<p>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.</p>

<p>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:</p>

<p><a href="http://www.jgiesen.de/astro/stars/roche.htm" rel="noreferrer">http://www.jgiesen.de/astro/stars/roche.htm</a> </p>

CC BY-SA 3.0
2018-03-13T18:36:55.540Z
2018-03-13T18:36:55.540Z
mysterium
<p>This question may be a little lazy, but can anybody give me a proof of the Hill sphere formula? Acording to <a href="https://en.wikipedia.org/wiki/Hill_sphere" rel="noreferrer">wikipedia</a>, the formula for the radius, $r$, is</p>

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

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

CC BY-SA 3.0
2016-10-24T22:26:38.687Z
Look at the introduction in [this paper](https://arxiv.org/pdf/1111.3843.pdf).
Dave
CC BY-SA 3.0
2016-10-24T23:03:00.100Z
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
CC BY-SA 3.0
2016-10-30T12:49:28.470Z
@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
CC BY-SA 3.0
2017-07-08T09:32:12.000Z
Because a(1-e) is periastron?
chris
CC BY-SA 4.0
2019-09-17T10:03:30.650Z
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
CC BY-SA 4.0
2019-09-17T10:07:00.357Z
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
CC BY-SA 4.0
2019-12-11T11:17:50.297Z
@userLTK: Well no, tidal forces don't play into that becasue those don't exist in the circularly restricted three-body problem, which is where $r_H$ is coming from. The 3 comes from coriolis(2x) and centrifugal(1x) forces. Check Murray&Dermott, Solar system dynamics, Chapt 3.6, has a nice derivation of this.
AtmosphericPrisonEscape
<gravity><celestial-mechanics>
The deduction of the Hill Sphere formula
Carlos Vázquez Monzón
# The deduction of the Hill Sphere formula

(Asked by: username_0 on 2016-10-24)

This question may be a little lazy, but can anybody give me a proof of the Hill sphere formula? Acording to [wikipedia](https://en.wikipedia.org/wiki/Hill_sphere), 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 $e$.

**Tags:** gravity, celestial-mechanics

> **Comments:**
> **username_1**: Look at the introduction in [this paper](https://arxiv.org/pdf/1111.3843.pdf).  
> **username_2**: 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?  
> **username_3**: @username_1 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.  
> **username_4**: Because a(1-e) is periastron?  
> **username_5**: 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)  
> **username_5**: 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)  
> **username_2**: @userLTK: Well no, tidal forces don't play into that becasue those don't exist in the circularly restricted three-body problem, which is where $r_H$ is coming from. The 3 comes from coriolis(2x) and centrifugal(1x) forces. Check Murray&Dermott, Solar system dynamics, Chapt 3.6, has a nice derivation of this.  

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## 1 answers

### Answer by username_6 on 2018-03-13. Score: 7

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:

<http://www.jgiesen.de/astro/stars/roche.htm>


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