Encyclopedia of the Solar System 2nd ed

Kuiper Belt Objects: Physical Studies 613

example, the KBO Varuna hasm=0.42±0.02 magni-
tude. If we assume we are seeing Varuna with an “equator-
on” aspect (i.e., an aspect angle of 90◦), which corresponds
to the largest possible lightcurve amplitude, we can relate
the amplitude of the lightcurve to an axial ratio,

m= 2 .5 log

a

b

.

Such an assumption gives a / b=1.5 for Varuna. Since we
don’t know if they are viewing Varuna “equator-on,”m
and a / b are lower limits. At the present time, all KBO axial
ratios are lower limits.

8.4 Density

Besides periods of rotation and shapes, lightcurves allow us
to estimate densities for KBOs and Centaurs. If nonspheri-
cal shapes are the result of rotational deformation, then we
can use the formalism developed by Chandrasekhar that
relates the period of rotation and shape to the density of a
strengthless ellipsoid. Application of Chandrasekhar’s for-
malism to Pholus, the object with the best shape estimation
(a/b=1.9 and c/b=0.9) and period of rotation (Prot=
9.980 hr), gives an average density ofρavg=0.5gcm−^3 .It
is interesting to note that the similar-sized Saturnian satel-
lites Janus, Epimetheus, Prometheus, and Pandora have
average densities of 0.61, 0.64, 0.42, and 0.52 g cm−^3 .In
the case of KBO Varuna, its assumed shape (a / b=1.5 and
c/b=0.7) and period of rotation (Prot=6.3442 hr) yield
ρavg=1.0 g cm−^3. Average densities≤1gcm−^3 suggest
that KBOs and Centaurs likely have ice-rich and porous
interiors.

8.5 Porosity

Porosity is the fraction of void space in a KBO or Centaur.
If a KBO is some mixture of ice, refractory material (dust),
and empty space, the average density of a KBO is given
by

ρavg=fiρi+frρr,

wherefiandfrare the fractional volumes occupied by icy
and refractory material andρiandρrare the densities of
icy and refractory material. In addition, the sum of the parts
must equal the whole, so

fi+fr+fv= 1 ,

where fvis the fraction of void space or the porosity. The
fraction of total mass locked up in refractories,ψ, is given
by

ψ=

ρrfr

ρrfr+ρifi

.

By combining the above three equations, it is possible to

obtain an algebraic expression for the porosity of a KBO or

Centaur,

fv= 1 −

ρavg

ρi

[

1 +ψ

(

ρi

ρr

− 1

)]

.

Assuming reasonable values ofρavg=1gcm−^3 ,ρi=1g

cm−^3 ,ρr=2gcm−^3 , andψ=0.5 for Varuna, Jewitt and

Sheppard estimate that Varuna has a porosity∼0.25. For

comparison, beach sand has fv∼0.4 and basaltic lunar

regolith has 0.4<fv< 0 .7.

9. Composition

9.1 Surface Color

An early expectation was that all KBOs should exhibit a sim-

ilar red surface color. Why? Initially, KBOs were thought to

form over a small range of heliocentric distances where the

temperature in the young solar nebula was the same. The

similar temperature suggested that KBOs formed out of

the solar nebula with the same mixture of molecular ices

and the same ratio of dust to icy material. In addition,

their similar formation distance from the Sun suggested

that KBOs should experience a similar evolution. Specifi-

cally, the irradiation of surface CH 4 ice by solar ultraviolet

light and solar wind particles should have converted some

surface CH 4 ice into red, complex, organic molecules. By

their nature, the complex organic molecules were expected

to absorb more incident blue sunlight than red sunlight.

Therefore, the light reflected from the surfaces of KBOs

was expected to consist of a larger ratio of red to blue light

than the incident sunlight. It was a surprise to find KBOs

exhibit a range of surface colors rather than just red colors.

At one extreme, some KBOs reflect sunlight equally at all

wavelengths (i.e., exhibit neutral or gray surface colors). On

the Johnson-Kron-Cousins photometric system such KBOs

have B-R=1.0. At the other extreme, some KBO have

extraordinary red colors, i.e., B-R=2.0.

Because it’s a painstaking process to measure the color

of a KBO or Centaur, taking as much as three hours of tele-

scope time to obtain an accurate color for a single object, the

first color surveys consisted of only 10 to 20 objects. These

small samples lumped KBOs and Centaurs together and

resulted in a controversy. Some groups found their samples

to exhibit a uniform distribution of colors from gray (B-R

=1.0) to extraordinarily red (B-R=2.0). These groups

suggested that KBOs and Centaurs experienced a steady

reddening of their surfaces by solar radiation, and occa-

sional impacts by smaller objects punctured the red sur-

faces and excavated gray, interior material. Such a radiation-

reddening and impact-graying mechanism would explain

the uniform distribution of colors. Another group found