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This question was triggered by a discussion regarding the computer game Elite: Dangerous, where spaceships routinely operate in close proximity to stars (two or three light seconds away), at which point heat becomes an issue. Someone claimed that you would hardly feel any heating effect since there is so little matter around, indeed that you couldn't really tell if you are inside a star or not. I found that very dubious.

So let us assume we have spaceships and could fly around the solar system at will. Our ships have a certain degree of ruggedness to them, so people might get adventurous. What would it actually be like to approach, or even enter, a star?

We know that the corona is hot, hotter even than the surface. But it is also much less dense.

Q1: Would there actually be enough matter in the corona to make our ship heat up, or would we be able to ignore the very hot but few particles around us? What of the photosphere, or the upper convection zone -- those are still much less dense than earth's atmosphere at sea level. How much would the heating affect us?

Assuming we could pass through the corona, we would reach the photosphere, the point at which the (mostly hydrogen) body of the sun becomes opaque.

Q2: Does this opaqueness only apply to a certain layer (i.e. would we be able to "see" again once we go below the photosphere), or are all lower layers of the sun opaque? How opaque is that -- would we be able to see a couple of meters, some kilometers, or nothing at all?

Q2b: Would the matter directly in front of our cockpit window be bright, or dark?

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  • $\begingroup$ Interesting question, but maybe asking for too much. Is it possible to narrow this all down to one or two very closely related questions? $\endgroup$ – Sean Mar 25 '15 at 13:12
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    $\begingroup$ Question Q2, based on what I've read on the cosmic background radiation, hydrogen ionizes at about 3,000 degrees, and anything ionized is opaque. The entire sun is hotter than that so it would be Opaque pretty quickly. I can't give you an exact number but I would think, just a few inches of thick, gaseous ionized hydrogen would be impossible to see through. $\endgroup$ – userLTK Mar 25 '15 at 13:49
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    $\begingroup$ I'd be more concerned about radiation than I would be about matter. $\endgroup$ – Jerry Schirmer Mar 25 '15 at 14:21
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    $\begingroup$ Tell that someone who said you'd feel nothing that they are stupid. Earth is 8 light minutes from the Sun and we get all of our heating from it even with such little matter from it hitting us. Radiative heating is the key. When you get a sunburn, that's not because of heated matter, it's radiation from the Sun. At a few light seconds away, heating happens rapidly. In fact, the lack of matter prevents you from cooling down faster. So yeah, don't listen to them. Too much heating is an issue even at the distance of Earth's orbit when you're in space $\endgroup$ – Jim Mar 25 '15 at 15:24
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    $\begingroup$ @Jimnosperm: "and we get all of our heating from it" -- no we don't. $\endgroup$ – DevSolar Mar 25 '15 at 15:34
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How opaque is that -- would we be able to see a couple of meters, some kilometers, or nothing at all?

The photosphere of our sun is somewhere on the order of 500 km thick. For a quick ballpark, you can imagine an exponential decrease in the transmission of light which about this characteristic thickness. It might be a little less, but it's still order-of-magnitude accurate. That means that you would (theoretically) be able to see local objects just fine within the photosphere.

It's difficult to compare this to Earth's atmosphere, because scattering of light is both Raleigh and aerosol, and the concentration of aerosols differs at different times. However, even Raleigh scattering attenuates more than half of the light within 100 km. That means that you can actually see better in the sun's photosphere than you can in Earth's atmosphere.

Would the matter directly in front of our cockpit window be bright, or dark?

Now for the qualifier, being inside the photosphere would render human eyes useless instantly, because it would be like painting the sun's brightness over your entire field of vision. A spacecraft 100 km away could be identified with a telescope (if this telescope is not actively melting), but it would not be any brighter than the background of photosphere gas. Nonetheless, it would still be identifiable by color, contrast, and so on. In fact, if the spaceship was mostly absorbing of radiation, it would be a distinct black spot among a bright background.

I could get more specific to speculation about spacecraft. They will need to have stored thermal mass in order to absorb more heat than they emit. Reflection is a good option, but it's not perfect. Active cooling of the spacecraft's surface is likely in order to avoid melting. Thus, spacecraft would be slightly "camouflaged" by thermal necessity, but not perfectly.

Does this opaqueness only apply to a certain layer (i.e. would we be able to "see" again once we go below the photosphere), or are all lower layers of the sun opaque?

Your vision will certainly decrease as you reach lower layers - since they have higher density and give off more radiation. The photosphere is only unique in that the radiation emitted commonly makes it out into space where it can continue indefinitely. For the lower convection layer zones, radiation is given off at a much higher rate, but it is almost entirely reabsorbed.

Stars above a certain mass threshold actually transmit their heat through the interior zones through a radiation gradient. So I hope this helps to illustrate that the inner zones are not "less bright". They would be blindingly bright (well the photosphere is already blindingly bright, but these would be worse), but possibly at higher temperatures, which go beyond the visible range.

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  • $\begingroup$ Ah... I am not sure I get the picture: Upon entering the photosphere, visibility wouldn't be the problem (the part about Raleigh scattering), but everything would be bright as... well, the sun. And as I go deeper, the wavelengths would go into the UV spectrum, so instead of being blinded by visible light, I'd be microwaved instead. Is that about right? $\endgroup$ – DevSolar Mar 25 '15 at 14:02
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    $\begingroup$ @DevSolar That is exactly what I had in mind. However, I was hesitant to elaborate on one detail, and upon further research, my doubts are warranted. As you go deeper into the sun, the visible wavelengths will not decrease in intensity, because increasing temperature only adds more in shorter wavelengths, but doesn't subtract from long wavelengths. So you'll be microwaved, but you'll still be blinded in the visible spectrum. $\endgroup$ – Alan Rominger Mar 25 '15 at 15:04
  • $\begingroup$ Sounds like lots of fun. :-D Thank you very much! $\endgroup$ – DevSolar Mar 25 '15 at 15:17
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Well, I can give you a definitive answer to Q1, but my answer to Q2 would only be educated speculation. Perhaps some of the astrophysicists on here can be more help with that one.

However, before I tackle Q1, a very important disclaimer: Temperature is a measure of the average kinetic energy of the particles of an object, and cannot be used all by itself to determine the amount of heat that will be transferred to another object. The total thermal energy possessed by the corona will depend not only on the temperature, but also on the total amount of matter.

Using temperature all by itself can be very misleading, because if you have a gas with very low density, and very few total particles, you can easily have a measured temperature of several thousand Kelvin, especially if there's just a few very energetic particles in the mix. However, if the total number of particles is small, the total thermal energy of the substance will also be small.

The exact mechanism that allows the corona to have a much higher temperature than the surface of the sun is considered an open problem, so I will not speculate as to how the corona would heat to your spaceship, since we're not even certain how the corona is heated. However, the corona actually has little bearing on the answer to your question. Regardless of what the corona does (or doesn't do) to your spaceship, you're still getting fried.

The reason is due to heat transfer through radiation. You see, it's not necessary that two objects physically touch to exchange thermal energy. Objects called black bodies emit electromagnetic radiation. The frequency and amount of radiation emitted depends on the temperature of the black body.

Black body curve for sun

The top curve in the above picture roughly represents the black body curve for the Sun. The y-axis shows intensity, and the x-axis shows the wavelengths emitted, which go from ultraviolet, to visible light, to infrared from left to right.

Turns out that the sun emits about 44% of its radiation as visible light, and about 48% as some type of infrared radiation. From 93 million miles away (8 light minutes), this is still enough radiation to light and heat the earth. If you want to travel around the sun from 1 light second away (roughly 1/500th the distance) imagine how much more heat you will receive! It turns out the the energy you receive from the sun is inversely proportional to the square of the distance, so if you are 500 times closer to the sun, you will receive $500^2$ or $250 000$ times as much radiation! As a result, the thermal radiation emitted from the sun is more than enough to bake you at that distance, regardless what the corona does (or doesn't) do to you.

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    $\begingroup$ Re "objects called black bodies": All physical objects emit thermal radiation, but the efficiency varies depending on the nature of the surface. "Black body" is the name of a theoretical 100% efficient radiator. I'm not a physicist, and I don't understand the theory, but it says that any 100% efficient radiator would also have to absorb 100% of incident light. It would therefore appear to be perfectly black (when too cold to emit visible light, of course), hence the name. $\endgroup$ – Solomon Slow Mar 25 '15 at 13:52
  • $\begingroup$ Good point. It would be accurate, however, to say that we often model objects (including stars) as black bodies. $\endgroup$ – Sean Mar 25 '15 at 13:54
  • $\begingroup$ Gee... yes, the idea of radiation heating came up, but since I am fuzzy on the details, I decided to not mention it. I also wasn't sure how much of the earth's temperature is due to the sun directly, and how much are other effects (radioactive decay in the core, greenhouse effects etc.). After all, Venus is hotter than Mercury despite being further away. Thanks for settling the issue of Elite: Dangerous being spot-on about the heat issue (which was my main question). $\endgroup$ – DevSolar Mar 25 '15 at 13:57
  • $\begingroup$ Hmm... excuse my notoriously weak maths, but is the calculation "1/500th the distance == 500^2 times as much radiation" actually correct? $\endgroup$ – DevSolar Mar 25 '15 at 15:39
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    $\begingroup$ @DevSolar - Generally, radiation falling on an object changes inversely to the square of the change in distance. Move something twice as far away, the radiation drops to 1/4. Move something half as far away, the radiation increases by 4. Move something to 1/500 the distance, the radiation increases by 500^2 = 250,000 times. Note that the actual temperature doesn't scale linearly... the hotter something gets, the faster it radiates heat away. FYI, Mercury is .387 AU from the sun (Earth is 1 AU), so each square meter on Mercury receives 1/(.387^2) = 6.7 times as much radiation as Earth. $\endgroup$ – kbelder Mar 25 '15 at 18:22

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