“Say your halfway between the Earth and the Sun. Shouldn’t the Sun be giving off enough heat the keep you from freezing?”
My brother asked me this. I said of course not. But I was unable to come up with a good explaination. I just guessed and said that it was because your surface area is not large enough to absorb enough radiant energy to keep you warm. Am I right?
Wearia
P.s This is of course disregarding that you can’t breathe and are being streched apart…
You need not go halfway to the sun. Why not just stay right here around earth? On the moon the temperature ranges from roughly -250 F to +250 F. That’s the difference between the sunlit side and the dark side.
Unfortunately I am not qualified to answer whether you would be half frozen and half boiling or if you’d or if you’d go all one way or the other. Note, however, that this whatever happens will take awhile. Space is cold but it is surprisingly hard to lose heat in space. Even in interstellar space (far from any sun) you would take longer to freeze than you might expect. Once you died from suffocation (and thus stopped generating heat) the only way for heat to escape your body is via radiant energy. This is a pretty inefficient way to lose heat. I could cool a dead body off much faster throwing it into some snow than by dumping it into space. I mention all of this because I don’t know at what rate you would collect energy from the sun vs. lose it out your ‘dark’ side (that’s begging for a joke).
BTW…you’d actually hold together quite nicely. You’d swell a bit but that’s about it. Your skin is plenty strong enough to hold you together against any internal pressure to expand.
At halfway to the sun, you’d absorb energy at a rate of about 5500 W per m[sup]2[/sup] of your surface area exposed to sunlight. I don’t know how fast you’d lose energy… Presumably, treating a human being as a blackbody radiator would be the first stab at this. If we assume this, then the rate of heat loss would be [symbol]s[/symbol]AT[sup]4[/sup], where A is your surface area, T is your temperature, and [symbol]s[/symbol] is the Stefan-Boltzmann constant, 5.6710[sup]-8[/sup] W/m[sup]2[/sup]K[sup]4[/sup].
(The number 5500 W/m[sup]2[/sup], by the way, comes from noting that the solar intensity at the earth is about 1370 W/m[sup]2[/sup] and that the entire thing follows an inverse square law. You can get the 1370 quoted above using the equation I gave, plugging in the sun’s size and temperature and the earth’s radius and distance from the sun. I haven’t done this so my memory could be off on the exact numbers, but I think I’m pretty close.)
head bursts
Right, thats exactly what I was thinking. nods dumbly
So does that mean hot? Or very cold?
Whack-a-mole is correct. Even at Earth’s distance, in full sunlight you will heat up, not cool off. You would be bathed in solar radiation of all wavelengths. Even though this is the only way to heat you (besides internal body heat), it is also the only way to cool off (radiation), and because of the light you are absorbing more than emitting. This is where optical properties come in to determine the actual numbers, and why space suits use white - it reflects the most energy.
If you remained facing the sun and didn’t turn around, it is possible you would get cold spots along the skin on the back side, though I would think your blood would help keep you more isothermal.
So, I was wrong then? Damn.
Wearia
Sorry, sorry… Umm… here. Assume a spherical human (woohoo! I’ve always wanted to say that!) weighing about 80 kg and being essentially a big bag of water. Then I get that you’d radiate at 4.536*10[sup]-9[/sup]*T[sup]4[/sup] watts and would absorb radiation at an energy rate of 1250 watts. Equating the two means you’d end up at about 450 degrees Centigrade or some 840 Farenheit. Needless to say, you are quite overcooked.
This is ridiculously crude, mind you (but hey, I got to assume a spherical human!), but it’s probably in the right general ballpark, give or take a factor of 2 or something like that. My general conclusion: this would be a bad bad bad plan.
Another note… (I’m trying to be helpful here, really!). If you want to try this, I suggest that you orient yourself so that your feet are pointing to the sun and your headaway, rather than it being directly in front of you. This will reduce the surface area that gets hit by radiation enormously and you’ll be much happier because you’ll absorb less sunlight. I think you’ll still cook, but you won’t get AS hot. And feet-first, you’ll get cooked from the feet up rather than from the head down, which I suppose is probably better or something.
You have to be careful here. The amount of heat radiated by a human is important, but in space the albedo is important to: the amount of light absorbed. The Earth, for example, reflects about 40% or so of the light that hits it, and if you put that in the heat balance equation you get the average temperature of the Earth is about 0 Celsius, the freezing point of water. The greenhouse effect is what keeps us above freezing.
I don’t know what the albedo of a human is. I am white, so mine is probably fairly high, around 50%. I would freeze solid in space, given enough time to radiate my internal heat.
Yeah, there’s definitely that. If you reflect half the heat directed at you, that would mean a factor of what, 15%, 20% lower temperature? This is, of course, a big deal. I think my math was probably somewhat off above anyway. Mostly, I just wanted to assume a spherical human and give a vaguely ballpark factor of 2 type figure, which probably isn’t enough to be helpful.
A totally unrelated question… if we didn’t have cloud cover, we wouldn’t have the greenhouse effect to keep us from freezing, but wouldn’t we also get a lot more direct sunlight here at ground level? To what extent would these two effects cancel?
A mistake I made initially in the problem was equating the surface area of the object with the surface area for purposes of absorbing light. The absorbtion area should only be the cross-section of the object, which for a sphere would be [symbol]p[/symbol]r[sup]2[/sup], as opposed to the total surface area which would be 4[symbol]p[/symbol]r[sup]2[/sup]. You don’t really care so much about the actual surface area or cross-sectional area as much as you care about the ratio of the two.
Plugging in those values, I get that, for a sphere with an albedo of 50%, assuming the incoming energy density at 5500 W/m[sup]2[/sup], that T = 332K = 59C = 138F.
Of course, most humans are not sphere-shaped, so your temperature would probably be lower than that. Also, this ignores that one side of you would be blisteringly hot, and the other side would be freezing cold.
Yeah, I did something wrong, but I don’t know what. I realized the same thing you did, Puno, and and I still don’t know how I got the number I did. I knew the cross section was [symbol]p[/symbol]r[sup]2[/sup], but I must have just put the wrong numbers in my calculator. Your answer looks a lot better.
Well, I might as well show my work, so others can double-check that I set it up correctly. I have:
[symbol]s[/symbol]AT[sup]4[/sup] = bCI
where [symbol]s[/symbol], A, and T are as above, b is the fraction of absorbed light, C is the cross-sectional area, and I is the incoming energy per unit area. From that, I get:
T = (bCI/[symbol]s[/symbol]A)[sup]1/4[/sup]
Since C/A = 1/4, then,
= [(.5 * 5500 W/m[sup]2[/sup]) / (5.67e-8 W/m[sup]2[/sup]K[sup]4[/sup]* 4)][sup]1/4[/sup]
= 121e8[sup]1/4[/sup]K
= 11.0e4[sup]1/2[/sup]K
= 332K
But I gotta give you muchos props for working the spherical human assumption into the problem.
Oh no, look what I’ve started…
You are all starting to scare me. But thanks anyhoo.