**The Hill Sphere**

In theory, an object’s gravity is infinite, but its influence decreases at the rate of the inverse square of the distance:

Below are a table illustrating the intensity fall-off as a function of the square of the distance:

... showing that the gravitational attraction becomes negligibly weak at significant distances from the center of mass. Thus, there is a practical limit to the distance at which an object’s gravity is effective. The Hill sphere radius is a relative value dependent on the mass of the body in question and the mass of the nearest more massive body with which it has a gravitational interaction. For example, the Hill Sphere of a moon orbiting a planet is limited by the mass of the planet it orbits; the Hill Sphere of the planet is limited by the mass of the star it orbits; and, finally, the Hill sphere of the star is limited by the nearest more massive object to it (usually another star or other massive object in its neighborhood.) Beyond an object's Hill sphere (or Hill radius), its gravity is overwhelmed by that of the neighboring more massive object. So, another planet with the same mass as Earth but whose star is two times as massive as the Sun would have a different Hill sphere radius than Earth’s. |

The Hill sphere is named for George William Hill, who first defined it, based on the work of Édouard Albert Roche (see below).

The equation for calculating the Hill sphere radius of a body is:

The equation for calculating the Hill sphere radius of a body is:

Where:

If we express both masses in terms of the mass of the primary, the equation becomes:

*α*= the distance between the two objects in whatever units are desired (usually*AU*);*M*= the mass of the primary (more massive) body, in whatever units are desired;*m*= the mass of the secondary body (in the same units as*M*);*H**S*= the outer limit of the Hill sphere (in the same units as*α*)If we express both masses in terms of the mass of the primary, the equation becomes:

Where

*n**=**m*/*M.*

**Earth's Hill Sphere**

Let's calculate the Hill sphere radius for Earth, using the full equation. The mass of the Earth and Sun are in grams and the distance between them is in astronomical units.

Sun:

Earth:

Let’s have the result returned in AUs, so

Sun:

*M**= 1.989*⨉*10**33*Earth:

*m*=*5.97*⨉*10**24**α*=*1*AU =*1.496*⨉*10**11*m

Let’s have the result returned in AUs, so

*α**= 1*for this equation.… or almost exactly ⅟₁₀₀ of an Astronomical Unit.

If we multiply this result by the number of meters in an astronomical unit (1.496 × 1011), we get Earth's Hill sphere radius in meters:

If we then divide this number by the Earth's radius in meters (6.371 ⨉ 106 m), we get Earth's Hill sphere radius in terms of Earth radii:

Note that if we had originally set

The distance to the Moon is about 60.336 Earth radii; if we divide Earth's Hill sphere radius by this value, we see that

Put another way, the farthest distance an object can orbit the Earth is almost four times farther than the distance to the Moon; so, the Moon is in no danger of being stolen away from Earth by the Sun. (In fact, in one sense, the Moon, itself, can be said to be directly orbiting the Sun, but more on that elsewhere.)

*α**= 1.496*⨉*10**11*m instead of 1.0 in the original equation, our result would have been in meters, eliminating the need for the first conversion equation. However, if we had set*α*=*2.348*⨉*10**4*meters (1 AU in Earth radii), then our result would have been in Earth radii, eliminating the need for the second conversion.

The distance to the Moon is about 60.336 Earth radii; if we divide Earth's Hill sphere radius by this value, we see that

*234.853 ÷ 60.336 = 3.892*, so the Moon orbits at just over ¼ of the Earth’s Hill sphere radius.

Put another way, the farthest distance an object can orbit the Earth is almost four times farther than the distance to the Moon; so, the Moon is in no danger of being stolen away from Earth by the Sun. (In fact, in one sense, the Moon, itself, can be said to be directly orbiting the Sun, but more on that elsewhere.)

Note that the mass of the Moondoes notalter the calculation of the radius of Earth’s Hill sphere.

As we've defined, the calculation of a body’s Hill sphere is a function of the mass of the body and the

Thus, it is theoretically possible for the Moon to have its own moon, so long as any such body orbited within the limits of the Moon's own Hill sphere radius. We can determine what this critical distance is by using the same equation as above.

If we define both masses in terms of the mass of the primary, then:

Earth:

Moon:

Using the simplified equation, we calculate the Moon's Hill sphere:

*next nearest more massive body*. Since the Moon is less massive than the Earth, its*only*relationship to Earth’s Hill sphere is whether or not it orbits within that distance, which we’ve shown above that it does certainly do.Thus, it is theoretically possible for the Moon to have its own moon, so long as any such body orbited within the limits of the Moon's own Hill sphere radius. We can determine what this critical distance is by using the same equation as above.

If we define both masses in terms of the mass of the primary, then:

Earth:

*M**= 5.97*⨉*10**27*Moon:

*m**= 7.342*⨉*10**22**n*=*m*/*M*=*7.342*⨉*10**22**÷ 5.97*⨉*1027 = 0.000123*terran*α*=*384,399 km ÷ 1731.1 km = 222.05*lunar radii (Earth-Moon distance)

Using the simplified equation, we calculate the Moon's Hill sphere:

… 3.538 lunar radii, or about 6.1245 million meters, or about 6.28% the distance between the Earth and the Moon.

There will be more on this topic in a later blog.

There will be more on this topic in a later blog.

**The Roche Limit**

Above, we determined the Hill sphere—maximum distance an object's gravity can extend before being overwhelmed by the gravity of a more massive neighbor. Essentially, this equates to the farthest orbit any natural satellite of that object can have. Conversely, the Roche limit (named for Édouard Albert Roche, who first described it in explaining Saturn’s rings) tells us the closest orbit any object can have to the body it orbits. However, four points should be noted:The Hill Sphere1. The Hill sphere is calculated for a body by comparing its mass and that of its gravitational primary; 2. The Hill sphere limit is true for all bodies orbiting the primary object: that is to say that any satellite of the body which moves beyond the primary’s Hill sphere radius will escape the primary’s gravity. |

**The Roche Limit**

1. The Roche limit, however, is calculated for individual satellites which are held together

*only*by their own gravity, and which are orbiting the body in question;

2. The Roche limit is

*entirely dependent*on the

*densities*--

*not the masses*—of the bodies involved; the Roche limit will be different for all bodies orbiting the primary, as long as they all have different densities.

So, while the Hill sphere for a body is "constant" (in the sense of applying equally to all less massive bodies which orbit it), the Roche limit is dependent on the density of the orbiting body, and is different for every orbiting body (unless they have exactly the same density, of course).

Generally speaking, if the primary and the satellite are of similar densities, the Roche limit will be about 2.44 times the radius of the larger body.

The basic equation is:

Where:

Note that

Note, also, that the densities may be expressed in any units desired, so long as both densities are in the same units.

When all the values are expressed in terms of the primary’s mass and density, the equation simplifies to:

*R**L*= the Roche limit;*R**P*= the radius of the primary;*ρ**P*= the density of the primary;*ρ**S*= the density of the satelliteNote that

*R**P*and*R**L*are always in the same units; if*R**P*is in meters, then*R**L*will also be in meters, whereas if*R**P*is in Earth radii, then*R**L*will also be in Earth radii, etc.Note, also, that the densities may be expressed in any units desired, so long as both densities are in the same units.

When all the values are expressed in terms of the primary’s mass and density, the equation simplifies to:

We can use this equation, for instance, to calculate how closely the Moon could orbit the Earth before it would be destroyed by gravitational tidal stresses.

For simplicity, let's express the densities in terms of Earth masses and the radius in terms of Earth radii:

For simplicity, let's express the densities in terms of Earth masses and the radius in terms of Earth radii:

*R**P**= 1*terran*ρ**P**= 1*terran*ρ**S**= 0.606*(density of the Moon, expressed in terrans)… which shows that the Moon cannot come closer than about 2.9 times the radius of the Earth (≈ 18476 km) without being pulled apart by gravitational tidal forces.

The Moon currently orbits at about an average of 384,399 km; if we divide this by the radius of the Earth (

The Moon currently orbits at about an average of 384,399 km; if we divide this by the radius of the Earth (

*384399 ÷ 6371 = 60.3357*), we see that the Moon’s current orbit is just over 60 Earth radii, which is 20.9252 times times farther out than the distance we’ve just calculated. Thus, the Moon is currently quite safe from destruction by tidal forces.Thus, combining the Hill sphere and the Roche Limit calculations, we can say that the "legal" limits of the range of orbits the Moon could inhabit runs from ≈ 2.9 Earth radii out to ≈ 235 Earth radii.

**A Fictional Example**

What if Mars' moon, Deimos, were orbiting Earth? How close could it safely approach?

Keeping the values for Earth the same, our inputs are:

Keeping the values for Earth the same, our inputs are:

*R**P**= 1*terran*ρ**P**= 1*terran*ρ**S**= 0.2668*(density of Deimos, expressed in terrans)This is ≈ 1.315 times farther than the Roche limit for the Moon; because of its lower density, Deimos could not get as close to Earth before being destroyed by tidal forces.

Let's reverse the scenario: What if the Moon were orbiting Mars; what would its closest possible orbit be, then?

We'll set the values for the primary in terms of Mars:

Let's reverse the scenario: What if the Moon were orbiting Mars; what would its closest possible orbit be, then?

We'll set the values for the primary in terms of Mars:

*R**P**= 1*(Martian radius)*ρ**P**= 1*(Mars density)*ρ**S**= 3.344*(*ρ**Moon*)*÷ 3.9335*(*ρ**Mars*)*= 0.8501*Martian densities**Converting Orbits Into Altitudes**

An object’s altitude is the distance it is above the surface of the primary body, as compared to its distance from the primary’s center of gravity.

We can determine the altitude at which the Moon would be orbiting Earth and Mars if it were at its Roche limit for each body.

The Moon’s Roche limit for Earth is 2.8834 Earth radii, and its Roche limit for Mars is 2.5757 Mars radii; if we subtract 1.0 from each value, we get the number of planetary radii

We can determine the altitude at which the Moon would be orbiting Earth and Mars if it were at its Roche limit for each body.

The Moon’s Roche limit for Earth is 2.8834 Earth radii, and its Roche limit for Mars is 2.5757 Mars radii; if we subtract 1.0 from each value, we get the number of planetary radii

*above the surface*the Moon is (would be) orbiting. Then, we multiply by each planet’s radius in meters to get a result in meters above the surface (called the object's*altitude*).So, the Moon in its closest possible orbit would have to orbit 2.247 times higher in altitude above Earth than above Mars in order to remain intact.

**The Earth As A Moon**

How closely could the Earth orbit Jupiter before being destroyed by tidal forces?

Jupiter has a much larger radius than Earth (71492 km vs 6371 km), but a lower density—1.33 g/cm³ compared to 5.51 g/cm³, so:

Jupiter has a much larger radius than Earth (71492 km vs 6371 km), but a lower density—1.33 g/cm³ compared to 5.51 g/cm³, so:

*R**P**= 11.209*terran*ρ**P**= 0.241*terran*ρ**S*=*1.0*terranSo, Earth could orbit Jupiter no closer than 108,435.283 km without being destroyed.

This would put it closer than the orbit of Jupiter's innermost moon, Metis, and 3.889 times closer to Jupiter than Io, the innermost of the Galilean moons.

At that orbital distance, Earth would orbit Jupiter in ≈ 5.55 hours, and would be tidally-locked. However, this orbit is also about one-quarter of the innermost major moon orbit for a Jupiter-sized giant planet, according to Dermott's Law (which I will discuss in a later blog).

This would put it closer than the orbit of Jupiter's innermost moon, Metis, and 3.889 times closer to Jupiter than Io, the innermost of the Galilean moons.

At that orbital distance, Earth would orbit Jupiter in ≈ 5.55 hours, and would be tidally-locked. However, this orbit is also about one-quarter of the innermost major moon orbit for a Jupiter-sized giant planet, according to Dermott's Law (which I will discuss in a later blog).