Possibly nasty astronomy question

[Calilasseia]

I've been trying to find out how the equation for the radius of a Hill Sphere is derived.

Trouble is, thus far I've drawn a LOT of blanks.

I tried deriving it myself using the notion that a test mass at the edge of the Hill Sphere would experience the same gravitational pull from the small-mass body and the large-mass body, fired up Newton's Law of Universal Gravitation, and obtained a nice looking quadratic equation ... which of course was hopelessly wrong as comparison with the actual Hill Sphere radius equation later demonstrated. Moreover, I could determine that it was wrong even before the comparison because if the masses of the two bodies were equal, my expression, which involved M-m in the denominator, became infinite (which of course is physically nonsensical).

Trouble is, all I can find is a lot of indirect references to what is required, followed by a lot of wild goose chasing of links before running into blind alleys.

Anyone know how it's done?

[Preno]

For a circular orbit (I'm too lazy to do the computation for the general case) with the radius a, you get the condition that for the perihelion, the attractive force of the planet + the centrifugal force should balance out with the attractive force of the sun.

If you can't read latex, you can plug the following latex code into this generator (http://dammserv.loopback.nu/latex/latex.php).

\frac{m}{r^2} + \frac{g}{a}(a-r) = \frac{M}{(a-r)^2}\\
(g = m \omega^2 a = \frac{M}{a^2})\\
a^3 \frac{m}{M} = r^3 + ar^2 \left(\frac{1}{(1-\frac{r}{a})^2}-1\right) \approx r^3 + ar^2(2r/a)

[VoxRat]

I've been trying to find out how the equation for the radius of a Hill Sphere is derived.
I think I've pinpointed your problem:
Hills are not spherical; they're roughly conical, or pyramidal.

VoxRat
-who promises to google Hill Sphere when he gets a moment.

[ofro]

Oh you meant the Roche sphere!?

[/euro-chauvinism]

[Preno]

For a circular orbit (I'm too lazy to do the computation for the general case) with the radius a, you get the condition that for the perihelion, the attractive force of the planet + the centrifugal force should balance out with the attractive force of the sun.You could also try solving a similar equation for the case where the test particle is on the planet-sun line but at the other side of planet test whether it really is a sphere. You get:

\frac{g}{a}(a+r) = \frac{M}{a^3}(a+r) = \frac{m}{r^2} + frac{M}{(a+r)^2}\\
\left(1+\frac{r}{a}\right)^3 = \frac{m}{M}\left(\frac{a}{r}+1\right)^2+1\\
3\frac{r}{a} \approx \frac{m}{M}\frac{a^2}{r^2}

i.e. the same expression for r.

[lpetrich]

If the Moon is at the Earth's Hill sphere, then the differential gravitational pull of the Sun on it would have the same typical size as the gravitational pull of the Earth on it.

I say "differential" here, because the Sun pulls on both the Earth and the Moon, though by different amounts because of the Moon's position. This differential effect is what makes tides.

The Earth's gravitational acceleration on the Moon is GME/aM2

for Earth mass ME and Moon mean distance from the Earth aM.

The Sun's differential acceleration on the Moon is typically around

GMS/(aE+aM)2 - GMS/aE2 ~ GMSaM/aE3

for Sun mass MS and Earth mean distance from the Sun aE.

The Hill sphere are where the two are approximately equal, or

GME/aM2 ~ GMSaM/aE3

Dividing both sides by aM gives this elegant expression:

GME/aM3 ~ GMS/aE3

that is, the density of the Sun inside the Earth's orbit must equal the density of the Earth inside the Moon's orbit. Using Kepler's third law with period T gives us this additional elegant expression:

TM ~ TE

meaning that the Moon orbits with period 1 year at the Hill sphere.


Low-Earth-orbit satellites? Most of them have an average density much less than the Earth's average density of 5.52 g/cm3, meaning that they are larger than their Earth-orbit Hill spheres. However, a steel fragment will be inside its Hill sphere, but only by a small margin; the density of iron is 7.87 g/cm3. By comparison, glass and ceramics and rock have densities of armound 2.5 to 3 g/cm3, and water and organic materials (biological, plastic, etc.) around 1 g/cm3. Lead and gold would do much better, at densities of 11.3 and 19.3 g/cm3, respectively.