Expert: Philip Stahl - 8/27/2011

Question
I was reading a book online called Comets II and the following is a calculation of how much mass is lost from a comet at the distance of Earth's orbit. Volatile Mass loss rate of m ~ 10–4 kg m–2 s–1 at R = 1 AU. I thought you might be able to decipher this equation for me.
I am trying to figure out if the following paragraph is about right: (This was online somewhere)"For comets that travel within an astronomical unit of the Sun, this evaporation rate can reach as high as 1030 molecules per second—about 30 tons of cometary material lost for every second the comet spends near the Sun (within Earth's orbit, say)"

Also, what do you know about sizes of comet nuclei?
Thanks so much.

Answer
Hello,

The quantitative mass loss rate to which you referred is actually not an equation per se, but a number specifying the rate of mass lost at the Earth's mean distance from the Sun (1 AU or 1.49 x 10^11 m).

Thus, it means:

10^-4 or 0.0001 kg (e.g. 0.1 gram) lost per square meter (m^-2) per second, at that distance (1 AU)


In effect, the mass loss rate is being parsed based on the surface area of comet. The bigger its radius, the greater the surface area (say 4 pi r^2 if assumed spherical) so the larger the mass lost. If say, a comet spent a whole day (or 86,400 s) at around R = 1 AU then the mass lost per m^2  for a 10km wide comet nucleus would be:

(0.0001 kg) x 86 400 x (4 pi x 5 x 10^3 m^2) = 2.7 x 10^9 kg

or about 2.7 million metric tons

Which is not that much mass, given a typical comet has a total mass ~ 10^13 kg

Now as to your puzzling over that paragraph, let's see:

An evaporation rate as high as 10^30 molecules per sec or about 30 tons of cometary material lost per second.

Check this claim against the previous 'back of the hand' computation I just did, yielding 2.7 x 10^9 kg lost per m^2 over *86 400 secs* for a 10 km dia. comet.

To get the mass lost per m^2 sec, divide by 86400s:

(2.7 x 10^9 kg/m^2 s^-1)/86400 s = 31250 kg/m^2 lost per second

That is in kilograms.

Dividing that by 1000 (kg/tonne) yields metric tons:

= 31.2 metric tons

which is pretty good as an estimate and a fair approximation to what your selected paragraph claims.

Re: sizes of cometary nuclei, I recall reading someplace (can't recall where now) that 10 km is a median nucleus diameter, which is why I used it in the example I did.