Hill sphere

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v t e

The Hill sphere is a common model for the calculation of a

To be retained by a more gravitationally attracting astrophysical object—a planet by a more massive star, a moon by a more massive planet—the less massive body must have an orbit that lies within the gravitational potential represented by the more massive body's Hill sphere.^{[not verified in body]} That moon would, in turn, have a Hill sphere of its own, and any object within that distance would tend to become a satellite of the moon, rather than of the planet itself.^{[not verified in body]}

One simple view of the extent of the Solar System is that it is bounded by the Hill sphere of the Sun (engendered by the Sun's interaction with the galactic nucleus or other more massive stars).^{[4][verification needed]} A more complex example is the one at right, the Earth's Hill sphere, which extends between the Lagrange points L₁ and L₂,^{[clarification needed]} which lie along the line of centers of the Earth and the more massive Sun.^{[not verified in body]} The gravitational influence of the less massive body is least in that direction, and so it acts as the limiting factor for the size of the Hill sphere;^{[clarification needed]} beyond that distance, a third object in orbit around the Earth would spend at least part of its orbit outside the Hill sphere, and would be progressively perturbed by the tidal forces of the more massive body, the Sun, eventually ending up orbiting the latter.^{[not verified in body]}

For two massive bodies with gravitational potentials and any given energy of a third object of negligible mass interacting with them, one can define a

]

Definition

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The Hill radius or sphere (the latter defined by the former radius^[

better source needed

]

As described by de Pater and Lissauer, all bodies within a system such as the Sun's Solar System "feel the gravitational force of one another", and while the motions of just two gravitationally interacting bodies—constituting a "two-body problem"—are "completely integrable ([meaning]...there exists one independent integral or constraint per degree of freedom)" and thus an exact, analytic solution, the interactions of three (or more) such bodies "cannot be deduced analytically", requiring instead solutions by numerical integration, when possible.^[6]: p.26 This is the case, unless the negligible mass of one of the three bodies allows approximation of the system as a two-body problem, known formally as a "restricted three-body problem".^[6]: p.26

For such two- or restricted three-body problems as its simplest examples—e.g., one more massive primary astrophysical body, mass of m1, and a less massive secondary body, mass of m2—the concept of a Hill radius or sphere is of the approximate limit to the secondary mass's "gravitational dominance",^[6] a limit defined by "the extent" of its Hill sphere, which is represented mathematically as follows:^{[6]: p.29 [7]}

${\displaystyle R_{\mathrm {H} }\approx a{\sqrt[{3}]{\frac {m_{2}}{3(m_{1}+m_{2})}}}}$,

where, in this representation, major axis "a" can be understood as the "instantaneous heliocentric distance" between the two masses (elsewhere abbreviated rp).^{[6]: p.29 [7]}

More generally, if the less massive body, ${\displaystyle m2}$, orbits a more massive body (m1, e.g., as a planet orbiting around the Sun) and has a

${\displaystyle a}$

${\displaystyle e}$

${\displaystyle R_{\mathrm {H} }}$

${\displaystyle R_{\mathrm {H} }\approx a(1-e){\sqrt[{3}]{\frac {m_{2}}{3(m_{1}+m_{2})}}}.}$

When eccentricity is negligible (the most favourable case for orbital stability), this expression reduces to the one presented above.^{[citation needed]}

Example and derivation

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In the Earth-Sun example, the Earth (5.97×10²⁴ kg) orbits the Sun (1.99×10³⁰ kg) at a distance of 149.6 million km, or one astronomical unit (AU). The Hill sphere for Earth thus extends out to about 1.5 million km (0.01 AU). The Moon's orbit, at a distance of 0.384 million km from Earth, is comfortably within the gravitational sphere of influence of Earth and it is therefore not at risk of being pulled into an independent orbit around the Sun.

The earlier eccentricity-ignoring formula can be re-stated as follows:

${\displaystyle {\frac {R_{\mathrm {H} }^{3}}{a^{3}}}\approx 1/3{\frac {m2}{M}}}$, or ${\displaystyle 3{\frac {R_{\mathrm {H} }^{3}}{a^{3}}}\approx {\frac {m2}{M}}}$,

where M is the sum of the interacting masses.

Derivation

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The expression for the Hill radius can be found by equating gravitational and centrifugal forces acting on a test particle (of mass much smaller than ${\displaystyle m}$) orbiting the secondary body. Assume that the distance between masses ${\displaystyle M}$ and ${\displaystyle m}$ is ${\displaystyle r}$, and that the test particle is orbiting at a distance ${\displaystyle r_{\mathrm {H} }}$ from the secondary. When the test particle is on the line connecting the primary and the secondary body, the force balance requires that

${\displaystyle {\frac {Gm}{r_{\mathrm {H} }^{2}}}-{\frac {GM}{(r-r_{\mathrm {H} })^{2}}}+\Omega ^{2}(r-r_{\mathrm {H} })=0,}$

where ${\displaystyle G}$ is the gravitational constant and ${\displaystyle \Omega ={\sqrt {\frac {GM}{r^{3}}}}}$ is the (Keplerian) angular velocity of the secondary about the primary (assuming that ${\displaystyle m\ll M}$). The above equation can also be written as

${\displaystyle {\frac {m}{r_{\mathrm {H} }^{2}}}-{\frac {M}{r^{2}}}\left(1-{\frac {r_{\mathrm {H} }}{r}}\right)^{-2}+{\frac {M}{r^{2}}}\left(1-{\frac {r_{\mathrm {H} }}{r}}\right)=0,}$

which, through a binomial expansion to leading order in ${\displaystyle r_{\mathrm {H} }/r}$, can be written as

${\displaystyle {\frac {m}{r_{\mathrm {H} }^{2}}}-{\frac {M}{r^{2}}}\left(1+2{\frac {r_{\mathrm {H} }}{r}}\right)+{\frac {M}{r^{2}}}\left(1-{\frac {r_{\mathrm {H} }}{r}}\right)={\frac {m}{r_{\mathrm {H} }^{2}}}-{\frac {M}{r^{2}}}\left(3{\frac {r_{\mathrm {H} }}{r}}\right)\approx 0.}$

Hence, the relation stated above

${\displaystyle {\frac {r_{\mathrm {H} }}{r}}\approx {\sqrt[{3}]{\frac {m}{3M}}}.}$

If the orbit of the secondary about the primary is elliptical, the Hill radius is maximum at the

${\displaystyle r}$

To leading order in ${\displaystyle r_{\mathrm {H} }/r}$, the Hill radius above also represents the distance of the Lagrangian point L₁ from the secondary.

Regions of stability

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The Hill sphere is only an approximation, and other forces (such as radiation pressure or the Yarkovsky effect) can eventually perturb an object out of the sphere.^{[citation needed]} As stated, the satellite (third mass) should be small enough that its gravity contributes negligibly.^{[6]: p.26ff}

Detailed numerical calculations show that orbits at or just within the Hill sphere are not stable in the long term; it appears that stable satellite orbits exist only inside 1/2 to 1/3 of the Hill radius.^{[citation needed]}

The region of stability for

Further examples

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It is possible for a Hill sphere to be so small that it is impossible to maintain an orbit around a body. For example, an astronaut could not have orbited the 104

The following table and logarithmic plot show the radius of the Hill spheres of some bodies of the Solar System calculated with the first formula stated above (including orbital eccentricity), using values obtained from the JPL DE405 ephemeris and from the NASA Solar System Exploration website.^[14]

See also

• Interplanetary Transport Network
• n-body problem
• Roche lobe
• Sphere of influence (astrodynamics)
• Sphere of influence (black hole)

Explanatory notes