Freezing in Space II - Turn On The Sun!

Yesterday I considered how long it would take a human to freeze in space. However, I considered only what would happen if you were not absorbing any radiation from nearby sources. Today we consider what happens if you do have hot objects nearby. Namely, the sun. The sun provides a lot of energy, even as far away from it as we are. It keeps the earth at a comfortable \~20 C, good for us humans, and provides the energy for life on earth, also good for us humans. That’s a lot of energy. So maybe the sun can keep you alive when you’re adrift in space. Or at least keep you warm. I still think you’ll asphyxiate. From here on out we’re going to assume that we are adrift in space near earth. You were out for a joyride in that new spaceship of yours and something went horribly wrong. We could go through a whole song and dance of calculating how much power the sun delivers to the earth, but we won’t (if you’re interested, let me know, an I can do that later). Instead, we’ll quote the known result, that the sun delivers \~1370W/m^2 in the vicinity of the earth. To find out what temperature we would cool to we set the power we absorb from the sun equal to the power we radiate $$P_{sun}=P_{rad}$$ $$1370W/m^2*A_{ab}*e_{ab}=e_{rad}*A_{rad}*\sigma*T^4$$ Where A_ab is the surface area absorbing the suns power, e_ab is a factor between 0 and 1 that indicates how much of the incident power we actually absorb, and e_rad is the emissivity of us, while A_rad is our radiating area. Note that the emitting and absorbing areas are not the same! Take a simple example. If you put a sheet in space, and face the flat side towards the sun, it will only absorb energy from the sun on one side, but it will radiate energy from both sides. Likewise e_ab and e_rad are not necessarily equal because we are radiating and absorbing at different wavelengths. We can solve the above equation for T, giving $$T=\left(\frac{A_{ab}}{A_{rad}}\right)^{1/4}\left(\frac{e_{ab}}{e_{rad}}\right)^{1/4}\left(\frac{1370W/m^2}{\sigma}\right)^{1/4}$$ For a first pass, we’ll make the simplifying assumption that e_ab=e_rad. Given this, $$T=394K*\left(\frac{A_{ab}}{A_{rad}}\right)^{1/4}$$ Now, the absorption area of an object is just the shape of the object flattened into the plane the incident radiation is perpendicular to. That is, the absorption area of a sphere is a circle (a sphere projected to 2D is always a circle), while the absorption area of a cylinder could be a sheet or a circle, or something stranger. The best area ratio we can ever have is that of a flat sheet, which gives 1/2. For a sphere, like the earth, the ratio is 1/4. As an aside, this gives an equilibrium temperature of the earth as \~5C, which is too cold. It turns out that we shouldn’t neglect either the emissivity ratio or the natural greenhouse effect in the case of the earth. Now, we need to figure out the area ratio for us. In a previous post I modeled myself as a cylinder with height 1.8 m and radius .14 m. Let us assume we are facing the sun dead on, beating down on our chests. This gives the cross sectional area of a sheet with width 2*.14 m =.28 m and height 1.8 m. This is an area of .5 m^2, while my total surface area is 1.7 m^2. This gives an area ratio of \~.3, or an equilibrium temperature of $$T=394K*(.3)^{1/4}\approx292K$$ That is an equilibrium temperature of 19 C. Not too cold, but certainly not body temperature! So the sun will not save us. We also have to factor in the fact that we reflect better in the visual that we do in the infrared, so the emissivity ratio we set to 1 probably is less than that, reducing our equilibrium temperature even more. It is interesting to note, though, that if we model a human as a two sided sheet instead of a cylinder, we can bring our equilibrium temperature up to 331 K. That’s \~58 C! So in our model our geometrical assumptions change whether or not we freeze or die of heat stroke. Finally, since it looks like the sun may not save us, lets see how much it might slow down our temperature loss. Instead of a net loss of 860W at body temperature, as we calculated yesterday, sticking with our cylindrical human we’ll have a net loss of $$860W-1370W/m^2(.5m^2)=175W$$ Similarly at our lowered body temperature of \~30 C, we’ll be losing a net of \~105 W. Once again taking a geometric average gives an average power loss of \~135 W. Using the energy to cool we found yesterday it would take 16300 s, or 4.5 hours to freeze! Also note that if you’re getting too hot or cold, given how much the geometry plays into things, by changing your orientation to the sun you’ll be able to have a certain amount of control over how much you heat up or cool down. Also, make sure you rotate yourself so that you end up evenly heated, and not roasted on one side and frozen on the other!

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