# How spherical is the Hill Sphere?

...because I have a budding sense it's not quite spherical. Or at least the center isn't equal to the body's center.

For one thing, at the distant end (say, in case of Moon's Hill Sphere, the point opposite from Earth) Earth's gravity is weaker than on the Earth side (due to the distance), meaning Moon's sphere of influence should reach farther. Or in case of certain elliptic polar orbits (similar to Sun-synchronous Earth orbit) they could reach much farther as the distance from "the big body" doesn't change by much, and the smaller body's gravity should be able to pull the satellite back from a quite a distance "above" or "below".

So, what's the shape of the Hill Sphere? And what magnitude of eccentricity/irregularity in its shape can be expected?

• If you read the Wikipedia article, it mentions that the sphere is really just an approximation. I think you are right in principal, but also remember, an ideal orbit is circular. Thus, it doesn't matter much if it would be stable at a far end, if you aren't stable at the worst point, you aren't stable. – PearsonArtPhoto Jun 19 '15 at 13:22
• @PearsonArtPhoto: Is it only about circular orbits, or also about elliptic orbits with apoaxis at given distance? (in the latter case the "critical point" gets quite narrow). – SF. Jun 19 '15 at 13:50
• By definition, the Hill sphere is a sphere. – David Hammen Jun 19 '15 at 15:00

tl;dr answer, and my math might be a little off, but here goes:

The Hill Sphere is an approximation which takes just 4 things into account:

### $$r \approx a (1-e) \cdot \sqrt[3\ \ \ ]{\frac{m}{3 M}}$$

• the Semi Major Axis $$a$$
• the eccentricity of the orbit $$e$$
• and the mass of the 2 objects, in this case the Earth and the Moon.

The Hill Sphere formula doesn't, by the way, take into account gravitational imperfections due to varied density, so it doesn't rise and fall with every mountain on Earth and it approximates a radius, which by definition, represents a perfect sphere, so, the Hill Sphere is a sphere, calculated by a formula. By definition, as David Hammen said.

So ...

...because I have a budding sense it's not quite spherical. Or at least the center isn't equal to the body's center. For one thing, at the distant end (say, in case of Moon's Hill Sphere, the point opposite from Earth) Earth's gravity is weaker than on the Earth side (due to the distance), meaning Moon's sphere of influence should reach farther. Or in case of certain elliptic polar orbits (similar to Sun-synchronous Earth orbit) they could reach much farther as the distance from "the big body" doesn't change by much, and the smaller body's gravity should be able to pull the satellite back from a quite a distance "above" or "below".

What you're describing works exactly like that with planetary magnetic fields. It doesn't work with gravitational orbits hardly at all because orbits don't switch from being stable to unstable as they go from the Earth side of the Moon to the far side of the Moon. The entire orbit is considered one or the other, though in a certain sense, all orbits are ultimately unstable given enough time, so there's some fuzziness to those definitions as well.

So, what's the shape of the Hill Sphere? And what magnitude of eccentricity/irregularity in its shape can be expected?

Well, we know the Hill Sphere is a perfect sphere, and what we might call a region of orbital stability, generally considered 1/3rd to 1/2th the Hill Sphere's radius (also in the Wikipedia article), that's pretty much a perfect sphere too, though it's also not definable with precision.

But, is there any form of gravitational influence that is stretched on the far side and compacted on the near side. Maybe. Any objects gravitational influence pretty much extends to any distance at the speed of light, so gravitational influence doesn't work either, but lets try this;

Lets make up a term and call it gravitational capture zone, or capture zone for short, which I'll define by a region where a planet can capture an object, usually an asteroid or comet that flies close to it, and lets define this as a situation where the object has to complete at least 1 full, 360 degree orbit. One is enough. Now, obviously, capture also depends on the speed of the asteroid or comet, so this is kind of rough, but the Earth does have, in a sense, a capture zone - so does Mars, in fact, Mars' 2 moons are believed to be captured asteroids. In fact, all the outer planets have captured asteroids orbiting them as moons.

Another way of looking at this, is, an asteroid that flies at a set speed, 1,000 km above the Moon's surface will be influenced by the Moon's gravity more than the Earth's, at least, when it flies that close (before it gets that close, it's more affected by the Earths - but the math gets complicated).

So, an asteroid flying 1,000 km above the Moon on the far side of the Earth would bend more than an asteroid flying 1,000 km above the Moon but in between the Moon and the Earth, where, in that position, the Moon's and Earth's gravity on the asteroid would be pulling in opposite directions - much less bend, and if we take that in mind, the influence where the Moon might capture an asteroid is greater on the away side of Earth than the near side of the Earth.

So, lets return to Hill sphere, and I'm going to simplify the math by ignoring eccentricity and assuming circular orbits - from the same Wikipedia link:

### $$r \approx a \cdot \sqrt[3\ \ \ ]{\frac{m}{3M}}$$

Lets do the Earth-Sun first;

$$a$$ (semi major axis, or, lets just call it distance), and I hate math so we'll call $$a$$ 1, for 1 AU, and mass, lets call little $$m$$ 1 and big $$M$$ 333,000 because the Sun is 333,000 times more massive than the Earth:

$$1 AU \cdot \sqrt{1/333,000\cdot3\ }$$ is just about 1/100th of an AU, so from the near end of the Earth's Hill Sphere to the Sun and the far end away from the Sun, the gravitational effect of the Sun at 0.99 AU to 1.01 AU (we're basically just talking tidal effects at this point), but, gravity decreases by the square of the distance, so this is about a 4% change in the Sun's gravitation from the far end to the near end of the Earth's Hill sphere.

The sphere is still a sphere, but the Earth should be slightly more able to capture an asteroid on it's far side from the Sun than the near side. In that sense, the term I made up, capture zone, would be stretched by a little bit. 4% might be upper limit of the stretching and it's probably less than that, because a gravitational capture zone would be smaller than the Hill Sphere. (if somebody wants to correct that, feel free, but this is ballpark.)

The Earth actually does this, by the way - capture asteroids. The Moon probably doesn't, because it's gravity is too weak and comparatively, asteroids move too fast. Asteroids captured by the Earth usually don't stick around very long though:

Now, lets look at the Earth-Moon system. The Earth weighs approximately 81 times as much as the Moon, so the Hill Sphere of the Moon is $$1/\sqrt{81*3}$$, or 16% of the distance between the Earth and the Moon. The Moon's theoretical gravitational capture zone would be significantly more elongated around it's far side than it's Earth side. The elongation of a capture zone depends on the ratio of size between the 2 objects, so the Sun being comparatively so large, makes Earth's "capture zone" much closer to a sphere, not much more stretched out as you suggest.

Now, in reality, the Moon's Hill sphere is all but irrelevant, because there is no stable orbit around the Moon due to tidal effects from the Earth and to a lesser extent, orbital perturbation from the Sun. But if the Moon was to capture an asteroid - that would need to just about exactly match the Moon's speed, in that hypothetical scenario, I think the Moon would be more likely to capture an asteroid on it's far side than it's near side - so there is in a sense, an elongation for temporary capture of a passing object.

But this is all quite vague, because anything that orbits the Moon also orbits the Earth, and any elongation of this theoretical capture zone is pretty irrelevant in terms of stable orbits, because a stable orbit has to account for all parts of the orbit, not just the part where the tidal effect is the lowest. So, I don't think it's a good way of looking at orbital mechanics, but I think you're correct in thinking there's something about a planet's gravitational sphere of influence not being quite spherical.

Loosely related on Moon's capturing asteroids: http://www.universetoday.com/109666/can-moons-have-moons/

• Thank you for the lengthy answer. One thing I'm going to disagree with: "It doesn't work with gravitational orbits hardly at all because orbits don't switch from being stable to unstable as they go from the earth side of the moon to the far side of the moon. " - that's for circular orbits. Take two different elliptical orbits, same semi-major and semi-minor axis lengths, but stretching into two opposite sides of a body. One may be quite stable, while the other - not at all, coming into vicinity of a different body. Only (normally quite slow) orbital precession would destabilize the former. – SF. Jun 21 '15 at 13:23
• Well, I do tend to ramble on sometimes, so, apologies. But I wasn't talking about circular orbits at all, only circular regions of stability. All orbits are elliptical, and in a 3 body system, those ellipses are perturbed but a perturbed orbit can also be a stable one. Also, the semi-major and semi-minor axis spiral around the object they orbit. They're rarely in synch so that the Semi-Major axis is always on the earth side. So, to my thinking, an unstable elliptical orbit doesn't become stable at the semi-minor axis just cause it's inside the Hill Sphere. – userLTK Jun 21 '15 at 23:38
• Also, I'll just add this picture. This is what an orbit around the moon would look like: ccar.colorado.edu/asen5050/projects/projects_2007/slavkin/… Source: ccar.colorado.edu/asen5050/projects/projects_2007/slavkin orbits around the Moon would tend to spiral into the moon because of the enormous earth tidal force, not spiral away at the near point to the earth. Another related article: realclearscience.com/blog/2014/12/can_moons_have_moons.html – userLTK Jun 21 '15 at 23:47