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One of the mechanisms for the heating of asteroids in the early history of the Solar System is believed to be decay of the isotope ${}^{26}\mathrm{Al}$. This was created by the supernova that produced the dust cloud from which the asteroids formed. For example in this paper the astrophysicist G. Jeffrey Taylor wrote:

${}^{26}\mathrm{Al}$ was present when meteorites were forming (see PSRD article Using Aluminum-26 as a Clock for Early Solar System Events). It is a radioactive isotope with a half-life of only 700 thousand years, so its presence means that the solar system formed within a few half-lives of the formation of ${}^{26}\mathrm{Al}$ in an exploding star. It decayed by emitting a beta particle (an electron), creating ${}^{26}\mathrm{Mg}$ (magnesium-26) and releasing energy. The energy released is considerable. If ${}^{26}\mathrm{Al}$ made up only $5 \times 10^{-5}$ ($0.005\%$) of all the aluminum in a chondrite (most is aluminum-27, which is not radioactive), it would release enough energy to melt asteroids a few kilometers across and larger. Lower amounts of ${}^{26}\mathrm{Al}$ cause less melting.

(my emphasis)

I have been trying to estimate the minimum radius that the asteroid would need to have for it to melted due to radioactive decay but I do not know how this calculation is done. Can anyone describe how this calculation is done and how Taylor arrived at the result a few kilometers across and larger?

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    $\begingroup$ This question isn't clear. Are you saying that you have some radioactive material, and you want to make an asteroid out of it? You expect radioactive decay will produce heat. Heat will diffuse slowly out, leading to a rise in temperature. You have in mind some level of radioactivity, which translates to some level of heat production. You also have in mind some thermal conductivity, which will govern how quickly heat can escape. Is this right? $\endgroup$ – mmesser314 Dec 17 '20 at 3:04
  • $\begingroup$ yes more specificly I read somewhere that one can estimate the minimum radius of an asteorid that experienced melting of its rocky portion due to radioactive decay of 26Al. assuming thatthat the mean rate of energy production of Al has a specific value. If you could explain a little it would really help $\endgroup$ – Jokerp Dec 17 '20 at 3:17
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    $\begingroup$ @Jokerp I have tried to rewrite your answer to make it clearer what is being asked. If you don't like my edit please feel free to revert it. $\endgroup$ – John Rennie Dec 17 '20 at 16:47
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Planetary science is rife is order-of-magnitude estimates, and this is a classic problem for that kind of solution.

Step one is somewhat contradictory: start with a spherical asteroid. Make it homogenous, too. It has:

A radius: $R$

A melting point: $T_0$

A density: $\rho$

An $^{26}$Al content per unit mass: $n$

and a surface temperature: $T$

and maybe some other parameters (emissivity, mass, and so on)

In the zeroth order approximation, the temperature is uniform. Power is generated according to the total number of Al-26 decays:

$$ P_{in} \propto (\frac 4 3 \pi R^3)(n\rho)\frac{E_{Al}}{t_{\frac 1 2}}$$

which most importantly has a volume term, which goes as $R^3$.

In thermal equilibrium, that power is radiated from the surface though blackbody radiation. The Stefan-Boltzmann Law relates the total power radiated per unit area:

$$ j = \sigma T^4 $$

which we can sum over our spherical asteroid:

$$ P_{out} = (4\pi R^2)j $$

and the equilibrium condition:

$$P_{out} = P_{in} $$

has:

$$ T^4 R^2 \propto R^3 $$

and you can solve for the $R$ that yields $T=T_0$.

Deeper estimates might include surface emissivity, or a temperature gradient inside the asteroid. The former is a minor correction, the latter is an involved analysis, and if this estimate fails miserably (e.g., comes out at 50 km), may be required, as it points to an interior that is hotter than the surface.

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