Of the 793,918 asteroids and trans-Neptunian objects (TNOs) currently catalogued, only 98 are in retrograde orbits around the Sun. That’s just 0.01%.

By “retrograde” we mean that the object orbits the Sun in the opposite sense of all the major planets: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune. From a vantage point above the north pole of the Earth, all of the major planets orbit in a counterclockwise direction around the Sun.

But a retrograde object would be seen to orbit in a clockwise direction around the Sun, as is shown in the animation below for Jupiter retrograde co-orbital asteroid 514107 (2015 BZ509), with respect to Jupiter and its two “clouds” of trojan asteroids.

Of these 98 retrograde objects, only 14 have orbits well-enough determined to have received a minor planet number, and only one has yet received an official name (20461 Dioretsa).

 Semimajor Axis (a) between… Number of Retrograde Minor Planets Mars – Jupiter 3 Jupiter – Saturn* 20 Saturn – Uranus* 15 Uranus – Neptune* 20 TNOs 40

*asteroids between the orbits of Jupiter and Neptune are often referred to as centaurs

At least some of these objects may be captured interstellar objects.

Let’s now take a look at some of these 98 retrograde objects in greater detail.

20461 Dioretsa
The first retrograde asteroid to be discovered was 20461 Dioretsa, in 1999. The only named retrograde asteroid to date, Dioretsa is an anadrome of the word “asteroid”. It is a centaur in a highly eccentric orbit (0.90), ranging between the orbits of Mars and Jupiter out to beyond the orbit of Neptune. Objects in cometlike orbits that show no evidence of cometary activity are often referred to as damocloids. Dioretsa is both a centaur and a damocloid. Its orbital inclination (relative to the ecliptic) is 160°, which is a 20° tilt from an anti-ecliptic orbit. It takes nearly 117 years to orbit the Sun once. It is a dark object with a reflectivity only around 3% and is estimated to be about 9 miles across.

2010 EQ169
This retrograde asteroid holds the distinction (at least temporarily) of being the most highly-inclined main-belt asteroid (91.6°), relative to the ecliptic plane. It is also the retrograde asteroid with the smallest semimajor axis (2.05 AU) and lowest orbital eccentricity (0.10). Unfortunately, it was discovered after the fact by analyzing past data from the Wide-field Infrared Survey Explorer (WISE) space telescope, and has not been seen since. We have only a three-day arc of 17 astrometric observations of 2010 EQ169 between March 7-9, 2010 from which to determine its orbit. Nominally, 2010 EQ169 orbits the Sun at nearly a right angle to the ecliptic plane once every 2.9 years, between the orbits of Mars and Jupiter. However, our knowledge of its orbit is extremely uncertain, as shown below, and it has been lost. Our only hope will be to back-calculate the positions of future asteroids discovered to these dates to see if it matches the WISE positions.

 Element Value 1σ Uncertainty Inclination (i) 91.606° 18.177° Semimajor Axis (a) 2.0518 AU 2.2176 Orbital Eccentricity (e) 0.10153 0.90213 Orbital Period (P) 2.94y 4.765

2013 BL76
This retrograde TNO has the largest known semi-major axis of any of the retrograde non-cometary objects: 966.4274 ± 2.2149 AU. In a highly eccentric cometlike orbit (e = 0.99135), its perihelion is in the realm of the centaurs between the orbits of Jupiter and Saturn (8.35 AU), and its aphelion is way out around 1,924 AU. It takes about 30,000 years to orbit the Sun. Its orbit is inclined 98.6° with respect to the ecliptic.

2013 LA2
This retrograde centaur is in an orbit closest to the ecliptic plane (i = 175.2°), tilted 4.8° with respect to the ecliptic. It orbits the Sun about once every 21 years between the orbits of Mars and Uranus.

2017 UX51
The distinction for this retrograde TNO is that it has the highest orbital eccentricity of any non-cometary solar system object (e = 0.9967). Or is it an old inactive comet? 2017 UX51 orbits the Sun every 7,419 ± 2,883 years as close in as between the orbits of Earth and Mars (perihelion q = 1.24 AU)—classifying it as an Amor object—out to far beyond the orbit of Neptune (aphelion Q = 759.54 ± 196.77 AU). Its orbital inclination is 108.2°.

343158 (2009 HC82)
An Apollo asteroid, 343158 is the only known retrograde near-Earth asteroid (NEA), with an orbital inclination of 154.4°. It orbits the Sun every 4.0 years, between 0.49 AU (almost as close in as the aphelion of Mercury) out to 4.57 AU (between the orbits of Mars and Jupiter).

References
Conover, E., 2017. Science News, 191, 9, 5.

JPL Small-Body Database Browser, https://ssd.jpl.nasa.gov/sbdb.cgi, retrieved 31 March 2019.

Kankiewicz, P., Włodarczyk, I., 2018. Planetary and Space Science, 154, 72-76.

Minor Planet Center, https://minorplanetcenter.net/iau/MPCORB.html, retrieved 28 March 2019.

Namouni F., Morais M. H. M., 2018. MNRAS, 477, L117.

Wiegert, P., Connors, M., Veillet, C., 2017. Nature, 543, 687–689.

## Effective Diameter of an Irregularly-Shaped Object

A diameter of a circle in 2D is defined as any straight line segment that intersects the center of the circle with endpoints that lie on the circle.  Since all diameters of a circle have the same length, the diameter is the length of any diameter.

Likewise, a diameter of a sphere in 3D is defined as any straight line segment that intersects the center of the sphere with endpoints that lie on the surface of the sphere, and the diameter is its associated length.

But how do we define the diameter of an irregularly-shaped object such as a typical asteroid or trans-Neptunian object?

For a well-characterized object such as 951 Gaspra—the first asteroid to be photographed up close by a spacecraft—we’ll see the dimensions of the best fitting triaxial ellipsoid given in terms of “principal diameters”.  In the case of Gaspra, that is 18.2 × 10.5 × 8.9 km.

In certain circumstances, however, it would advantageous to characterize an irregularly-shaped object using a single “mean diameter”.  How should we calculate that?

There are two good approaches, provided you have enough information about the object.  The first is to determine the “volume equivalent diameter” which is the diameter of a sphere having the same volume as the asteroid.  This is particularly relevant to mass and density.

For purposes of illustration only, let’s assume Gaspra’s dimensions are exactly the same as its best-fitting triaxial ellipsoid.  If that were true, the volume of Gaspra would be

$V = \frac{{4\pi abc }}{3}$

where V is the volume, and a, b, and c are the principal radii of the triaxial ellipsoid.

Plugging in the numbers 9.1 km, 5.25 km, and 4.45 km (half the principal diameters), we get a volume of 890.5 km3.

The volume equivalent diameter is

$d_{vol} = \left (\frac{6V_{obj}}{\pi } \right )^{1/3}$

where dvol is the volume equivalent diameter, and Vobj is the volume of the object.

Plugging in the volume of 890.5 km3 gives us a volume equivalent diameter of 11.9 km.

The second approach is to determine the “surface equivalent diameter” which is the diameter of a sphere having the same surface area as the asteroid.  This is most relevant to reflectivity or brightness.

Once again using our triaxial ellipsoid as a stand-in for the real 951 Gaspra, we find that the general solution for the surface area of an ellipsoid requires the use of elliptic integrals.  However, there is an approximation that is more straightforward to calculate and accurate to within about 1%:

$S\approx 4\pi\left ( \frac{a^{p}b^{p}+a^{p}c^{p}+b^{p}c^{p}}{3} \right )^{1/p}$

where S is the surface area, p ≈ 1.6075 can be used, and a, b, and c are the principal radii of the triaxial ellipsoid.

Once again plugging in the numbers, we get a surface area of of 478.5 km2.

The surface equivalent diameter is

$d_{sur} = \left (\frac{S_{obj}}{\pi } \right )^{1/2}$

where dsur is the surface equivalent diameter, and Sobj is the surface area of the object.

Plugging in the surface area of 478.5 km3 gives us a surface equivalent diameter of 12.3 km.

You’ll notice that the surface equivalent diameter for 951 Gaspra (triaxial ellipsoid approximation) is 12.3 km which is larger than the volume equivalent diameter of 11.9 km.  The surface equivalent diameter is apparently always larger than the volume equivalent diameter, though I leave it as an exercise for the mathematically-inclined reader to prove that this is so.

References
Herald, David (2018, October 23).  [Online forum comment].  Message
posted to https://groups.yahoo.com/neo/groups/IOTAoccultations/conversations/messages/65158

Thomas, P.C., Veverka, J., Simonelli, D., et al.: 1994, Icarus 107The Shape of Gaspra, 23-26.