There was a recent update on ISRO's Mars Orbiter Mission's (MOM) Facebook page that it has crossed the Earth's Sphere of Influence (SOI):

MOM moves on!

MOM is going to cruise out of the sphere of Influence of Earth in about 4 hours from now! we will keep you posted.

Sphere of Influence Explained:

The Sun is much more massive than any of the planets and its gravity dominates the Solar System. Only very near to the planets, the planetary gravity become stronger than that of the Sun. This region around the planet is referred to as the Sphere of Influence (SOI) of that planet.

My questions are:

  • How to calculate the Earth's gravitational Sphere of Influence (SOI)?

  • How far does the Earth's SOI reach from the center of the Earth?


3 Answers 3


There are a couple of definitions, but the most useful one is called the Hill Sphere. Essentially this is the area around which one can orbit around an object, and not be pulled away by another object (Such as the Sun). As the linked article states, it can be calculated for objects that have one object much more massive than the other (Almost every case of interest) by using the following formula:

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

Where $e$ is the eccentricity of the orbit, $m$ is the less massive object, and $M$ is the more massive object, and $a$ is the semi-major axis (Distance between objects).

Assuming a circular orbit, which makes the math easier, gives that distance to be :

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

Earth's Hill Sphere is around 1,500,000 km, as shown on this graph from Wikipedia.

enter image description here

  • 1
    $\begingroup$ @FMaz008 $a$ is the semi-major axis of the celestial body (which is equal to the radius between it and the sun for circular orbits, $e=0$). And like PearsonArtPhoto said, $m$ and $M$ are the masses of the orbiting and the central celestial body (so the first would indicate the mass of a planet and the last the mass of the sun). $\endgroup$
    – fibonatic
    Commented Dec 6, 2013 at 14:34
  • $\begingroup$ But I do wonder how this sphere of influence is derived ($r_{SOI}=a\left(\frac{m}{M}\right)^\frac{2}{5}$). $\endgroup$
    – fibonatic
    Commented Dec 6, 2013 at 15:34
  • $\begingroup$ @fibonatic - See my answer. $\endgroup$ Commented Mar 9, 2014 at 6:32
  • $\begingroup$ Elsewhere you say "DSCOVR is at a Lagrange point, so I don't really consider that to be a satellite." Is DSCOVR a satellite orbiting earth or not? It's on the boundary of the Hill Sphere. An object at the edge of the Hill Sphere could quite easily drift into its own heliocentric orbit. $a(m/M)^{2/5}$ fits your criteria better. $\endgroup$
    – HopDavid
    Commented Oct 8, 2017 at 5:14

Sphere of influence

The gravitational sphere of influence asks which of two gravitating bodies should be used as the origin for purposes of modeling the behavior of some third body such as a spacecraft. This comes into play in at least two key places:

  • In a patched conic approximation, what's the right place to switch from one conic to another?

  • When a spacecraft is moving away from a large body and toward a smaller body, when should the spacecraft navigation switch from a large body centric to a small body centric point of view?

When looking at things from the perspective of a frame of reference with its origin at the center of the smaller body, the gravitational acceleration toward the larger body is calculated as a "third body effect" (sorry for the confusing nomenclature; it's not mine). The answer to the above questions (where should I switch patched conics / where should I switch my flight software) is the surface on which this perturbing third body acceleration is equal in magnitude to the gravitational acceleration toward the smaller body. This surface has a rather complex shape, approximately that of an oblate spheroid, but it cannot be expressed in terms of the elementary functions. However, the "right" place along the line connecting the two bodies can be expressed simply. Relative to the smaller body, this distance is $R\left(\frac m M\right)^{2/5}$, where $R$ is the distance between the two bodies, $m$ is the mass of the smaller body, and $M$ is the mass of the larger body.

Who developed this concept? That's a good question. Some aerospace textbooks call this the Lagrange sphere of influence after Joseph-Louis Lagrange, others call it the Tisserand sphere of influence after Felix Tisserand, and yet others just call it the sphere of influence, period.

Hill sphere

The Hill sphere asks a rather different question: Given a smaller body orbiting a larger body, can an even smaller body orbit the small body? The Hill sphere (aka the Roche sphere) looks at things from the perspective of energy rather than force. One of the key developments initiated by Lagrange was to switch from the Newtonian focus on force to instead focus on energy. Lagrangian physics, and later, Hamiltonian physics, were complete re-writes of classical mechanics. In many cases, particularly where energy is conserved, it makes a lot more sense to look at things from the perspective of energy rather than force.

So what determines whether an orbit is stable? The answer is very complex. In the three body problem, if that third object stays within an extremely complex boundary called the Roche lobe, the orbit of that third object about the smaller body will be stable for at least some amount of time. The Roche lobe just touches the L1 and L2 points and fans out from there. George Hill used the L1 point to define a sphere that approximated the Roche lobe. This is still intractable; the L1 point is defined by a fifth order polynomial that cannot be solved in terms of the elementary functions. Hill further simplified things by realizing that a simple cubic equation yields a very good approximation of that intractable fifth order equation. The result is $R\left(\frac m {3M}\right)^{1/3}$.

So which is "right"?

So which is "right", the Lagrange/Tisserand sphere of influence or the Hill sphere? First off, it's important to note that both are approximations. This muddies the waters as to which is "right". More importantly, the two concepts attack very different questions. If you are doing mission planning to send a spacecraft to the Moon or some other planet, you should be using the sphere of influence. Similarly, you should probably be using the sphere of influence if you are building the guidance and navigation subsystem for that spacecraft. On the other hand, if you are bringing an asteroid back to Earth for mining, you probably want to place it in a distant selenocentric retrograde orbit. Now the Hill sphere is the correct choice with regard to the orbit in which you place your asteroid.

As far as the question is concerned, the sphere of influence is the correct answer simply because the questioner asked about the sphere of influence, not the Hill sphere. Whether it was Lagrange or Tisserand, both have "sphere of influence" naming precedence over Hill simply because both predate Hill.

I will admit that that is being ridiculously nit-picky. The "right" answer is that the two concepts address two different questions. Both are "right."


Sphere of influence: http://en.wikipedia.org/wiki/Sphere_of_influence_(astrodynamics), or practically any textbook on the astrodynamics field within aerospace engineering. For example, both Vallado (Fundamentals of Astrodynamics and Applications) and Bate, Mueller, and White (Fundamentals of Astrodynamics) write about the sphere of influence.

Hill sphere: http://en.wikipedia.org/wiki/Hill_sphere, or practically any textbook or journal article that addresses invariant manifolds as applied to space exploration. For example, Koon, Lo, Marsden, and Ross (Dynamical Systems, the Three-Body Problem and Space Mission Design) write about the Hill sphere.

  • $\begingroup$ According to both Battin and Bate & Mueller & White, the sphere of influence is due to Laplace. $\endgroup$
    – user10716
    Commented Mar 20, 2017 at 14:23
  • 1
    $\begingroup$ From messing with the sun and planets I used to think the SOI was always smaller than the Hill Sphere. But the SOI and Hill Sphere seem to have the same radius if the central body is 243 times more massive than the orbiting body. $243 = 3^5$. The Moon's SOI is a little larger than the Hill Sphere. $\endgroup$
    – HopDavid
    Commented Oct 8, 2017 at 5:58

This is what I use in class, these are standard formulas from any text book. Define 2 functions

sphereOfInfluence[dominantMass_, minorMass_, distanceBetween_] := 

sphereOfGravitation[dominantMass_, minorMass_, distanceBetween_] := 
   (*valid only for  minorMass<<dominantMass*)

For example, to find earth SOI w.r.t sun

sunMass = 1.989*10^30;
earthMass = 5.944*10^24;
earthSunDistance = 1.495978*10^8;
earthSOI = sphereOfInfluence[sunMass, earthMass, earthSunDistance]

(* 922790.  in km *)

To find moon SOI w.r.t. earth

earthMass = 5.944*10^24;
moonMass = 7.3483*10^22;
moonEarthDistance = 384400;
moonSOI = sphereOfInfluence[earthMass, moonMass, moonEarthDistance]

(* 66317.3  in km *)

To find earth gravitational sphere of influence w.r.t. sun

earthgSOI = sphereOfGravitation[sunMass, earthMass, earthSunDistance]
(* 258611. km *)

As expected, the gravitational sphere of influence is much smaller than the SOI.

  • 1
    $\begingroup$ +1. Strictly speaking, this is the correct answer. The Hill sphere and the sphere of influence are distinct concepts. I'll add details in a separate answer. $\endgroup$ Commented Mar 9, 2014 at 3:52

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