# 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
• The scaling of the hill radius is such that the effect is so dominant at periapse as to be a good approximation to the limiting radius over the whole orbit (i.e. even though it's only at periapse for a short while it's where most mass loss occurs) – Zephyr Sep 17 at 10:03