Now we're going to get a little sneaky.

The density of a body is equal to its mass divided by its volume, and since the volume of a sphere is expressed as:

[math]V = \frac{4 \pi R^3}{3}[/math]

we can express the density of the primary as

[math]\rho M = \frac{3M}{4 \pi R^3} [/math]

If we now insert this expression into our orbital period equation, we get:

[math]T = 2 \pi \sqrt{\frac{2R^3 \frac{3M}{4 \pi R^3} }{GM \rho m}}[/math]

After canceling and simplifying, we get:

[math]T= \sqrt{ \frac{6\pi}{G \rho m}}[/math]

Since [math]\pi[/math] and

**G**are constants, the only variable left is [math]\rho m[/math].

This means that the minimum period for a moon that is subject to break up from passing within the Roche Limit is determined by the density of the moon alone and no other factor.

DISCLAIMER

The above works under the following conditions: That the moon is much less massive than the planet it revolves around and that the planet is nearly spherical and not too much of an oblate spheroid.