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I have just read 2001: A Space Odyssey and despite the fact that the author was quite optimistic about the space exploration capabilities, communication between a spacecraft and Earth was not possible while a planet was obstructing the line of sight. So, a Non-line-of-sight signal propagation was not possible.

I am trying to find out if there are any solutions (or at least promising ones) to solve such an issue and found about the neutrino-based communication. However it requires very large and expensive equipment.

Is it possible for a spacecraft to communicate with Earth when a planet is in the way? Or at least are there any promising solutions to solve this?

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    $\begingroup$ It would make far more sense to use a relay satellite. $\endgroup$
    – GdD
    Jan 6, 2022 at 16:19
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    $\begingroup$ We do not have non-line-of-sight communications over long distances. But we do have a cluster of relays around Earth that can relay, so the Earth "target" is much wider than just the rock of Earth is. And at Mars, for example, we also have relay satellites. Still, when Cassini went behind Saturn, it lost its link briefly. And when the Sun is between Earth and the probe, the interference can extend for several days to weeks. $\endgroup$
    – CuteKItty_pleaseStopBArking
    Jan 6, 2022 at 16:31
  • $\begingroup$ @PM2Ring I have removed it as it is a poor example. $\endgroup$
    – Alexei
    Jan 6, 2022 at 17:30
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    $\begingroup$ @RogerWood eek. conceptually, that could work. But you would need the sort of power levels that synthetic aperture radar need, and at interplanetary distances the power needs would be beyond scary. We are talking giga-to-terawatt pulses, at least. Not practical. $\endgroup$
    – CuteKItty_pleaseStopBArking
    Jan 9, 2022 at 6:40
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    $\begingroup$ @CuteKItty_pleaseStopBArking yes, probably better to wait for the planet to move out of the way $\endgroup$
    – Roger Wood
    Jan 9, 2022 at 7:50

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If the hindering planet has an atmosphere, refraction can be considered. You aim your radio beam(s) toward planetary limb and use it as a lens. On the receiver side across solar system you'll get much weaker but still potentially detectable signal.

Extremely low frequency band allows to communicate "around" planets. There are technical obstacles though. The transmitting spacecraft would have to unfurl huge antenna, with tethers thousands of kilometers long, and have very high power to be detectable over long distances. On Earth, the Earth itself and its magnetosphere can help as antenna. The bitrate is extremely low.

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Broadly, no.

On a planet, the equivalent mountain in the way would block most radio, all visual and any other electro-magnetic media.

On a planet with atmosphere, the mountain might have little effect on purely sonic comms… though circumstances alter cases. "Most radio" allowed for the several frequencies which could work by bouncing up and down the atmosphere.

In space, the planet will block all EM transmissions, including radio, and lack of atmosphere will rule out sonics, too.

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Just for curiosity, this is a back-of-the-envelope calculation of what would be required to communicate between Earth-Sun Lagrange points L1 and L2 by bouncing signals off the moon. (The Earth lies exactly between L1 and L2 and prevents direct communication)

Assume laser communications at 1 $\mu m$ wavelength with a 1 meter aperture so we can get a nice tight transmit beamwidth (and receiver resolution) of about 1 microradian. We're aiming at the Moon which is about 1.5E9 meters away from both L1 and L2. The Moon is 3.5E6 meters in diameter so it subtends about 2 milliradians which is a big target compared with the system beamwidths. Assume the albedo of the Moon is 0.1 and the reflection is isotropic. For an isotropic reflection the fraction of reflected photons entering the 1 meter receiver aperture will be roughly (1 meter/1.5E9 meters)^2 ~= 5E-19.

With these assumptions, 100% of the transmitted photons arrive at the moon, 10% get reflected and finally about 5E-20 of them reach the receiver. Say we'd like a million photons/second to establish a ~100 kbit/s communications path, we need to transmit 5E26 photons per second. A 1$\mu m$ photon is 2E-19 Joules, so we'll need a 10 MegaWatt CW laser, - give or take a few factors of $\pi$, etc.

10 MW is a very big laser to put in space! But a 1 kW laser to provide a slow telemetry link of a few bits per second might be quite practicable - though I'm still not sure why anyone would want to do this.

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    $\begingroup$ Unfortunately the wide distribution in path lengths will introduce a real challenge for your encoding scheme. If there are say a dozen dominant paths there are complex multipath algorithms that can help, but in this case I think you’ll be limited to the 300 to 1200 baud, VT100 + dialup range. $\endgroup$
    – uhoh
    Jan 12, 2022 at 5:57
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    $\begingroup$ @uhoh good point. The beam illuminates a 1.5 km patch on the moon,. That's about 5 microseconds. So transmission would start getting difficult above ~200 kbaud. But maybe 1 microradian is a bit optimistic? $\endgroup$
    – Roger Wood
    Jan 12, 2022 at 7:57
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    $\begingroup$ Oh, yes I didn't read carefully, for the Lagrange point geometry and a million wavelength aperture (sounds fine) you'll have a few km footprint only, and the path length differences will not be as large as the worst case scenario (fraction of a lunar hemisphere) I'd imagined. Now if you could just lay down a thousand square meters of aluminum foil... $\endgroup$
    – uhoh
    Jan 12, 2022 at 8:11
  • $\begingroup$ @uhoh It would be better to paint the Moon white. I don't think anyone would object. $\endgroup$
    – Roger Wood
    Jan 12, 2022 at 8:28
  • $\begingroup$ I'm thinking that near-specular reflection will give you a big gain boost and the smaller spread in path lengths will give more bandwidth (if not noise limited) but I have't applied any rigor to it so no idea if it works or not. Hard part is keeping the aluminum foil fairly flat, and slowly tilting to keep the spacecraft in one halo orbit as specular to the spacecraft in the other. $\endgroup$
    – uhoh
    Jan 12, 2022 at 9:58

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