## 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.