## mattfeury 3 years ago What is the optimal distance from the Sun to cook a pizza in outer space? Totally serious question. Things to consider: - Optimal temperature at which a pizza cooks - How heat diminishes over distance (does the Inverse Square Law apply in the vacuum of space?) - Would the lack of ozone make the pizza full of radiation, i.e. inedible?

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Totally serious question will not get totally serious answer from me so I should probably leave now haha

2. apoorvk

Okay.. Hmm.. So, optimum temperature to cook a pizza is generally about 200$$^o$$C Now, what governs variance of heat delivered with distance, some concept sregarding solid angles, regarding heat being directed to the pizza, some funda regarding solar constant, and a good good calculator.

3. mattfeury

So I know: "The sun has a surface temperature of 9,980 °F (5,526 °C)," (via http://arcana.wikidot.com/temperature-of-outer-space ) Does a vacuum follow the Inverse Square Law? Then we can tie radiation decay to distance. According to wikipedia: "the intensity of radiation from the Sun is 9126 watts per square meter at the distance of Mercury (0.387 AU); but only 1367 watts per square meter at the distance of Earth (1 AU)—an approximate threefold increase in distance results in an approximate ninefold decrease in intensity of radiation." So how many watts / sq meter would cook a pizza?

4. legoauto220

Well the temperature of the space a bit further from mercury is 450 C http://www.esa.int/esaMI/Space_Engineering/SEMCL2BE8JG_0.html If you were to go into space just here around the Earth in direct view of the sun you will reach 123 C So I think you have to go a bit closer towards sun, in the orbit of mars

5. apoorvk

you mean in the orbit of venus right?

6. msayer3

since it's out of space you should consider radiative heat transfer

7. legoauto220

@apoorvk No, orbit of mars. Venus would be too hot

8. Carl_Pham

The key information you need is the Stefan-Boltzmann Law, $E = \sigma T^4$ which relates the energy radiated by a black body to its temperature. The energy that the fully cooked pizza must radiate is given by: $E = L \frac{A}{4 \pi R^2}$ where L is the luminosity of the Sun, A is the area of the pizza, and R is the distance of the pizza from the Sun. Using these two equations you can find the proportion of the temperature of the pizza to the temperature of the surface of the Sun: $\frac{R_{cook}}{R_{sun}}= \left(\frac{T_{sun}}{T_{cook}}\right)^2$ Now we need data. Assume you want T_cook = 200C = 573 K, and T_sun = 5500 K and R_sun = 7.0e-08 m, we have R_cook = 6.4e+10 m or about 0.42 AU, a smidge outside the orbit of Mercury.

9. Carl_Pham

Rats, I used the temperature of the Sun from memory, and I'm off a little. T_sun turns out to be 5780 K. I also converted from C to K wrong, 200C = 473 K. So the right answer should be 1.7e+11 m or 1.1 AU, outside the orbit of the Earth. Oh well, in astrophysics being correct within an order of magnitude is usually good enough.

10. mattfeury

That is excellent. I don't really see the correlation between equations 2 and 3 (where did the Time ratio come from?), but I figured a lot of this would be over my head. Regardless, thank you. This is exactly what I hoped to see. I'm very glad to see that it's close to Earth's orbit. This makes my plan to setup an orbiting pizza parlor one step closer to fruition.

11. asnaseer

:) I believe the T in the equations represent temperature rather than time.

12. walne

Something that should be considered is the type of pizza. Are you talking Deep Dish or Neopolitan? If it's thin crust, you're looking for higher temp (800 degrees), shorter cook time.