# The deduction of the Hill Sphere formula

This question may be a little lazy, but can anybody give me a proof of the Hill sphere formula? Acording to wikipedia, the formula for the radius, $r$, is

$$r\approx a(1-e)\left(\frac{m}{3M}\right)^{1/3}$$

where a body of mass $m$ is orbiting a much more massive body of mass $M$ with a semi-major axis $a$ and eccentricty $e$.

• Look at the introduction in this paper. – Dave Oct 24 '16 at 22:26
• Place a test mass between two masses, assume the origin is in the bigger mass and calculate where the magnitudes to both forces are equal? – AtmosphericPrisonEscape Oct 24 '16 at 23:03
• @Dave that's a pretty cool paper (I'd planned on getting something done today, but now...), and I am sure it's in there; $R_H=3^{-1/3}$ and "unit of length is scaled by the factor µ${}^{1/3}$" but I don't see how to get the (1-e) in the front so easily. – uhoh Oct 30 '16 at 12:49
• Because a(1-e) is periastron? – chris Jul 8 '17 at 9:32
• It seems they've actually added a derivation to the wikipedia page - interestingly something not mentioned on the wikipedia page is that this surface is not spherical, it refers to when a particle on-axis is lost (during a single event at least - multiple non-resonant events eventually strip all material outside of the Hill radius leaving a sphere) – Zephyr Sep 17 '19 at 10:07

Hill sphere is named after John William Hill (1812–1879) and its simple logic follows from the presence of three bodies (let's assume Sun is the largest mass with Earth as the secondary mass and a satellite of negligible mass orbiting the Earth as the third mass), where the radius of the Hill sphere will be the largest radius at which a satellite could orbit the secondary mass (Earth in this case). If its orbit exceeds the Hills radius, then it will fall to the gravitational influence of the first body (sun) and hence will no longer be a satellite of the secondary body.

One could write Newton's equations using the idea that the satellite has the same angular velocity as the secondary object. This is that, the angular velocity of the Earth around the sun equals to the angular velocity of the satellite around the sun. A demonstration about the derivation is given in the following link as well as that of the Roche limit:

The Hill sphere is defined slightly differently to the Roche lobe, but the radius is approximated by the distance to the Lagrange point L1.

For circular motion with angular velocity $$\omega$$ around the origin, we have:

$$\ddot{\mathbf{r}} = - \omega^2 \mathbf{r}$$

The acceleration due to gravity from a point mass on another mass at position $$\mathbf{r}$$ is given by the usual inverse square law:

$$\ddot{\mathbf{r}} = -\frac{Gm}{\left\|\mathbf{r}\right\|^2}\hat{\mathbf{r}}$$

Now consider a two-body system with masses $$m_1$$ and $$m_2$$, separated by a distance $$r$$ orbiting their common centre-of-mass (c.o.m.) at distances $$r_1$$ and $$r_2$$ respectively.

This is a one-dimensional system, so we can switch from vectors to scalars. From the definition of the centre-of-mass, we have:

$$r_1 = \left(\frac{m_2}{m_1 + m_2}\right) r$$ $$r_2 = \left(\frac{m_1}{m_1 + m_2}\right) r$$

For the orbit of $$m_2$$ around the centre-of-mass, equating the gravitational acceleration with the required acceleration for circular motion gives:

$$\omega^2 r_2 = \frac{G m_1}{r^2}$$

And then expressing $$r_2$$ in terms of $$r_1$$ gives Kepler's Third Law:

$$\omega^2 = \frac{G\left(m_1 + m_2\right)}{r^3}$$

Next we find the distance to the L1 point, where the gravitational forces of the primary and secondary combine to provide the required acceleration for circular motion. Equating the acceleration for circular motion with the gravitational forces gives:

$$\omega^2 \left(r_2 - h\right) = \frac{G m_1}{\left(r - h\right)^2} - \frac{G m_2}{h^2}$$

And substituting $$\omega$$ results in:

$$\frac{\left(m_1 + m_2\right)\left(r_2 - h\right)}{r^3} = \frac{m_1}{\left(r - h\right)^2} - \frac{m_2}{h^2}$$

Then rewrite this in terms of the mass ratio $$q = \frac{m_2}{m_1}$$ and the relative distance $$z = \frac{h}{r}$$, giving:

$$1 - z\left(1 + q\right) = \left(1 - z\right)^{-2} - qz^{-2}$$

This results in a quintic equation for $$z$$, which must be solved numerically as general quintics do not have algebraic solutions (I'm not going to pretend to understand the proof of this).

Provided we are in a situation where $$m_1 \gg m_2$$, which is a good approximation for the Solar System planets, we can make approximations to avoid solving the quintic. In this case the Hill sphere is much smaller than the separation between the two objects, which means we can approximate:

\begin{aligned} 1 + q &\approx 1 \\ \left(1 - z\right)^{-2} &\approx 1 + 2z \end{aligned}

Where the second line is the binomial approximation. This gives:

$$1 - z \approx 1 + 2z - qz^{-2}$$

Rearrange to solve for $$z$$:

$$z^3 \approx \frac{q}{3}$$

And then using the definitions of $$z$$ and $$q$$ this becomes

$$h \approx a \left(\frac{m_2}{3 m_1}\right)^{1/3}$$

Which is the usual formula for the size of the Hill sphere. This works for circular orbits. For eccentric orbits, the usual approach is to simply replace the distance $$a$$ with the pericentre distance $$a\left(1 - e\right)$$. A more rigorous approach would be to use the angular velocity at pericentre and derive from there, but I'll leave that as an exercise for the interested reader :-)