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This answer mentioned thermal cycling made me think of this question: Are there any points in the solar system, such as Lagrange points, where a spacecraft could reside in perpetual shade, protected from heat and perhaps even from the worst of solar radiation damage? I mean specifically in space, excluding the interior of solid bodies.

L₂ would be a candidate, however, Wikipedia points out:

It is, however, slightly beyond the reach of Earth's umbra, so solar radiation is not completely blocked.

Are there any points in the Solar system where solar radiation would be completely blocked?

Orbits would be fine, too, but I can't imagine how any orbit could be in perpetual shade.

Edit: in an earlier version, this question mentioned stable points in particular, but I realised later I don't mean the technical definition of stable at all

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Great question! Outer planets' L2 points could be candidates. Large planets and long distance to the Sun should help put L2 in the shadow. Their large mass however puts the L2 further away, if I understand it correctly. And the moons maybe disturb it. Not so easy to find data on outer planets' L2 points. – LocalFluff Feb 3 '14 at 12:19
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Or go to 50,000 AU, and arguably everywhere is in perpetual shade ;-) – gerrit Feb 3 '14 at 12:22
up vote 14 down vote accepted

To the extent of my creativity, L2 is the only location that can satisfy this for a given planet. We would also need to constrain the discussion to planets because, again, my creativity fails me to think of any other option.

Thus, we must use the method described in the other answer to ask "does any planetary L2 fall within the planet's umbra?"

To start with, the location of the L2 point can be approximated by the following, where M1 is the sun's mass and M2 is the planet's mass. R is the distance of the planet from the sun. r is the distance of the L2 point from the planet.

$$ M_2 \ll M_1 \\ r \approx R \sqrt[3]{\frac{M_2}{3 M_1}} $$

So for L2 to be close to the planet, we would want the planet's mass to be small. However, a smaller planet has a smaller radius. Let's move on to calculating the termination point of the umbra. This can be done with geometry, using similar triangles defined by the end of the umbra with the planet and with the sun.

$$ r_u \approx R \frac{ D_{planet} }{ D_{sun} } $$

Our basic criteria for an eternally shaded L2 point is then simple, that $r>r_u$ in the above equations. Here are my calculations of the ratio between the two values using planetary data:

  •  MERCURY  0.920628441
  •  VENUS  0.931131053
  •  EARTH  0.916884469
  •  MARS  1.02662275
  •  JUPITER  1.505844946
  •  SATURN  1.897836869
  •  URANUS  1.505423974
  •  NEPTUNE  1.382215047
  •  PLUTO  1.322002729

For all values above 1, the L2 point is fully shaded. If the value is very close to 1, there are likely to be some other complications, since you might not be in exactly that point, so some sunlight will still get through. You'll also still need some station-keeping at that point to hold in place. Also, note that a space elevator is possible (in theory) for a point like the sun-Mercury L2 point, but the counter-weight would be further away than the L2 point itself, meaning that it would not be perfectly shaded.

I repeated these calculations for planet's perihelion and aphelion. They were exactly the same, so there must be some definitional reason why (unless I messed it up). But managing the L2 point within an elliptical orbit is a complicated maneuver, so this couldn't be so simple in real life.

But otherwise, the conclusion is pretty clear. Once you get out of the inner solar system, the sun-planet L2 points lie fully within the planet's umbra. I lack a general technical explanation for why this should be. I just used the data.

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Given that mass is density times volume, i.e. Proportional to the third power of mass, it all boils down to: the density of he planet must be low. Lower densities mean a wider radius, and hence a larger shadow. – oefe Feb 3 '14 at 21:21
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@oefe I think the better observation is that the distance from the Sun dominates this equation. Rocky planets exist on both sides of the "1 line," but all planets beyond Earth have ratio > 1. As you get further from the Sun, it is better approximated by a point source, and is therefore easier to mask with a planet of any density. – codeMonkey Aug 21 '15 at 17:55
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I agree with @oefe , if you sort them from lowest to highest ratio, you get the same list as if you sorted them from highest to lowest density. – Hohmannfan Dec 23 '15 at 14:36

Other than Lagrange points (since you asked "such as Lagrange points"), another option would be to enter a retrograde orbit around a moon with the same period as the prograde orbital period round its planet.

In the case of the Earth/Moon system; go "backwards" round the Moon and take 27⅓ days doing it.

You can work out the distance your satellite would be from the Moon with this handy equation: From https://en.wikipedia.org/wiki/Orbital_period

Where:

  • G = 6.67 x 10-11 Nm2kg2
  • ML = 7.34 x 1022 kg
  • T = 2.36 x 106 s (27⅓ days)
  • a = distance (m)

Turns out you need to be 88,436 km away. This will be in shade as we know the umbra extends to the Earth (384,400 km) during a total eclipse.

I'm not sure how stable this would be when it passes between the Moon and Earth. It may be stable enough for long enough for the mission and it's less of a trek than Mars L2 ... use the fuel saved to keep it in orbit. The Earth/Moon neutral point (where the gravity of each cancels out) is 70,000 km from the Moon.

Or course, there may be other systems where the numbers work out better ...

For example Phobos: - G = 6.67 x 10-11 Nm2kg2 - MPh = 1.066 x 1016 kg - T = 27,554s s (7 h 39.2 min)

This works out at 23.9 km ... better, but Phobos' gravity is so weak, I'm sure it'd only be stable for ¾ of a month. For Earth's moon, that's a 3 week total eclipse which some astronomers would get excited about.

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Nice idea, but I worry the gravitational influence of Earth would render this orbit very instable. – gerrit Mar 9 at 10:30

First thought. We need to compare the radius of the L2 from object centre to the length of the Umbra. The Umbra is determined (partly) by diameter of the object while the mass determines the L2 radius. So mass rises with the cube of object radius, while area is a square function of radius. So, density is be key. see http://www.merlyn.demon.co.uk/gravity4.htm#Ap1

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