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Considering how low the power density is at the sun's core, I seem not to be able to expect what would happen to matter in case it was thrown inside the sun's core. For example, let's assume an Earth-like planet is placed at the center of the Sun's core, and with the power density of 276.5 $watts$/$m^3$ which is very low to even raise the temperature of the Earth by any noticeable degree. That was one thought. The other thought was that, the temperature in the core is already 15 million degrees K, so any matter there should get close to this temperature quickly enough.

So now I'm very confused, like will the planet stay there intact for thousands or even millions of years until it has accumulated enough energy to melt or vaporize ? Or there are other types of energy absorption the planet would experience and thus have a shorter time staying as one piece in the core ?

Just for trying to get a practical and numerical answer to this question, let it be : how long can the Earth as a whole survive in the core of the sun ?

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    $\begingroup$ What happens when you take an ice cube and stick it in a cup of water? Does it melt entirely at once or does it melt layer by layer? $\endgroup$
    – Kyle Kanos
    Commented Aug 29, 2014 at 14:26
  • $\begingroup$ I think what I mean is a bit different. I understand how an ice cube will melt, but considering the bizarre environment in the core of the sun (very high temperature with very low energy density), I can't seem to imagine how this would happen to a planet there. $\endgroup$ Commented Aug 29, 2014 at 14:38
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    $\begingroup$ Why would it be different? Heat transfer still occurs due to temperature differences, right? It seems to me that the "low energy density" is irrelevant. $\endgroup$
    – Kyle Kanos
    Commented Aug 29, 2014 at 15:06
  • $\begingroup$ Using Stefan-Boltzmann law, with e=1, area of the Earth and the ambient temperature of the core of the sun, it yields energy dissipation to the earth at 10^36 watts, or about 4 order of magnitudes more than the energy required to evaporate the mass of the earth. Even assuming the earth is heated to 14 million degrees on its way to the core still gives something in the 10^35 watts, which is a billion times the sun's luminosity!! isn't that weird ? $\endgroup$ Commented Aug 29, 2014 at 15:30
  • $\begingroup$ Earth would be not only quickly turned to plasma but also crushed by extreme pressure (up to 26.5 million gigapascals!). $\endgroup$
    – Mithoron
    Commented Sep 25 at 11:51

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As you say, the power produced per cubic metre of the Sun's core is surprisingly low. This is because proton-proton fusion is a very slow process, as has been discussed hereabouts before. The core is so hot because conduction of heat through the core is slow. The average speed with which a photon escapes the core is the astonishingly low value of about 30$\mu$m/s.

However the photon net speed is so slow because the dense plasma at the Sun's core scatters photon extremely efficiently. If you were to magic the Earth into the Sun's core then the Earth would start receiving energy at the rate predicted by the Stefan Boltzmann law. I make this around $10^{21}$ W/m$^2$ of the Earth's surface, so the Earth would start boiling away pretty quickly. The Earth would cool the plasma around it, and as that plasma cooled and recombined it would become transparent to the next layer of plasma out, and so on. To a first approximation the heat flux entering the Earth would remain at around $10^{21}$ W/m$^2$ until the material of the Earth became hot enough to form a plasma itself. At this point that plasma would start scattering photons and the heat flow would start slowing.

Actually calculating the rate at which the Earth would vaporise would be a difficult task. You could treat it as the heating of a sphere with a known thermal conductivity, but as mentioned above once the temperature gets hot enough to ionise the material from the Earth this would strongly affect the heat flow from the plasma around it.

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    $\begingroup$ Any order of magnitude accuracy possible to calculate ? Also, won't the ionized material from the Earth work as the plasma did ? I mean these material will have enough temperature to become plasma, so what does it differ from the plasma from the sun itself ? in other words, why would the heat flow slow although the temperature difference is the same ? $\endgroup$ Commented Aug 29, 2014 at 19:24
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    $\begingroup$ This page estimates the heat of vaporization of crust as $10 GJ/m^3$ From Wikipedia we find the heat of vaporization of iron to be about $11 GJ/m^3$ dominated by heating the material to boiling. With the volume of the earth $10^{21} m^3,$ we need $10^{40}$ J to vaporize it. That comes very quickly. $\endgroup$ Commented Aug 29, 2014 at 21:15
  • $\begingroup$ @RossMillikan you mean $10^{31}$ J to vaporize it, don't you ? $\endgroup$ Commented Aug 29, 2014 at 21:40
  • $\begingroup$ @AbanobEbrahim: Oops, you are right. I counted the $9$ twice. It is $10^{31}$ J That comes even more quickly. $\endgroup$ Commented Aug 29, 2014 at 22:14
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Imagine you have a hot tub, and you heat it up to a nice toasty temperature. Then the power goes out. The metabolism of the hot tub environment is now zero. It won't get any warmer.

But if you get in the tub, you'll still warm up. The thermal mass of the water won't be cooled much by you entering. You're taking advantage of the heat that was produced earlier.

The earth would be heated by the existing thermal mass of the core.

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  • $\begingroup$ At what rate ? and by what means of heat transfer ? $\endgroup$ Commented Aug 29, 2014 at 21:33
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    $\begingroup$ I had interpreted your question as asking more about the difference between the high temperature of the core and the low metabolism there. My answer meant to point out that the low power rate does not matter because the heating is dominated by the existing hot mass present. John Rennie's answer already gives quantitative estimates about the heat transfer. $\endgroup$
    – BowlOfRed
    Commented Aug 29, 2014 at 21:58
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If you were to "magically" place a planet in the sun's core, I'm fairly sure that is would not be there long. The ambient temperature of the sun's core is somewhere around 15.7 million K, as you said. You should think about why its so how there before you think about melting planets. The density of the core is something like 150 times denser than water. When you compress something that much, it gets hot. (Think of compressing a gas. What you compress a gas, it gets hot.) And when something like that is that compressed and that hot, a really huge fusion reaction occurs. So on one hand, placing an Earth inside the sun would increase the mass of the sun, negligibly increasing the pressure, increasing the heat a tiny bit. On the other hand, introducing something with so many heavy metals might do something to the fusion reaction of the sun (I don't know what, but it might.)

The appropriate question to ask yourself would involve the specific heat capacity of Earth, and the ambient core temperature of the sun. (Also, the melting point of Earth). Though at 15.7 million K, I cannot think of anything that could stay in a solid form for long.

Just think of the boiling point of iron. (3134K) some simple division tells us that the sun is over 5000 times hot enough to boil iron. The iron would never even touch the sun before it boiled.

Another one, according to Wikipedia, granite melts at 1533K. More division tells us that the sun could melt granite more than 10,000 times over.

I hope I was able to answer your question, at least partially.

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    $\begingroup$ I believe the OP's question is mostly about how long it would take to heat and vaporize the earth, not whether it would happen at all. We know the sun is hot - but how quickly would that heat transfer to our teleported planet? $\endgroup$
    – user20846
    Commented Aug 29, 2014 at 21:20
  • $\begingroup$ @JonofAllTrades I cannot say. There are too many different elements, with different heat conductivity values. One could, however, probably say that the Earth would be vaporized within a week, or something like that. Nothing with much accuracy I don't think though. $\endgroup$
    – CoilKid
    Commented Aug 31, 2014 at 5:16

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