As described in Ellipses And Orbits, it is possible for two objects to share an orbit if their locations are according to the Lagrange configuration described. (There

In fact, there are two sets of Lagrangian triples right here in our own back yard: Saturn's moons Tethys-Telesto-Calypso and Dione-Helene-Polydeuces.

Lagrangian moons and planets are able to share an orbit in a stable configuration, located 60° apart. While it is inviting to imagine a system of six worlds equally spaced around a given orbit—and thus all in one another’s L4 and L5 points—studies have shown that without some other strongly stabilizing influence, such a system would not remain stable, especially for highly elliptical orbits <1>.

*are*other orbit "sharing" configurations, such as horseshoe orbits, not covered here).In fact, there are two sets of Lagrangian triples right here in our own back yard: Saturn's moons Tethys-Telesto-Calypso and Dione-Helene-Polydeuces.

Lagrangian moons and planets are able to share an orbit in a stable configuration, located 60° apart. While it is inviting to imagine a system of six worlds equally spaced around a given orbit—and thus all in one another’s L4 and L5 points—studies have shown that without some other strongly stabilizing influence, such a system would not remain stable, especially for highly elliptical orbits <1>.

While all that follows is equally applicable to moons, for simplicity, I'll only discuss Lagrangian planets.

There are a number of possible configurations of planet types as Lagrangian sets, such as an ice giant with two terrestrials at its L4 and L5; a gas giant with two ice giants each leading and trailing in its orbit; a Tellurean with two dwarf-planet companions, etc. Sean Raymond at Planetplanet.net <2> has written a truly fascinating series of blog posts in which he explores the various possibilities.

Lagrangian pairs are not double planets (see below), as they do not orbit a barycenter common only to the two of them, but each independently orbits their central star(s), yet both on the same orbit. Also, double planets orbit their common barycenter fairly close to each other—typically within a few tens of their own radii. Lagrangian pairs are much farther apart: indeed, since they are separated by 60° on the same orbit, they are as far from one another as each is from the center of the star system. Inhabitants of small enough Lagrangian-pair planets might not even be aware of the other planet until an indigenous culture on one or the other invents telescopy. Asteroid 2010 TK7 <3>, which orbits at Earth's L4 point, was only discovered in 2010, and then only by the space-based WISE <4> telescope.

This is compounded by the distances involved. As it turns out, large enough (or bright enough) objects orbiting at Earth's L4 and L5 points would be visible from Earth only just before sunrise and/or just after sunset, in much the same way Venus is. As shown in the diagram below, those areas of Earth's surface in the shaded area would see the companion at L5 as a "morning star", rising before the Sun, and fading from view as the brightness of the daylight sky drowned it out of view. If a non-luminous object were very large (see below) and had a high albedo, it might be faintly visible in a clear daylight sky. Also, the object would become visible to about half of the daylight side of the Earth during total solar eclipses, when the Sun's light is blocked. If the companion were at L4, visibility would be as an "evening star", descending below the horizon shortly after sunset.)

Lagrangian pairs are not double planets (see below), as they do not orbit a barycenter common only to the two of them, but each independently orbits their central star(s), yet both on the same orbit. Also, double planets orbit their common barycenter fairly close to each other—typically within a few tens of their own radii. Lagrangian pairs are much farther apart: indeed, since they are separated by 60° on the same orbit, they are as far from one another as each is from the center of the star system. Inhabitants of small enough Lagrangian-pair planets might not even be aware of the other planet until an indigenous culture on one or the other invents telescopy. Asteroid 2010 TK7 <3>, which orbits at Earth's L4 point, was only discovered in 2010, and then only by the space-based WISE <4> telescope.

This is compounded by the distances involved. As it turns out, large enough (or bright enough) objects orbiting at Earth's L4 and L5 points would be visible from Earth only just before sunrise and/or just after sunset, in much the same way Venus is. As shown in the diagram below, those areas of Earth's surface in the shaded area would see the companion at L5 as a "morning star", rising before the Sun, and fading from view as the brightness of the daylight sky drowned it out of view. If a non-luminous object were very large (see below) and had a high albedo, it might be faintly visible in a clear daylight sky. Also, the object would become visible to about half of the daylight side of the Earth during total solar eclipses, when the Sun's light is blocked. If the companion were at L4, visibility would be as an "evening star", descending below the horizon shortly after sunset.)

**Big Enough To See?**

Let's say Earth had a Lagrangian partner orbiting at the L5 point—let's call it Danu. How easy would it be for inhabitants of Danu to see Earth in their sky? Using the equation for the small-angle approximation:

Distance between Earth and Danu: 1.496 × 10⁸ km

Actual diameter of Earth: 12742 km

Distance between Earth and Danu: 1.496 × 10⁸ km

Actual diameter of Earth: 12742 km

…a very tiny 0.000085 radians. To convert the answer to degrees:

The average apparent diameter of the Moon in Earth’s sky is 0.51784°, so we find that Earth would have a size in Danu's sky which is

*0.51784° ÷ 0.0049° ≈ 105.68*times smaller than the size of the full Moon as seen from Earth —and that would be for a body the size of Earth!In fact, any object at Earth's L4or L5point would have to be the size of the Sun to appear as large in the sky as the full Moon, because that object would be as far from the Earth as is the Sun.

Calculating the apparent brightness of our Earth as seen from Danu:

Earth’s albedo: 0.306

Luminosity of the Sun: 3.828 × 10²⁶ watts

Radius of the Earth: 6.374 × 10⁶ meters

Distance from the Earth to the Sun: 1.496 × 10¹¹ meters

Distance from the Earth to Danu: 1.496 × 10¹¹ meters

Earth’s albedo: 0.306

Luminosity of the Sun: 3.828 × 10²⁶ watts

Radius of the Earth: 6.374 × 10⁶ meters

Distance from the Earth to the Sun: 1.496 × 10¹¹ meters

Distance from the Earth to Danu: 1.496 × 10¹¹ meters

… shows that Earth would have an apparent brightness of ≈ 0.0000095543 watts/m², which is ≈ 0.02% the brightness of the full Moon, or a little more than ⅟7 the brightness of Venus. It would be 3.5 times brighter than Jupiter and 72 times brighter than Saturn, but it would be invisible in a daytime sky. Were it to be visible, it would be more likely to be noticed than either Jupiter or Saturn, because its position in the sky would never change. Were it to be visible at night, its position relative to the background stars would change on a nightly basis by 0.986°, almost twice the width of the full Moon. |

If such a twin had existed in Earth's orbit, it would surely have been noted by the ancients (who did observe the sky during total solar eclipses), and its special nature would have been quickly realized: though it would only appear during an eclipse, its orientation with respect to the Sun in the sky would never change. Once it came under observation using space-based telescopes, it would always be lit as a waning crescent, though its brightness would likely change due to its axial rotation, differing cloud patterns, and whether land or ocean was visible in the daylit portion (assuming it had oceans).

**Double-Planets**

There is currently no official definition of what constitutes a double (or binary) planet. (But there is for

One distinction that would certainly make sense is that both bodies orbit a common barycenter; this at least distinguishes them from co-orbital bodies such as Lagrangian pairs or orbit-exchanging pairs such as Saturn’s moons Epimetheus and Janus <5>.

Some suggested (and generally accepted) criteria for double-planets include:

*dwarf*planets … don’t get me started.)One distinction that would certainly make sense is that both bodies orbit a common barycenter; this at least distinguishes them from co-orbital bodies such as Lagrangian pairs or orbit-exchanging pairs such as Saturn’s moons Epimetheus and Janus <5>.

Some suggested (and generally accepted) criteria for double-planets include:

- Both bodies must qualify individually as planets, per the 2006 IAU definition <6>;
- Both orbit a common center-of-gravity (barycenter), which itself orbits their host star(s) <7>;
- The barycenter of their combined orbits
*must lie outside the radius of both bodies*<8>; - The ratio of their masses must approach a value of 1.0 <9>.

Where:

For simplicity, we’ll express

*α*= the average separation between the two bodies;*d**p*= average distance from the primary to the barycenter (in the same units as*a*);*M*= mass of the primary body;*m*= mass of the secondary body (in the same units as*M*)For simplicity, we’ll express

*α*in terms of the radius of the primary body, so that if the value returned for*d**p*≤ 1.0, then we’ll know that the barycenter lies within the physical volume of the primary.**Example: The Earth And The Moon**

Mass of the Earth: 1.0 terran

Mass of the Moon: 0.0123 terran

Mass of the Moon: 0.0123 terran

*α**= 60.366*Earth-radii… reveals that, because

What would the mass of the Moon need to be at its current distance for the barycenter to lie outside Earth’s radius?

Rearranging the above equation, we can find the mass of the secondary body based on the location of the barycenter and the distance between the bodies:

*d**p*< 1.0, the barycenter for the Earth-Moon pair lies within the radius of Earth, and so, technically, this pair cannot be defined as a double-planet. Note also that the ratio of their masses doesn’t even begin to approach 1.0—it is 0.0123.What would the mass of the Moon need to be at its current distance for the barycenter to lie outside Earth’s radius?

Rearranging the above equation, we can find the mass of the secondary body based on the location of the barycenter and the distance between the bodies:

To answer the current questions, we would set

*d**p**= 1.0*, meaning that the barycenter of the Earth-Moon system would be located precisely at the surface of the Earth. The math then becomes:… so, the Moon would have to be

If we leave the Moon’s mass as it is, at what distance would it have to orbit to pull the system barycenter outside of Earth’s radius?

Again, rearranging the equation, we can calculate the necessary value of a when we want

*at least*1.369 times more massive at its current distance for the barycenter of the system to be 1.0 Earth-radius from the center of the Earth. Bear in mind, this means that 0.0168 terran is the*minimum*mass for the Moon at this distance; any mass larger than 0.0168 terran will ensure that the barycenter of the system always falls outside the radius of the Earth.If we leave the Moon’s mass as it is, at what distance would it have to orbit to pull the system barycenter outside of Earth’s radius?

Again, rearranging the equation, we can calculate the necessary value of a when we want

*r**P*to be*at least*1.0 Earth radius:Here,

*r**P**= 1.0*, so the equation simplifies to:… where

*α*is expressed in Earth-radii. Thus, the Moon would have to orbit*at least*1.363 times farther away than currently to ensure that the system barycenter always lies outside Earth’s radius.The two ratios are both close to1.37 : 1—the difference between them is due to rounding errors in calculations.

There is an argument that the Earth and Moon

*do*constitute a double-planet….**The "Tug-Of-War" Ratio**

In 1963, Isaac Asimov <10> suggested that a ratio (which he called a “tug‑of‑war" value) can be calculated to determine whether the star or the primary body has more gravitational effect on the satellite: Where: FT = tug-of-war force value;Pm = mass of the primary in solar units;Sm = mass of the star in solar units;D = distance between the smaller body and the star in astronomical units;d = distance between the primary and the smaller body in astronomical units |

The result is interpreted as follows: the larger the number, the greater the ratio of the gravitational pull of the primary on the secondary as compared to the pull of the star. The magnitude of

*F**T*, then, determines whether or not the pair are a double planet. Below is a table listing the tug-of-war values for various planet-moon pairs: The Earth-Moon figure is interesting: it is less than 1.0, which means that the Sun actually has a stronger gravitational pull on the Moon than does the Earth. That means—by Asimov's reckoning—that the Moon is not a satellite of the Earth, but a companion planet (even though we showed earlier that they are not a double-planet system based on the location of their barycenter—who says astronomy is an exact science?).In Asimov’s words the Moon is, “…neither a true satellite of the Earth nor a captured one, but is a planet in its own right, moving about the Sun in careful step with the Earth.” <11> Of course, by modern IAU standards, the Moon would have to be called a dwarf planet…. Asimov’s argument is further strengthened by the fact that at no point in its journey around the Sun does the Moon ever travel “backward” relative to the Sun: "It is always ‘falling toward’ the Sun. All the other satellites, without exception, ‘fall away’ from the Sun through part of their orbits, caught as they are by the superior pull of their primary planets – but not the Moon.” <12> In other words, all of the other satellites in the Solar system have loops in their orbit which, when viewed from above the plane of the Solar system, means that for part of their orbits they travel in the opposite direction of their primary’s orbital motion. The Moon’s orbit does not have any such loops, because the speed at which the pair are orbiting the Sun is faster than any "backward" velocity the Moon has in its orbit when it is between the Earth and the Sun. A very helpful explanation of this is given by Helmer Aslaksen at his website Heavenly Mathematics & Cultural Astronomy <13>. |

What happens to the Earth-Moon tug-of-war value if we increase the Moon’s mass to 0.0168 terran as we did above to ensure a barycenter outside of Earth’s radius?

Our known values are:

… which generates a tug-of-war value of:

Our known values are:

*M**p**= 3.003 × 10*̄⁶ solar;*M**s**= 1.000*solar;*d**s**= 1.0*AU;*d**p**= (60.366 × 6371*km*) ÷ 1.496 × 10*⁸ km*= 2.571 × 10*̄³… which generates a tug-of-war value of:

… which is not appreciably different than the value given above.

**Variation In The Barycenter**

It can happen that the barycenter migrates between lesser-than and greater-than the radius of the primary. Applying the following relation can reveal if this is the case:

Where:

If this relation evaluates to true, then the barycenter may sometimes migrate between less-than and greater than the physical volume of the primary body; otherwise, it always remains either within or outside the body of the primary, as calculated using the equation above.

*r**P*= the average distance from the primary to the barycenter (in terms of the radius of the primary);*R**P*= the radius of the primary;*e*= the eccentricity of their orbitsIf this relation evaluates to true, then the barycenter may sometimes migrate between less-than and greater than the physical volume of the primary body; otherwise, it always remains either within or outside the body of the primary, as calculated using the equation above.

*r**P**= 0.73312*terran*R**P**= 1.0*terran*e**= 0.0549*The second term is less than the third but the first term is not less than the second term, so the relation as a whole does not evaluate as true; thus, the Earth-Moon barycenter

So, is the Moon a satellite of the Earth, or a companion planet?

I suggest the following criteria to help answer the question:

*never*lies outside the body of the Earth. This argument can also support the contention that the pair do not constitute a double planet.So, is the Moon a satellite of the Earth, or a companion planet?

I suggest the following criteria to help answer the question:

* The mass of the secondary body must be < ¹/₁₀₀th the mass of the primary in order to be considered a moon, and ≥ ¹/₁₀₀th the mass of the primary in order to be considered a companion planet.

For the Earth-Moon system:

Tug-of-war Ratio (

Mass Ratio (

Barycenter Locus (

Barycenter Motility (

Tug-of-war Ratio (

*F**T**):*0.455Mass Ratio (

*f*): 81.3 : 1 (1 : 0.0123)Barycenter Locus (

*r**P*): InteriorBarycenter Motility (

*σ*): FixedTherefore, by these criteria, the Earth-Moon system

is a Class 1 double-planet system.

We can use the left two terms from above to determine the eccentricity necessary for the barycenter motility relation to always evaluate to true:

… which in the case of the Earth and Moon works out to:

So, the Earth-Moon system would need an orbital eccentricity of

We can also rearrange the left two terms, thus:

*greater-than*0.364 to ensure that the barycenter motility expression always evaluates to true.We can also rearrange the left two terms, thus:

… to calculate the distance (in terms of the radius of the primary) which is necessary to ensure that the barycenter motility relation always evaluates to true for a given eccentricity value. For the Earth-Moon, this distance is:

… which tells us that the barycenter of the Earth-Moon system cannot lie farther than 94.8% of the distance from the center of the Earth to its surface; otherwise, the barycenter motility relation will

Keeping the masses of Earth and Moon the same, the distance the Moon would have to orbit to make

*not*evaluate true.

Keeping the masses of Earth and Moon the same, the distance the Moon would have to orbit to make

*r**p**= 0.948*would be:… about 1.292 times farther out than currently.

The only other significant example in the Solar system is the Pluto-Charon system. Ignoring for the moment the fact that Pluto is not a "full-fledged" planet — their mass ratio is 8.197 : 1.

Their tug-of-war value is:

The only other significant example in the Solar system is the Pluto-Charon system. Ignoring for the moment the fact that Pluto is not a "full-fledged" planet — their mass ratio is 8.197 : 1.

Their tug-of-war value is:

… which indicates that Pluto has a stronger gravitational effect on Charon than does the Sun.

Their barycenter locus is:

Their barycenter locus is:

*M**P**= 1.0**M**S**= 0.122**a*=*19591km ÷ 1188.3*km*= 16.4886*Their barycenter motility calculates as:

*e**= 0.0002**r**P**= 1.793**R**p**= 1.000*… showing that the barycenter of the system consistently remains outside the physical volume of Pluto.

Thus, by our criteria above, Pluto-Charon constitutes a Class 2 planet-moon system.

Thus, by our criteria above, Pluto-Charon constitutes a Class 2 planet-moon system.

**Maximum Possible Separation**

Though I have previously discussed calculating minimum and maximum orbital separations, here I wish to discuss the maximum possible separation for a double-planet pair.

The following will be true:

The maximum possible separation can be calculated by the equation:

The following will be true:

- The more massive of the pair is the primary;
- They will orbit a barycenter which, itself, will orbit the central star(s);
- The barycenter of the system will lie outside the radius of either body;
- Their orbital eccentricities around the barycenter will be identical;
- They may be closely or widely spaced, limited by the calculation below.

The maximum possible separation can be calculated by the equation:

Where:

This is related to the Hill sphere; if the two bodies are farther apart than the value of

For a double-planet system of two Earth-mass bodies at 1.0 AU from the Sun:

*D**m*= maximum separation in astronomical units;*M*= mass of the primary in solar masses;*m*= mass of the secondary in solar masses;*S*= mass of the star in solar masses;*R*= distance between the star and the barycenter in astronomical unitsThis is related to the Hill sphere; if the two bodies are farther apart than the value of

*D**m*, then they escape one another’s gravity and orbit the central mass independently (or escape the star system, altogether).For a double-planet system of two Earth-mass bodies at 1.0 AU from the Sun:

…

At a distance of 902.8 million meters, a twin Earth would subtend 0.81° in the sky (about 1.6 times the size of the full Moon); it would orbit ≈ 2.35 times farther away than the Moon orbits the Earth, and have an orbital period of

≈ 70.132 days.

*D**m*is ≈ 0.0061 AU, or ≈ 902.8 million meters; just over 0.6 times Earth’s Hill sphere radius, or ≈ 141.7 times the radius of the Earth. This means that two Earth-mass bodies in orbit around one another would have to remain closer than 902.8 million meters or the gravity of the Sun would overpower their gravitational attraction on one another. They would then orbit the Sun independently as individual bodies.

At a distance of 902.8 million meters, a twin Earth would subtend 0.81° in the sky (about 1.6 times the size of the full Moon); it would orbit ≈ 2.35 times farther away than the Moon orbits the Earth, and have an orbital period of

≈ 70.132 days.

The inner limit for such a system would, of course, be the Roche limit, which for two Earth-density objects works out to 2.44 Earth radii. At a distance of 2.44 Earth radii, using the small angle approximation:

Radius of the Earth:

Distance to twin:

Radius of the Earth:

*6731 km (12742 km in diameter)*Distance to twin:

*6731*⨉*2.44 = 15545.24*km… we calculate that the other Earth would subtend ≈ 46.964° in the sky, or 92 times bigger in apparent size than the full Moon! However, these twin Earths would be tidally locked, each always showing the same hemisphere to the other, and they’d orbit one another in ≈ 3.8 hours (see below for how to calculate this). They’d also be at one another’s Roche limits, so if they moved

*any*closer to each other, they would start to break up from gravitational tidal stresses.**Orbital Period For Double-Planets**

The orbital period of double planets—as that of moons—can be found by using the following equation:

Where:

𝜒

𝜒

Bear in mind that all of the units must be in Earth units for the equation to give results in Earth time units. Thus,

*P**S*= period of the system’s orbit;*R**S*= radius of the secondary’s orbit in Earth radii;*M*= mass of the primary in Earth masses;*m*= mass of the secondary in Earth masses;𝜒

*= 0.0588*to have*P**S*expressed in days;𝜒

*= 1.4112*to have*P**S*expressed in hoursBear in mind that all of the units must be in Earth units for the equation to give results in Earth time units. Thus,

*M**P*must be the actual mass of the planet divided by Earth’s actual mass;*M**S*must be the actual mass of the secondary divided by Earth’s actual mass; and*R**S*must be the actual radius of the secondary’s orbit divided by the actual radius of the Earth.**Example**

Let’s specify a double-planet system, Koryn and Kor.

Everything must be in Earth units, so we divide each mass by Earth’s mass in grams (5.972⨉10²⁷ g), and the distance by Earth’s radius in km (6371 km)

We want the orbital period in Earth days, so we’ll use 𝜒

*= 0.0588*:… we obtain an orbital period of just over 7 days and 16 hours.

**Orbital Speed For Double-Planets**

Should there be a need to discover the orbital speed of the members of a double-planet system, it is sadly not as straightforward as finding the orbital speed of a planet around a star.

As the website MathIsFun.com states: “Rather strangely, the perimeter of an ellipse is very difficult to calculate!” <14> They list several approximations, but I’ve chosen to go with the first one they list, which has as its primary limitation the requirement that the semi-minor axis of the ellipse be no less than one-third the length of the semi-major axis.

Since a 3 : 1 ratio between the semi-major and semi-minor axes results in an eccentricity of 0.94281, and that is an

The following is an equation for calculating the approximate perimeter of an ellipse:

As the website MathIsFun.com states: “Rather strangely, the perimeter of an ellipse is very difficult to calculate!” <14> They list several approximations, but I’ve chosen to go with the first one they list, which has as its primary limitation the requirement that the semi-minor axis of the ellipse be no less than one-third the length of the semi-major axis.

Since a 3 : 1 ratio between the semi-major and semi-minor axes results in an eccentricity of 0.94281, and that is an

*extremely high*eccentricity for an elliptical orbit (the highest eccentricity of any known moon in the Solar system is that of Nereid, at 0.75 <15>), I consider that the inaccuracies of using this method are acceptable. (If for some reason you have a situation where such an insanely high eccentricity exists in your double-planet or planet-moon pairing, please see the mathisfun.com site for more accurate approximations.)The following is an equation for calculating the approximate perimeter of an ellipse:

Where:

Notice that I have not specified units for

Now that you know the distance your object is traveling, use the next equation to calculate the approximate orbital speed:

*p*= the approximate perimeter of the ellipse;*a*= the semi-major axis of the ellipse;*b*= the semi-minor axis of the ellipse

Notice that I have not specified units for

*a*or*b*; this is because the units used will be determined by the desired units of the ultimate answer. If you are using days for the orbital period and you use meters for the axes, your answer will be in meters per day. If you want kilometers per second, it will be necessary to convert the orbital period in days into seconds and use kilometers for the axes.Now that you know the distance your object is traveling, use the next equation to calculate the approximate orbital speed:

Where:

Let's crunch the numbers for the Moon.

The semi-major axis of the Moon’s orbit in kilometers is 3.84399 × 10⁸ meters. The eccentricity of the Moon’s orbit is

*v**O*= the approximate orbital speed;*P**S*= the orbital period calculated above;*p*= the approximate perimeter of the elliptical orbitLet's crunch the numbers for the Moon.

The semi-major axis of the Moon’s orbit in kilometers is 3.84399 × 10⁸ meters. The eccentricity of the Moon’s orbit is

*e**= 0.0549*. The semi-minor axis of the Moon’s orbit calculates as:The approximate perimeter of the Moon’s orbit:

*a*=*3.843*⨉*10*⁸ meters*b**= 3.838*⨉*10*⁸ meters… just over 2.4 billion meters.

The orbital period for the Moon is calculated as:

Radius of the Moon’s orbit in Earth-radii: 60.336

Mass of the Moon in Earth-masses: 0.0123

The orbital period for the Moon is calculated as:

Radius of the Moon’s orbit in Earth-radii: 60.336

Mass of the Moon in Earth-masses: 0.0123

… represented in Earth hours. Converted to seconds, this becomes: 2.3665⨉10⁶ seconds.

Finally:

Finally:

… just less than 1020 meters-per-second. Converted to kilometers, this is 1.020 km/sec, which is very close to the official value of 1.022 km/sec <16>.

## Notes

- "Klemperer Rosette," Wikipedia, July 21, 2018, , accessed July 26, 2018, https://en.wikipedia.org/wiki/Klemperer_rosette.
- "Building the Ultimate Solar system," PLANETPLANET, April 04, 2018, , accessed July 26, 2018, https://planetplanet.net/2014/05/13/building-the-ultimate-solar-system/comment-page-1/.
- "2010 TK7," Wikipedia, July 16, 2018, , accessed July 26, 2018, https://en.wikipedia.org/wiki/2010_TK7.
- Tony Greicius, "Wide-field Infrared Survey Explorer (WISE)," NASA, February 18, 2015, , accessed July 26, 2018, https://www.nasa.gov/mission_pages/WISE/main/index.html.
- "Epimetheus (moon)," Wikipedia, July 24, 2018, , accessed July 26, 2018, https://en.wikipedia.org/wiki/Epimetheus_(moon).
- "International Astronomical Union," IAU, , accessed July 26, 2018, https://www.iau.org/news/pressreleases/detail/iau0603/.
- "Double Planet," Wikipedia, July 27, 2018, , accessed July 27, 2018, https://en.wikipedia.org/wiki/Double_planet.
- Ibid.
- Ibid.
- Isaac Asimov,
*Asimov On Astronomy*(New York: Crown Publishers, 1976), pp 125-139. - "Double Planet," Wikipedia, July 27, 2018, , accessed July 27, 2018, https://en.wikipedia.org/wiki/Double_planet.
- Ibid.
- Helmer Aslaksen, "The Orbit of the Moon around the Sun Is Convex!" Stonehenge and the Sun, , accessed July 27, 2018, http://www.math.nus.edu.sg/aslaksen/teaching/convex.html.
- "Perimeter of an Ellipse," Definition of, , accessed July 27, 2018, http://www.mathsisfun.com/geometry/ellipse-perimeter.html.
- "Appendix 1a: Solar system Data," Nine Planets, , accessed July 27, 2018, http://nineplanets.org/data.html.
- "Moon," Wikipedia, July 25, 2018, , accessed July 27, 2018, https://en.wikipedia.org/wiki/Moon.