How close of an orbit to the Sun could an iron meteor get before it melts into a ball?


As this source stats, galvanized building iron has an albedo of 0.351.

In general, we can assume, that a half of its total surface is shined by the Sun, with an average 45% angle. That means that it gets the radiation of the 6000 K Sun on $\frac{1}{2} \cdot \frac{1}{\sqrt{2}} \approx 35\%$, as if it would be shined directly on its whole surface. It radiates back 35% of it, the remaining 65% heats it. That means, in its equilibrial temperature, the iron meteor radiates $\approx 23\%$ in all direction due to its temperature, than it gets from the Sun.

Furthermore, being $n$ solar radii away from the Sun, it also gets the per-surface thermal energy of $\frac{1}{n^2}$. Here I committed yet another "approximation", because the Sun is not a pointlike radiator if $n$ is small, and the final result will be probably small, but it won't be too big.

The thermal radiation is proportional roughly with the 4th power of the temperature.2 Thus, $n$ solar radii away, it gets $\frac{1}{n^2}$ per-surface solar heating, it radiates away $0.23\frac{1}{n^2}$, what makes its equilibrial temperature $\sqrt[4]{0.23 \cdot \frac{1}{n^2}} = 0.69\frac{1}{\sqrt n}$ of the Sun's.

The average surface temperature of the Sun is 5778 K, the meltig point of iron is 1881K. Thus,

$$5778\ \mathrm K \cdot 0.69\frac{1}{\sqrt n} = 1881\ \mathrm K$$

$$\frac{1}{\sqrt n}=\frac{1881\ \mathrm K}{5778\ \mathrm K\cdot 0.69}=0.47$$

$$n=\frac{1}{0.47^2} \approx \underline{\underline {4.5}}$$

The answer is: 4.5 solar radius from the solar center, which is $\approx$ 1.8 solar diameter from the surface. It is far deeper than the orbit of the Mercury. As a comparison, the Parker Solar Probe will near the Sun 8 solar radii away. A 2000 K maximal temperature is predicted on its heat shield (note, that probe won't have an equilibrial temperature).

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Another remark: liquid things in vacuum evaporate. The evaporation for molten metals near their melting point is slow, but it exists. Such an object, if it does not have something to keep the iron vapor around it (f.e., gravitation), will slowly disappear.

1Surprisingly, I did not found better data with google.

2Also this is not exact, because it is true only for absolute black body radiators, but the spectrum of most ordinary materials does not cause a huge error in the calculation.

  • $\begingroup$ like a magnetic field $\endgroup$ – Muze Apr 14 '19 at 3:09
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    $\begingroup$ @Muze Yes, it would work to keep away the solar wind. But magnetic field keeps only moving charged particles away. It can affect neutral gases, like iron vapor, at most diamagnetically, which is a hundred million times weaker. And there is no way for a molten iron ball to have a natural magnetic field, even solid iron is no more ferromagnetic over its Curie point ($\approx$ 900C). Some artificial electronic should operate in the shadow of the iron ball. $\endgroup$ – peterh Apr 14 '19 at 3:11
  • $\begingroup$ True but it could be given a field by running electricity through it? $\endgroup$ – Muze Apr 14 '19 at 3:38
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    $\begingroup$ @Muze Yes it could work. The current could be also induced by an external EM field from the back. $\endgroup$ – peterh Apr 14 '19 at 3:41

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