# Is there a way for a permanent Lunar eclipse, or another Object to always be in Earth's shadow?

I have a question regarding worldbuilding and haven't found a question like this yet, so I created an account and hope that someone can answer it :D

As a starting point, imagine our real-life Sun, Earth, and Moon.

1. Is it possible for the Moon to always be in Earth's shadow, to basically create a permanent Lunar eclipse?

2. Which Values (for example Earth-Moon-Distance, Moon's speed around the Earth, Moon's Mass, etc. (sorry for the non-scientific words, I'm not good at astronomy and physics)) would have to change for it to be possible?

3. If it is not possible for the actual Moon (or a Moon with different Values) to always be in Earth's shadow, would it be possible to have another Moon-like or Planet-like object that orbits the Sun to always be in Earth's shadow?

4. How big, small, dense, close earth could that other object be to be stable in Earth's shadow?

EDIT: As I understand the answers so far, basically my idea would not be possible in the real world. I know this forum is not really for discussions, but I had another idea: What about, instead of a circular Moon orbit with Earth at the center, an oval-ish Moon orbit that has its center somewhere in Earth's shadow.... something like: (SUN)--------(Moon)-(Earth)-----(Moon) Oh dear, I hope you can understand what I mean, it's not to scale of course. My new question is basically, would a Moon orbit be possible where the moon is like more than 90% (maybe even like 99%) of the time "behind" the Earth? I guess there would be a lot of implications, tides come to my mind and I am sure you smarter people can think of a lot more implications... but the basic question stands if it would be at all possible.

If that is not possible as well, then I guess I will have to give the Moon some kind of thrust, that seems to be the only way to give it a stable "orbit" in Earth's shadow, if I understand the answers correctly.

• The shadow of the Earth has a conical shape; it extends only over a relatively small distance. (That's because the Sun is so much larger than Earth.) An object more than 1.4 million km (870,000 miles) away will never be in Earth's shadow; for comparison, that is about 1% of the distance between the Earth and the Sun, or about 3.5 times the distance between the Earth and the Moon, or 2.7% of the absolutely smallest possible distance between Earth and Mars. – AlexP May 25 at 22:51
• @AlexP please don't use comments to post answers. – Trioxidane May 26 at 5:43

It is not possible for an object to be in the shadow of Earth permanently - although it could be in the semi-shadow called penumbra, the full shadow (umbra) does not extend that far. There is a special place called the L2 Lagrange point that allows an object to orbit around the Earth at about the same rate as the Earth orbits around the Sun - thus always remaining in the same relative place in terms of shadow.

However, L2 is not a super stable point to orbit in, and thus objects tend to fall away. This is not a problem for artificial satellites, though, and the James Webb Space Telescope, when launched, will take advantage of this constant half-shadow to make better celestial observations.

So:

1. Yes, it is possible, although unlikely. This "moon" is more like the Death Star, and will need some kind of course correction to stay in orbit. This could potentially be "explained" by matching the moon diameter and the umbra diameter, meaning if it started to move out of position the sun would warm part of the surface, leading to off-gassing pushing the moon back into place. Probably not 100% scientifically valid, but otherwise this needs to be a spaceship, not a moon.

2. To calculate the L2 point: $$d_{Earth-L2} = d_{Earth-Sun}\sqrt[3]{\frac {M_{Earth}}{3M_{Sun}}}$$ Where $$d$$ is respective distance, and $$M$$ is the respective mass. Note the cube root.
To calculate the Umbra Diameter at the L2 point:

• First calculate the umbral distance: $$d_{umbra}= \frac{d_{Earth-Sun}}{\frac{r_{Sun}}{r_{Earth}} - 1}$$ Where $$r$$ is the respective radius of the body.
• Then, if the umbra distance is more than the L2 distance, you can calculate the size of the umbra (max size of your moon) with this equation:
$$r_{umbra}=\frac{r_{Earth}}{\frac{d_{umbra}}{d_{umbra}-d_{Earth-L2}} + 1}$$

For the non-mathematically-inclined, I combined this all into a Google Spreadsheet, make a personal copy to edit.

1. So, the Lagrange point kind of orbits both.

2. Normal densities (3,000-8,000 kg/m^3) should work. Mass of the object at the L2 point does not matter significantly. However, some kind of automatic correction will be necessary.

• L2 is not fully in Earth's shadow astronomy.stackexchange.com/q/13585/26216 – BMF May 25 at 20:43
• @BMF Totally forgot about the Umbra, I'll do some calcs on a spreadsheet and make some edits. – IronEagle May 25 at 20:53
• Thanks! I have edited my original question and added another approach to my idea, I would appreciate it if you gave it a go as well. I have one more idea that involves a geo-centric system, but I think that's too far removed from the original topic and requires a different question... – Arimeris May 25 at 22:34
• @Alexander - The shadow of the Earth is so small, staying at the L2 point actually makes the most sense for the greatest % of darkness. The problem is that it is a "saddle", not a stable "cone". I'm doing some more research, but the "same size, light exposure pushes it back on track" idea is probably enough to make it plausible. – IronEagle May 26 at 1:54
• Weird idea: Could the planet and moon be in some stable configuration between two + large bodies? Say a star and a super-Jupiter or small black hole? The star and black hole could orbit each other, with planet/moon wedged between. No idea how to do that math, though. If @Nuclear Hoagie has an idea, would love to hear what you all think. I'm creative, but a biologist. – DWKraus May 26 at 22:46

Not possible with current configuration

To have an object perpetually in the earth's shadow, it must circle the earth at the same rate that the earth circles the sun, in order to keep the earth between it and the sun. To have a satellite with an orbital period of 1 year, it needs to be very far from the earth, approximately 2.1 million kilometers away (according to this orbital calculator). Unfortunately, that distance is outside the Hill Sphere of the earth, which is the region in which the force of the earth's gravity is the dominant force. At a distance of 2.1 million km, an object is simply not orbiting the earth anymore. Even if it was, the earth's umbra (region of total shadow) only extends 1.4 million kilometers, so an object at such a distance would never see the sun completely occluded by the earth.

You could put an object at the L2 Lagrange point, which would allow the object to orbit the sun with the same period as the earth, even though it has a greater orbital altitude than the earth. Unfortunately, the L2 point is unstable, and it exists at about 1.5 million km away from the earth - too far for the earth's umbra to reach. Even if you could get an object to stay at the L2 point, it would never be in complete shadow from the earth.

I'm not sure how you could achieve such a configuration by modifying the solar system. If you increased the mass of the sun significantly, you could shorten the earth's year while staying at the same distance from the sun, which would in turn allow the eclipsed satellite to also have a shorter period and a lower orbital altitude, although the Hill sphere of the earth gets smaller as the sun's gravity increases, so the math might not work out.

• You could also increase the physical diameter of Earth by 9%, making the umbra end at L2. – BMF May 25 at 21:00
• @BMF - although the moon would then need to be very small, as the diameter of the umbra would not cover it. – IronEagle May 25 at 21:07

What you're trying to do is maintain a satellite in Earth's shadow while it travels around the Sun. You can't have the satellite orbiting the Sun with the same orbital period as Earth and not be in Earth's orbit. (Unless you're at L2 [or any other L-point, but only L2 is "behind" Earth], which is just barely outside Earth's umbra and unstable in the long-term anyway.)

The only solution I can see is that the satellite needs to be on a powered trajectory, perhaps a circular orbit "behind" Earth, in between Earth and L2 and within Earth's umbra. Such an orbit can't exist naturally and would need constant thrust to maintain.
If the Object is alien in origin, this can probably just be handwaved away.

I think other answers are on the money, but we can play with the values a bit to make it work.

So the Earth's umbra falls just short of an object at the Sun-Earth L2 point. Now what if the "moon" is the big object? you'll want to see it like a moon from Earth even though it'll be much farther away than the actual Moon. So the moon is a large planet, and the Earth is a smaller planet at the Sun-moon L1 point between the Sun and the moon. This is also an unstable configuration, but handwave that for now. You can make the Sun a smaller star and move the whole system closer together to keep Earth habitable, extend the umbra, and move the moon deeper into the umbra.

Tada! The stability thing is still a problem, only the L4 and L5 points in these three body systems are stable (Jupiter has it's own pseudo asteroid belts in those spots called Trojans). However... you can probably just ignore that issue if it's problematic to the setting. It's not like Larry Niven's Ringworld is super-accurate from a physics perspective.

• The star is going to be much larger than the planet, so the umbra shrinks as it gets further from the planet that casts the shadow - it's not possible for a smaller planet to cast a larger one completely in shadow. If you have Sun - Small Planet - Large Planet, the small planet's shadow is nowhere big enough to cause a complete solar eclipse from the large planet. A planet can't cast a shadow bigger than itself (under the normal conditions that it is smaller than its local star). – Nuclear Hoagie May 26 at 13:19

## Cover it in corner reflectors

Others have covered why a permanently eclipsed Moon is hard, requiring major changes to the Earth-Moon system. But the Moon could be in its own shadow if it returns all the light striking it to sender. I don't think I can post animation here, but the second image at the Wikipedia article on corner reflectors illustrates how that works. However you move the light beam, it strikes three perpendicular mirrors and goes back exactly the way it was sent. Think of markers on the road at night.

(We actually have a few corner reflectors on the Moon that send laser light back to astronomers so they can measure its precise distance. But not enough to change the color of the entire surface as of yet!)

Such a moon might never reflect the Sun, or at least, only when it is nearly full if there is some imprecision (in which case it would be nearly as bright as the Sun for a bit). Most of the time, you would see a Moon the approximate shade of your own patch of Earth, dark at night, brighter during the day, always featureless and homogeneous. If the reflectors are precise enough, on Saint Patrick's Day zealous folks would try to get everyone in the city to display green lights until the Moon turned green.

• Interesting idea, however re your last point about St Patricks Day - no, see mandatory xkcd what if for why: what-if.xkcd.com/13 – KerrAvon2055 May 26 at 22:40
• That cartoon illustrates what happens if the red light strikes the lunar surface and is scattered in all directions. Nonetheless, NASA can see a laser strike a corner reflector on the Moon, because that sends the entire light signal back to the source. – Mike Serfas May 27 at 11:25
• Interesting, I like this. A "natural" source for corner reflectors would be some kind of spherical glass beads, which could form if the moon has some kind of wind or something. This would lead to the moon always being illuminated with a little bit of Earthshine, but a lunar eclipse would be terrifying - the moon would illuminate like the sun as it got close! – IronEagle May 27 at 18:30