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Could large craters on the moon be used as reflective lenses for radio signals?

Acting like a large radio telescope reflecting radio waves to a satellite positioned over the crater.

enter image description here

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Interesting idea. I think the answer is both yes and no -- yes with a manufactured dish but no in the crater's raw state.

The Arecibo telescope sits in a natural crater, but adds a dish which has a couple of important things required by a radio dish:

  1. a radio-reflective surface
  2. a specific curvature, classically parabolic, but also shaped
  3. low surface roughness typically in the order of mm to µm for radio frequencies we're interested in. See http://en.wikipedia.org/wiki/Ruze%27s_Equation
  4. a receiver close to the reflective surface (focal length), a satellite would probably be beyond the focal point
  5. probably some some other things, but that's probably enough.

Edit: Added #4

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    $\begingroup$ Could rovers feasibly resurface a naturally nearly spherical Lunar crater to an Arecibo type telescope, by moving the Lunar soil around a bit and then burning it to a precision glassy surface by focusing Solar light on it at incredible heat? Does that sound as a possible technology this century, or is it conceptually physically just wrong? Does it make more sense to just fold out or build our own parabolic dishes from scratch over there? $\endgroup$
    – LocalFluff
    May 5, 2014 at 16:57
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    $\begingroup$ Actually the Arecibo dish is spherical, not parabolic, because it's steered by moving the detector, not the dish. Reference: en.wikipedia.org/wiki/Arecibo_Observatory $\endgroup$
    – Keith Thompson
    May 6, 2014 at 22:15
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    $\begingroup$ LocalFluff -- another interesting idea, but it may be easier to just install a pre-fab surface. Have a look at Spektr-R. $\endgroup$
    – user1563
    May 9, 2014 at 0:05
  • $\begingroup$ A satellite couldn't be used. There are no Luna Stationary orbits. There are the Earth-Moon Lagrange points, but all are much to high, and inconveniently positioned. The receiver would be part of the telescope, just as at Arecibo. A satellite could be used as a relay, to transmit data back to Earth. $\endgroup$
    – James K
    Sep 20, 2015 at 7:55
  • $\begingroup$ Arecibo does not sit in a crater, it's a sink hole, and irrelevant to the question, which specifically mentions [impact] craters on the moon, and "their" suitablility. $\endgroup$
    – tahwos
    Dec 16, 2018 at 20:20
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Could large craters on the moon be used as reflective lenses for radio signals?

You'd have to line the surface with something reflective to microwaves, like a metallic mesh, or similar materials.

Secondly, the shape of the crater is probably not quite ideal, so it would have to be adjusted a little, carved up a bit in various places. But it's a good start, and definitely better than starting with a flat ground.

There is also the question of stability - you need to make sure that whatever changes you make (carving a different shape, lining it with mesh) do not affect the stability of the crater, or else various parts may slide or collapse. This is an engineering problem.

Acting like a large radio telescope reflecting radio waves to a satellite positioned over the crater.

Not possible unless the crater is exactly on the equator, and even then it would be tricky.

But a crater like the one in your picture is so strongly curved, the focal length is about the same as the diameter. In other words, if the diameter of the hole is X, the altitude of the receiver is pretty close to X - give or take something like 50% or so, depending on the exact curvature. It might be easier to just build a giant arch over the crater. Again, this is a matter of engineering.

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If it works for Arecibo it could be made to work on the Moon, Ceres, and maybe even Mars. The point about sending a prefab is a good one. And using two radio telescopes in tandem separated in space would allow for astronomers to take advantage of the improved angular resolution this affords provided they were arranged so as to allow them to simultaneously focus on the same point. That would be a neat problem in astrophysics! Radio telescopes aimed at the same point simultaneously from the Earth and Mars while they are on opposite sides of their orbits would have incredible resolution.

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Yes, Frank Drake looked into this - see page 91 here http://www.lpi.usra.edu/lunar/strategies/objectives/ast_observations_moon.pdf

He calculated that you could make a telescope with diameter 30 km using steel, and up to 60 to 90 km with stronger materials. And no problem with wind loading on the Moon.

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I agree that the crater needs to be lined, but you also have the problem of maintaining a satellite in stationary orbit above the crater. Nearly impossible unless the crater is on the equatorial plane. Also, a stationary satellite around the moon would be influenced by the Earth, so you would need to burn fuel to keep the satellite in position.

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  • $\begingroup$ "a stationary satellite around the moon would be influenced by the Earth" - if the orbit is close enough to the Moon, the orbit would be stable. Look up the notion of Hill sphere. Orbits inside the Hill sphere tend to be stable. $\endgroup$
    – Florin Andrei
    May 6, 2014 at 0:16
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    $\begingroup$ @FlorinAndrei The moon only rotates once every 29 days, that's a pretty high orbit. I think it would easily be outside the hill sphere. $\endgroup$
    – LDC3
    May 6, 2014 at 1:53
  • $\begingroup$ A calculation on Wolfram|Alpha says that selenosynchronous orbit would be 92 040 km from the Moon's centre. This seems to be about the same as the size of the Moon's Hill sphere (if I've understood Wikipedia's formula correctly). $\endgroup$
    – Warrick
    May 7, 2014 at 6:03
  • $\begingroup$ @Warrick I get the Hill Sphere to be 61530 km and the stationary orbit at 88460 km. That means the satellite would be in an unstable position. $\endgroup$
    – LDC3
    May 8, 2014 at 1:35
  • $\begingroup$ @LDC3: I'm not familiar with Hill spheres so I might be misapplying the formula. There seems to be a factor of $(1/3)^{1/3}$ in the first formula but not the last. Either way, I'm on your side that selenosynchronous orbits would need effort to maintain. $\endgroup$
    – Warrick
    May 8, 2014 at 5:40

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