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My question:

I have a similar question to Is Jupiter bright enough to be seen in color by the naked eye from Jupiter orbit? but from a different perspective: (Sorry in advance if I use any incorrect terms or get the science slightly wrong)

Assume we wanted to generate 100mW electricity above the atmosphere of Io using a Sterling engine (a satellite) and parabolic mirrors (more satellites). I'm trying to calculate the amount of heat I could generate with a comparable number of mirrors in the same position above Earth's atmosphere.

From reading through the responses on the above posted similar question, I understand that sunlight has brightness with a unit measurement in lux. Sunlight also has radiation that we experience as heat, measured in Kelvin (and some others)

In the 130,000 lux environment described above, let's say the temperature we feel is 311 Kelvin (100°F).
Assuming IO has an atmosphere, what's the corresponding temperature (work with the assumption that the atmosphere has the exact same diluting factors as ours does, same insulating properties, same reflectivity, and if breathable air is necessary, etc)

Maybe stated another way if, as described in an earlier answer, the luminosity is 1/25th of Earth's at Jupiter, does that mean that the total thermal impact would be about 1/25th, too?

Thanks for any help in understand how to approach this problem.

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  • $\begingroup$ It is not possible to convert lux to temperature. Lux may be converted to Watt if you know the spectrum of your light source. $\endgroup$
    – Uwe
    Commented Dec 11, 2022 at 8:33
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    $\begingroup$ @Uwe Lux can't be converted to Watt either. It could be converted to W/m². $\endgroup$
    – asdfex
    Commented Dec 11, 2022 at 12:52
  • $\begingroup$ I'm also confused if you are asking about an Earth or Io satellite. You say "I'm trying to calculate the amount of heat I could generate with a comparable number of mirrors in the same position above Earth's atmosphere" but then you seem to go on and talk about Io's atmosphere. What exactly are you asking? If Earth, what does Io have to do with it? $\endgroup$ Commented Dec 11, 2022 at 13:16
  • $\begingroup$ I guess what I'm trying to calculate is the joules output by the sun at a specific distance from the sun, whether that be 300 miles above the earth, above Venus, above Pluto, above Jupiter, or at any point in space. $\endgroup$ Commented Dec 11, 2022 at 20:21

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Lux is not what you want for when dealing with heat and temperature.

Lux measures illuminance, which is just irradiance compensated for how sensitive the human eye is to various wave lengths. To a piece of equipment, like a Stirling engine, a watt is a watt after the the light has been absorbed.

What you want is irradiance, measured in $W/m²$, power per area. If what you have is lux, 128000 lux of sunlight corresponds to 1361 $W/m^2$ at Earth distance from the Sun. Sunlight has the same frequency distribution at Jupiter, so we can simply scale this by the inverse square law to 4700 lux or 50 $W/m²$ out there.

With a power input of 50 $W/m²$ you can multiply by the cross section area of the satellite, subtract what gets reflected or transmitted, and end up with the total power input to the system.

This is still not enough to determine temperature.

For any system at thermal equilibrium, the following equation holds:

$$P_{in} = P_{out}$$

Power in equals power out.

We have $P_{in}$ from irradiance and area. Heat is lost in one of 3 ways:

  • Convection: Not relevant in space since there's no medium to carry heat away
  • Conduction: Not relevant in space since there's no medium to conduct heat to.
  • Radiation

Since the only mode of heat transfer away from the satellite is via radiation, we can relate power to temperature with the Stefan-Boltzmann law

$$P_{out} = A\sigma T^4$$

Or in the order we want it:

$$T = \sqrt[4]{\frac{P_{out}}{A\sigma}}$$

Where $A$ is the surface area of the spacecraft, $\sigma$ is the Stefan-Boltzmann constant and $T$ is the surface temperature (in Kelvin).

This however comes with caveats:

  • Radiated power is assumed to just go away. But your spacecraft's geometry could make some of this radiation hit other parts of the spacecraft. This gets messy.
  • This assumes that the temperature is uniform over the spacecraft. This doesn't need to be the case since the irradiance is not necessarily uniformly distributed (night side of the Earth is colder than the day side). To accurately model the surface temperature you have to model internal convection and conduction too, and this is very messy.
  • The Stefan-Boltzmann law holds for perfect black bodies. Real materials may behave slightly differently.
  • To extract energy from all of this with a Stirling engine, you want your spacecraft to have a hot and a cold side. This means that modelling internal convection and conduction is something you can fundamentally not avoid. (You also have to deal with the efficiency of heat engines)
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