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