# Asteroid Cooling

1. Jun 10, 2008

### Kreizhn

1. The problem statement, all variables and given/known data
Consider an asteroid with an iron core $\rho_m = 8000 kg\; m^{-3}$ covered by a thin silicate mantle $\rho_m = 3500 kg\; m^{-3}$ with a thickness of 20% the raidus R of the asteroid.

Assume that the internal temp is $T_i = 600K$ and is constant throughout the core due to high thermal conductivity of iron. Take the thermal energy in the core to be $3k T_i$ per atom and assume that the thermal conductivity is $k_c = 2 W\; m^{-1} K^{-1}$. Ignore the heat capacity of of the mantle. If the surface has a temp of $T_s = 200K$, find the value of R for which the cooling rate is about 1 K per million years

2. Relevant equations
$$\frac{dQ}{dt} = -k_c A \frac{dT}{dx}$$
A is the surface area

3. The attempt at a solution

There's a few things that are confusing me right off the bat. What is the k is the thermal energy of the core $3k T_i$? There is no reference to it in any of the literature, so I find it rather ambiguous which doesn't help my situation.

I started by saying that $\frac{dT}{dx} = \frac{ T_i - T_s}{0.8 R}$ in an approximating sense, using that the iron core is 80% of the asteroid. Then I assumed a spherical asteroid and subbed this into the above equation to get

$$\frac{dQ}{dt} = 5 \pi R k_c (T_i - T_s)$$

Now I'm kinda stuck. I want to say something about the heat production like

$$\frac{dQ}{dt} = ML$$ where M is the mass and L is the energy production, however I'm not sure how to find energy production (something to do with the $3k T_i$?). And then where does the 1K per million years come in?

2. Jun 10, 2008

### rl.bhat

k is the Boltzmann constant = 1.38*10^-23J/K. By using the molecular weights of iron and silicate and their masses find the number of atoms of them. That will give you the energy content of them.

3. Jun 11, 2008

### dynamicsolo

Admittedly, this can be confusing when you're first learning how people notate these quantities. And, unfortunately, the letter k is somewhat overused (since it comes from konstant in German, I believe...). rl.bhat has sorted this one out for you already; for future reference, the product kT comes up a lot in expressions relating to thermal energy -- the 'k' is Boltzmann's k in such situations.

The use of the equation

$$\frac{dQ}{dt} = -k_c A \frac{dT}{dx}$$

is a little crude for this purpose, since it describes conductive heat transfer in what is called the "plane-parallel" case, such as heat passing through flat walls and windows. There is an alternative version of this for conduction through a uniform sphere (which the iron core isn't quite...), but this equation should be good enough for an estimate of heat transfer rates.

Again, rl.bhat has told you how to calculate the heat content for the iron core. Note that your heat transfer equation gives a rate of heat flow in units of energy per time. What you will need now is a value for the heat capacity of iron, in order to find out how much heat would be released by a mass of iron cooling by 1 K (or 1º C). You will thus want to find how much heat the entire iron core releases to cool by 1 K and, consequently, how much energy per second a decline of 1 K per 10^6 years represents. That gives you the number for the left-hand side of your heat transfer equation, now permitting you to solve for the radius R that would have that cooling rate.

Last edited: Jun 11, 2008
4. Jun 11, 2008

### Kreizhn

Thanks a lot guys

5. Jun 11, 2008

### Kreizhn

So I got that the heat content of the core was $4.598e9 R^3$ Joules. I also found that the amount of heat energy required for the core to cool by 1K is $2.458e6 R^3$ Joules. Now if a million years is 31.56e12 seconds, then this corresponds to a heat drop of 7.787e-8 Joules/Second.

Now it seems like I would use the 7.787e-8 Joules/Second as $\frac{dQ}{dt}$. But then what was the point in calculating the heat content of the core? I don't see where it's used...

6. Jun 11, 2008

### dynamicsolo

I take it that this is your result at 600 K; I found substantially the same value
(I have 4.24·10^9 J).

This will turn out to be subject to more estimation. In checking up on the heat capacity of iron, it is temperature dependent (and there is some variation among sources on the value), as well as pressure-dependent. I found the value given typically at 449 J/kg·K at 0º C and 220 J/kg·K at
100º C. The change in heat capacities tends to "flatten out" as temperature continues to rise, so I just used 200 J/kg·K, with a sizeable uncertainty. (There are so many other approximations in this model that this won't bother things much more...)

So I have a result for the core of 3.2·10^6 · (R^3) J/K by this method; using the result for the heat content of the core above and dividing by 600 K, I have 7.1·10^6 · (R^3) J/K . As I say, there will be a range for the estimate on the size of the asteroid.

I agree with your result from your reckoning (although four significant figures is a bit excessive for the problem: I'd be happy if we have two. [It's probably more like one...])
From the two approaches I've described, I would have

heat content: 2.2·10^-7 · (R^3) J/K/sec
heat capacity: 1.0·10^-7 · (R^3) J/K/sec

All right, so the method that rl.bhat suggested requires that you find the total heat content (using the 3kT assumption in the problem) and further assume that the heat released per Kelvin of cooling is constant. It looks like that corresponds to an assumption of a uniform and larger heat capacity over the core's temperature range than iron may actually have. But this is the nature of physical estimation: if you state your assumptions and reasoning and make your calculations consistently, your estimate is acceptable on that basis. (It may not be accurate, but that's why we work to learn more about natural processes and to refine our models.)

Oh, and don't forget that your dQ/dt value is 7.787·10^-8 · (R^3) .

You can now set

$$\frac{dQ}{dt} = XXX · (R^3) = 5 \pi R k_c (T_i - T_s)$$ ,

where XXX stands in for the value you find from your chosen method, and solve for R.

I believe you'll find that you get credible asteroidal radii. (For reference, 1 Ceres has a radius of about 480 km., with 2 Pallas, 3 Juno, and 4 Vesta are in the 200-280 km range [they aren't spheres...].)

Last edited: Jun 11, 2008