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Imagine I put a floating probe inside the subglacial ocean of Encelado or Europa: how much power should my radio have to be able to communicate from external surface with the probe? Or, in different words, how much attenuation do 100 km of solid ice cause to a radio signal at, say, UHF frequency?

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Plan on packing a decent-sized reactor. –  Ignacio Vazquez-Abrams Apr 13 at 9:29
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A couple of alternative approaches come to mind. Drag an optical fiber behind the probe. Acoustic communication which can also act as a backup in case the fiber breaks. –  Nick Alexeev Apr 13 at 15:45

2 Answers 2

I can't answer that directly but Nasa are probing Greenland's ice sheets with aeroplane radar to find the depth of the bedrock. Here's what they say about ice and radio waves: -

Ice, on the other hand reacts differently depending on the radar's frequency. It reflects high-frequency radio waves, but despite being solid, lower frequency radar can pass through ice to some degree. This is why MCoRDS uses a relatively low frequency—between 120 and 240 MHz. This allows the instrument to detect the ice surface, internal layers of the ice and the bedrock below. "To sound the bottom of ice you have to use a lower frequency," said John Paden, CReSIS scientist. "Too high a frequency and signal will be lost in the ice."

This came from here and it's interesting to note that this is radar and requires a reflection from the bedrock to pass back through the ice to the receiving aeroplane. I would imagine that the reflected power is a fraction of the incident power reaching the rock so maybe you might get 10x this distance thru a solid ice sheet with a one-way transmission.

Here's the sort of image they are getting: -

enter image description here

It looks to me that +3km is possible with radar. I don't know what the radar beam angle is so it's impossible to calculate what the incident power at the surface of the ice is - the transmission from the aeroplane might be a 1MW pulsed radar with a very tight beam angle producing an incident power at the top surface of the ice of hundreds of watts. Also, the reflection from the bedrock will not be a tight beam - this means the power reflected back will be spread thinly as distance increases (see Friis equations). Also the power received at the aeroplane will be much smaller than that emitting from the surface of the ice - again see the Friis equations.

Addendum

I had a think about the link loss for the radar application: -

  • Link-loss from plane to ground. A 2m diameter dish will have a gain of \$\frac{\pi^2 D^2}{\lambda^2}\cdot 0.6\$ = 3.35 or about 10.5 dB. If the aircraft is 1km above the ice, the link loss to an identical dish (on the ice) will be -21 (antenna gains) +32.5 + 20log(MHz) + 20log(km) = 11.5 + 46 + 20 = 78dB link loss. If the radar output power is 1 MW (90 dBm), the received signal at the ground/icecap surface will be 12 dBm (16 mW).
  • It's the same problem for the reflection signal. At the surface, it is subject to the same attenuation up to the aircraft (78dB) that is 1km higher.

These losses won't be encountered by a simple transmission thru ice - transmitting and receiving antennas are sited either in the ice or at its surface. This all bodes well for being able to transmit in a single direction through large distances of ice.

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Assuming that it behaves similarly to water ice on Earth, there were some measurements done of the RF attenuation of the Ross Ice Shelf in Antarctica. The attenuation length was found to be 300-500m for frequencies from 75MHz to 1.25GHz.

(Attenuation length is the distance for the signal to drop to 1/e ~=0.368 ~= -4.3dB, somewhat analogous to time constant)

That's going to be a pretty intimidating amount of attenuation for 100km thickness (something like -950dB). Ain't gonna happen.

The power would, of course, depend on the bandwidth of signals that need to be transmitted.

To put it in perspective, the record for moon bounce communication is something like 3mW transmit power (~ -300dB attenuation). If we had 1GW, that would be another 115dB, but still well short of what's required.

enter image description here

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I'm unsure what "attenuation length" actually means. –  Andy aka Apr 13 at 17:45
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Edited.. or is it a deeper question? –  Spehro Pefhany Apr 13 at 18:03
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No, nothing deeper. Interesting article and I'm still trying to work out what the results mean - it seems they're trying to infer distance by looking at reflections. Maybe you have a better take on this. It does seem to contradict the NASA results in my answer and I'm genuinely scratching my head to the large disparity. –  Andy aka Apr 13 at 18:26
    
How do you "pass" from attenuation dB to needed power? (e.g. from 115 dB to 1 MW) –  jumpjack Apr 14 at 7:09
    
dB is 10*log(x/y) where log is base 10 and x/y is a ratio of powers. If x = 1 gigawatt and y = 0.003W, that's about 115dB. –  Spehro Pefhany Apr 14 at 8:21

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