What does the Sun-Earth-Moon system look like from the Sun-Earth L-2 point?

Question

The L-2 point of the Sun-Earth system is away from the Earth on the night-side of the Earth; i.e. it's always local midnight at the sub-satellite point. This is an attractive property for some spacecraft.

Illustration from Wikimedia Commons. Not to scale.

Fortunately for solar-powered spacecraft, the Sun is still visible, or those satellites couldn't be solar powered. But the disk of the Earth must be a sizeable fraction of the disk of the Sun (this should affect energy budgets too...). What does the Sun-Earth-Moon system look like from the Earth-Sun L-2 point? It should be extra interesting when Earth-based observers see a lunar eclipse. Have there been any photos of Earth taken from this point?

Answer 1

I treated this as a problem of geometry and came up with this:

• The sun is the large yellow disk.
• The earth is the largest black disk, obscuring most of the sun
• The left-hand dark-grey disk is the moon as it transits across the near-side of the earth, with respect to L2. In reality in this position, the disk of the moon would appear completely black. I made it dark-gray for illustrative purposes.
• The right-hand dark-grey disk is the moon as it transits behind the far-side of the earth, with respect to L2. In this position, the moon might not appear as a completely black disk - its possible it may be illuminated somwwhat by light reflected off the back of the earth. However this effect may well be minimal given the overwhelming power of direct light from the annulus of the sun.

My numbers look a little different to @PearsonArtPhoto's, so here they are:

Averages from wikipedia/google:

• Sun-Earth distance: 150 million Km
• Earth-Moon distance: 384,000 Km
• Earth-L2 distance: 1.5 million Km
• Sun radius: 696,000 Km
• Earth radius: 6370 Km
• Moon radius: 1740 Km

From these I calculated:

• Sun-L2 distance: 151.5 million Km

• Sun radius angle: 4590 μrad

• Earth radius angle: 4250 μrad

• Near moon-L2 distance: 1.12 million Km

• Near moon-L2 radius angle: 1560 μrad

• Far moon-L2 distance: 1.88 million Km

• Far moon-L2 radius angle: 924 μrad

Notes - my terminology may be inaccurate. Here are some clarifications:

• In all the above, I am using L2 to refer to the sun-earth L2, i.e. the location of the observer.
• radius angle is the angle subtended from the center of the given object to the edge of its disk
• Since angles are small, I used the approximation x ≈ tan(x)
• The sizes of the objects I have drawn are directly proportional to their calculated radius angles.

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Update:

I just saw this from NASA today. While this is taken from L1, and this question is about L2, these two Lagrange points are similar distances from the Earth (in opposite directions), so we can see that the proportions, at least of the Earth and moon, are more or less correct in my simulation.

Earth from L1 in photo (similar distance to L2, obviously lit differently though)

Answer 2

The spacecraft that orbit at L2 usually don't just stay at the L2 point, but do what's called a Halo Orbit, or related Lissajous Orbit. Essentially they orbit the L2 point, instead of right at it. As a result, they actually see the Earth and Moon as distinct from the Sun. And they usually orbit the point in such a way that they will not have a direct line from Earth to Sun, which makes communication much easier (The Sun produces a significant amount of RF noise)

As for the view, if at the exact L2 point, the Earth would show up as a disc about 0.48 degrees, and would thus actually be in partial eclipse perpetually, as the sun is a disc about 0.53 degrees from the L2 point. Thus, most (~82%) of the Sun would be covered all of the time by the Earth, with slightly more if the Moon was also covering the Sun. The atmosphere of Earth would also cover some, perhaps another few percent. This would be a significant annular eclipse. A typical annular eclipse on Earth of the Moon is over 90%.

Answer 3

A reasonably accurate¹ rendering can be made with Celestia. Here we are at 1,500,000 km from Earth's center, on August 26, 2018 (the next full moon as I write this), at approximately the Earth-Sun L2 point.

Earth mostly but not completely eclipses the Sun's disk. The Earth and Moon appear as completely black disks in comparison to the Sun's intensity.

Because it's a full moon, the Moon is closer to us than Earth in this image. It's barely visible as a small black disk near the bottom of the image.

Going forward a few weeks to Sep 9 we get to the new moon:

The Moon looks a little smaller due to the increased distance. We also see a few sunspots.

In Celestia, a lunar eclipse from this perspective doesn't look like anything, as the Moon is just a black disk atop another black disk. I suspect this is accurate: the illumination on the Moon is very faint relative to the Sun's disk which is directly visible, and besides at the L2 point we are on the wrong side of the Moon to observe even that.

Here's first quarter, Aug 16, 2018. I've enabled artificial ambient light so the dark side of the Moon is illuminated: otherwise it wouldn't be visible in this perspective. Earth is on the right, the Moon on the left. The brightest star near the middle is Regulus, and the fuzzy spot just above it is Leo I.

Zooming in on the Moon and disabling the artificial ambient light reveals a tiny crescent, with IC 613 at the top of the frame. This is as "full" as the Moon ever gets from this perspective.

I believe the Moon should be slightly illuminated indirectly by light reflected from Earth in this view. There's some talk of planetshine being implemented in 2008 in Celestia, but apparently it was either never committed or isn't working.

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¹ Celestia is very accurate with regard to geometry, but has its limitations. In some of these images we see stars and possibly deep space objects, the sun's disk, and sunspots at the same time. I'm doubtful any real imaging instrument would have sufficient dynamic range to generate such an image without compositing.