# Are there any (Lagrange) points in the Solar System in perpetual shade?

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 at 12:19
Or go to 50,000 AU, and arguably everywhere is in perpetual shade ;-) –  gerrit Feb 3 at 12:22
Just noticed: the question specifies "stable" points, and L2 points are not stable. So that makes it kind of trivial to answer "no" with those qualifiers. –  AlanSE Feb 3 at 14:53
@AlanSE Hm, oops. I think I didn't really mean stable. I will rephrase. –  gerrit Feb 3 at 15:15

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 at 21:21

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 diameter 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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This doesn't really answer the question, nor I believe it was meant to. Would you like it converted into a comment, or will you expand on it? –  TildalWave Feb 3 at 14:13
@TildalWave This answer reflects a correct method (and probably the only method) to accurately answer the question. It only lacks a final step of tabulating/calculating. I empathize with your point about comments, but my take is that anything novel with direct bearing on the answer should be an answer, because comments can be purged by moderators. An answer is the only way to preserve the commentary. –  AlanSE Feb 3 at 14:31
@AlanSE Since he started with "first thought" and doesn't have rep to post it as a comment, I thought to offer mod services if needed, nothing more :) –  TildalWave Feb 3 at 14:41
I have no rep! (or time to do the maths) However, thanks to AlanSE. I think r(L2) should be < r(u) to have L2 in shade? –  Hunter Feb 3 at 17:10