<p>As <a href="https://books.google.de/books?id=c-v7OSaxGSsC&amp;pg=PA112&amp;lpg=PA112&amp;dq=iron+albedo&amp;source=bl&amp;ots=XV_S83V-3f&amp;sa=X#v=onepage&amp;q=iron%20albedo&amp;f=false" rel="nofollow noreferrer">this</a> source stats, galvanized building iron has an albedo of 0.35<sup>1</sup>.</p>

<p>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 <span class="math-container">$\frac{1}{2} \cdot \frac{1}{\sqrt{2}} \approx 35\%$</span>, 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 <span class="math-container">$\approx 23\%$</span> in all direction due to its temperature, than it gets from the Sun.</p>

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

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

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

<p><span class="math-container">$$5778\ \mathrm K \cdot 0.69\frac{1}{\sqrt n} = 1881\ \mathrm K$$</span></p>

<p><span class="math-container">$$\frac{1}{\sqrt n}=\frac{1881\ \mathrm K}{5778\ \mathrm K\cdot 0.69}=0.47$$</span></p>

<p><span class="math-container">$$n=\frac{1}{0.47^2} \approx \underline{\underline {4.5}}$$</span></p>

<p>The answer is: 4.5 solar radius from the solar center, which is <span class="math-container">$\approx$</span> 1.8 solar diameter from the surface. It is far deeper than the orbit of the Mercury. As a comparison, the <a href="http://parkersolarprobe.jhuapl.edu/" rel="nofollow noreferrer">Parker Solar Probe</a> 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).</p>

<p><a href="https://i.sstatic.net/suABms.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/suABms.png" alt="enter image description here"></a></p>

<p>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.</p>

<p><sup>1</sup><sub>Surprisingly, I did not found better data with google.</sub></p>

<p><sup>2</sup><sub>Also 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.</sub></p>

CC BY-SA 4.0
2019-04-14T03:09:04.360Z
like a magnetic field
Muze
CC BY-SA 4.0
2019-04-14T03:11:57.477Z
@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.
peterh
CC BY-SA 4.0
2019-04-14T03:38:06.883Z
True but it could be given a field by running electricity through it?
Muze
CC BY-SA 4.0
2019-04-14T03:41:27.350Z
@Muze Yes it could work. The current could be also induced by an external EM field from the back.
peterh
CC BY-SA 4.0
2019-04-14T02:33:16.693Z
2019-07-18T12:34:05.433Z
2019-07-18T12:34:05.433Z
peterh
<p>How close of an orbit to the Sun could an iron meteor get before it melts into a ball? </p>

CC BY-SA 4.0
2019-04-13T23:21:47.610Z
Does https://astronomy.stackexchange.com/questions/10113/how-to-calculate-the-expected-surface-temperature-of-a-planet help?  I get about 0.02 AU
James K
CC BY-SA 4.0
2019-04-14T03:52:55.220Z
@JamesK You got the same :-)
peterh
<the-sun><orbit><meteor><roche-limit><thermodynamics>
Could the Sun have liquid iron in orbit?
Muze
# Could the Sun have liquid iron in orbit?

(Asked by: username_0 on 2019-04-13)

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

**Tags:** the-sun, orbit, meteor, roche-limit, thermodynamics

> **Comments:**
> **username_1**: Does https://astronomy.stackexchange.com/questions/10113/how-to-calculate-the-expected-surface-temperature-of-a-planet help?  I get about 0.02 AU  
> **username_2**: @username_1 You got the same :-)  

---

## 1 answers

### Answer by username_2 on 2019-04-14. Score: 2 | Accepted answer

As [this](https://books.google.de/books?id=c-v7OSaxGSsC&pg=PA112&lpg=PA112&dq=iron+albedo&source=bl&ots=XV_S83V-3f&sa=X#v=onepage&q=iron%20albedo&f=false) 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](http://parkersolarprobe.jhuapl.edu/) 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).

[![enter image description here](https://i.sstatic.net/suABms.png)](https://i.sstatic.net/suABms.png)

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.

> **Comments:**
> **username_0**: like a magnetic field  
> **username_2**: @username_0 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.  
> **username_0**: True but it could be given a field by running electricity through it?  
> **username_2**: @username_0 Yes it could work. The current could be also induced by an external EM field from the back.  

---

