Blog Tie-in:

• Planets and Worlds, Part 3: Moons

**Note: In the text sections, I have resorted to E-notation for representing large and small numbers, because of the inability of the blog text editor tool to represent superscripts.**## 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:

... meaning that it 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 equation for calculating the Hill Sphere radius of a body is:

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 equation for calculating the Hill Sphere radius of a body is:

If we use the mass of the primary as the standard unit, and express the mass of the secondary as a fraction of the mass of the primary (

*m = MS/MP*), then the mass of the primary becomes 1.0 by definition, and our equation simplifies to:*Note that if you do the equation without any value for*a

*, the result always returns the percentage of the distance between the two objects at which the Hill radius falls.*

__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.

… or just almost exactly 1/100 of an Astronomical Unit.

If we multiply this result by the number of meters in an Astronomical Unit (1.496E+11), we get Earth's Hill Sphere radius in meters:

If we multiply this result by the number of meters in an Astronomical Unit (1.496E+11), we get Earth's Hill Sphere radius in meters:

If we then divide this number by the Earth's radius in meters (6.371E+06 m), we get Earth's Hill Sphere radius in terms of Earth radii:

Note that if we had originally set a = 1.496E+11 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 a = 2.348E+04 (1 AU in Earth radii), 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 in another post).

Does the mass of the Moon alter the calculation of the Earth’s Hill Sphere?

In a word: no.

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

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; however, if we define both masses in terms of the mass of the primary, then:

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 in another post).

Does the mass of the Moon alter the calculation of the Earth’s Hill Sphere?

In a word: no.

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

*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 Earth’s Hill Sphere, which we’ve shown above that it does certainly do.

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; however, if we define both masses in terms of the mass of the primary, then:

... and using the simplified equation, we calculate the Moon's Hill Sphere as:

or about 6.1245 million meters, or ~6.28% of the distance between Earth and the Moon.

## The Roche Limit

In the section on the Hill Sphere, we determined the 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 tells us the closest orbit any object can have to the body it orbits. However, four points should be noted:

So, while the Hill Sphere for a body is "constant" (in the sense of applying equally to all less massive bodies orbiting 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 two bodies are of similar densities, the Roche Limit will be about 2.44 times the radius of the larger body.

The fundamental equation is:

Conversely, the Roche Limit tells us the closest orbit any object can have to the body it orbits. However, four points should be noted:

**Hill Sphere**- The Hill Sphere is calculated for a body by comparing its mass and that of its gravitational primary;
- The Hill Sphere limit is true for all bodies orbiting the object in question. That is to say, any satellite of the body which moves beyond its Hill Sphere radius will escape the its gravity

**Roche Limit**- The Roche Limit, however, is calculated for satellites, held together only by their own gravity, orbiting the body in question;
- The Roche Limit is also a relative value, but is entirely dependent on the densities—not the masses—of the bodies involved; the Roche Limit will be different for any two bodies of different densities.

So, while the Hill Sphere for a body is "constant" (in the sense of applying equally to all less massive bodies orbiting 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 two bodies are of similar densities, the Roche Limit will be about 2.44 times the radius of the larger body.

The fundamental equation is:

Note that RP and RL are always in the same units; if RP is in meters, then RL will also be in meters, whereas if RP is in Earth radii, then RL 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 planet's mass and density, the equation simplifies to:

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 planet'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 was 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:

The Moon currently orbits at about an average of 384,399 km; if we divide this by the radius of the Earth (6371 km), we see that the Moon’s current orbit is just over 60 Earth radii (60.336), or 20.93 times times farther out than the distance we’ve just calculated. That is, 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 ~2.88 Earth radii out to ~235 Earth radii.

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 ~2.88 Earth radii out to ~235 Earth radii.

What if, however, Mars' moon, Deimos, were orbiting Earth, instead?

How close could it safely approach?

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

How close could it safely approach?

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

This is ~1.315 times farther than the Roche Limit for the Moon; Deimos, because of its lower density, cannot 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:

Going a step further, we can determine how far above the surface of each planet the Moon would be orbiting.

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

For Earth and the Moon: (2.8834 - 1.0) × (6.371E+06) = 1.8834 × (6.371E+06) = 1.19991E+07 meters, or 11,999.14 kilometers

For Mars and the Moon: (2.5757 - 1.0) × (3.39E+06) = 1.5757 × (3.39E+06) = 5.341E+06 meters, or 5,340.84 kilometers

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 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 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*).For Earth and the Moon: (2.8834 - 1.0) × (6.371E+06) = 1.8834 × (6.371E+06) = 1.19991E+07 meters, or 11,999.14 kilometers

For Mars and the Moon: (2.5757 - 1.0) × (3.39E+06) = 1.5757 × (3.39E+06) = 5.341E+06 meters, or 5,340.84 kilometers

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 larger radius than Earth, but a lower density—1.33 g/cm³ compared to 5.51 g/cm³, so:

Jupiter has a larger radius than Earth, but a lower density—1.33 g/cm³ compared to 5.51 g/cm³, so:

Earth's radius is 6371 km, so Earth could orbit Jupiter no closer than 108435.283 km without being destroyed. This would put it closer than the orbit of Metis, Jupiter's innermost moon, 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.