The Parker Solar Probe’s trajectory will take it within 8.5 solar radii of the sun’s photosphere. Its instruments, hiding in the shadow of the alumina-coated composite sun shield, will bask in 29 °C comfort, even without a cooling system. (Only exposed solar panels are cooled).

The JWST uses the same strategy (at an orbital distance of 214 solar radii) to attain a temperature of 27 K in the shade.

enter image description here

From sketchfab.com,

If a probe were equipped with radiators on the entire anti-solar surface, how close could a probe approach the sun?

Porous alumina has a reflectivity of 99.0% for visible light and 99.4% for IR. On the radiator side, there are materials with emissivity of 97.0-98.5%.

This means a spacecraft which is highly reflective on the sunward side and highly emissive on the anti-sunward side should come to thermal equilibrium somewhere between the solar surface temperature of 5800 K and the cosmic background temperature 2.7 K, a rather large range. Any idea how to calculate this equilibrium temperature for a given solar distance?

As an example, this cube-shaped spacecraft has a 0.414 solar radii perihelion. It has heat pipes to keep the interior temperature the same as the radiator panels. The sunny side is 99% reflective and the radiators 98% emissive. What would be the temperature of the interior?

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Or, conversely, beyond what perihelion could the interior temperature be compatible with living astronauts? Functioning space-hardened electronics?

  • $\begingroup$ Could you clarify what you would consider "active cooling" and "passive cooling" in a spacecraft context? $\endgroup$
    – Dragongeek
    Commented Jan 2, 2023 at 16:01
  • $\begingroup$ @Dragongeek ... "active"=consumes power or expends mass $\endgroup$
    – Woody
    Commented Jan 2, 2023 at 16:29
  • $\begingroup$ Are we allowed to postulate a sun-shield that has a larger radius than the sun itself? $\endgroup$
    – MikeB
    Commented Jan 4, 2023 at 16:50

1 Answer 1


This simple example shows the effectiveness of a sunshade. Admittedly, a spherical black body may not be the best design for a sunshade, but it makes the calculation easy. In cold space, the temperature a cooler black body goes as the 4th root of the solid angle of the hot black body to which it is exposed. Cascading a calculation from the Sun to the sunshield and then to the spacecraft gives 10 solar radii as the minimum distance assuming 300K for the spaceraft.spherical black sunshield

[old answer]
An isothermal conical spacecraft provides another interesting example. Assume the blunt end is highly reflective (~99%) and faces the sun. The pointy end is very black (~99%) and is designed to just fit nicely inside its own shadow. Taking the sun as 6000K and assuming 300K as a comfortable temperature for the occupants, then it seems we will be able to orbit at about 18 solar radii. hot pointy spacship

  • 2
    $\begingroup$ A missing term in the calculation is the emissivity of the sunshade. Some alumina ceramics have low absorbance but high emissivity in IR. Since the sunshade is the hottest part of the spacecraft, it can emit a substantial portion of the absorbed flux. Can't do the math. $\endgroup$
    – Woody
    Commented Jan 2, 2023 at 4:44
  • $\begingroup$ The circle area is small compared to conical area. PS. You're reading far too much into this, @Woody. It's just a chunk of conductive metal with silver paint on the blunt end and black paint on the pointy end :-) You can presumably do much better if you put some structures in to prevent/control heat flow - multiple sunshades like Webb. $\endgroup$
    – Roger Wood
    Commented Jan 2, 2023 at 6:01
  • $\begingroup$ The logic of your answer makes good sense. But the result of the calculation doesn't match up with the Parker Probe. Parker is 8.5 radii from the sun, but is about the same temperature as your example which is 2x further away. I was searching for an explanation. $\endgroup$
    – Woody
    Commented Jan 3, 2023 at 8:05
  • $\begingroup$ @Woody the Parker probe cheats by using a sunshield (I modified the answer) $\endgroup$
    – Roger Wood
    Commented Jan 3, 2023 at 21:32

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