If you were 10,000 light years away from all radio stations and digital broadcasts from our world, what equipment would be necessary to detect and capture some information like a voice or a photo? If we found a similar planet in the galaxy, what could we do to listen to their radios?
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2$\begingroup$ Our technology wouldn't allow for it, and I doubt any alien technology 10,000 light years away could do that either. We haven't been broadcasting for 10,000 years, and radio waves travel at the speed of light. There are systems close enough to receive broadcasts from our past, but they would need equipment that could receive and decode those signals. $\endgroup$– Phil N DeBlancCommented Aug 2, 2019 at 7:52
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2$\begingroup$ Our radio transmissions haven't had time to travel 10,000 light years yet so it would be impossible to detect them. Now if the aliens were 100 light years away, they could use something similar to the square kilometer array to detect us. $\endgroup$– SurpriseDogCommented Aug 3, 2019 at 19:57
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3 Answers
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As the above commentator says, our technology isn't sufficiently sensitive to pick up everyday radio broadcasts from so far away, and even if we could it wouldn't be worth sending them a message. The earliest we could hope to receive a reply would be in 20,000 years time, by which time there might be no humans left on Earth to receive it, as a result of nuclear war or some other catastrophe. If this extra-terrestrial civilisation were much, much nearer, tens of light years rathe than thousands, prospects would be very much better, but it is extremely unlikely that an advanced alien civilisation is as close as that.
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For more fun, let's posit an experiment 10 000 years from now. I found some forum which provided this data for a transmitter:
The Taldom transmitter is a large facility for longwave and shortwave broadcasting located near Taldom, Russia. It transmits on two longwave frequencies, on 153 kHz with 300 kW and on 261 kHz with a power of 2500 kW, the latter is the most powerful broadcasting station in the world.
As a first-order model, let that 2500 kW be transmitted via a reflecting disk with some modest spread angle, say 0.5 degree cone angle (roughly the full moon), which is equivalent to 6E-5 steradian. The area subtended at any distance D is thus $ D^2 * 6E-5 $ square meters (if D is in meters). 1E4 light-years is 9.5E19 meters, so the subtended area is $ (9.5*10^{19})^2 * 6*10^{-5} = 5.4*10^{35}\space m^2 $ . This leads to $ 4.6*10^{-30} W/m^2 $
thanks to SteveLinton for pointing out my math error
I don't know what sort of antenna size would be required to collect enough signal to overcome electronics noise, let alone galactic background noise (which presumably we will use a PLL or some such to remove); hope some other gallant soul will help me out here.
answered Aug 2, 2019 at 16:51
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1$\begingroup$ The area over which the signal is spread is $(9.5\times 10^{19})^2 \times 6\times 10^{-5} m^2$ (square of radius times solid angle, so the flux is more or less $4.6\times 10^{-30} W/m^2$). Assuming a spoken voice signal with a $2.4 kHz$ bandwidth that is about $2\times 10^{-7} Jy$ (see wikipedia on the Jansky). This seems to about 1000 times too faint for the Square Kilometer array to even detect its presence (over a one hour observation) let alone recover the signal, so something pretty huge would be needed to listen in. $\endgroup$ Commented Aug 2, 2019 at 20:58
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1$\begingroup$ @SteveLinton Thanks, I put the R^2 into my answer. $\endgroup$ Commented Aug 4, 2019 at 21:59
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The book of Yosif Shklovskii and Carls Sagan, "Inteligent life in the universe" gives a formula for how far a signal could travel and still be detectable by an antenna, which is influenced by a certain degree of noise ( the signal should be larger that the noise (antenna temperature)). $$ R\le\frac{W^\frac{1}{2}}{k^\frac{1}{2}}\frac{\pi}{4}\frac{d_1 d_2}{\lambda}\frac{\left(\tau \Delta f \right)^\frac{1}{4}}{T^\frac{1}{2}}, $$ where $\tau$ is the signal accumulation time $d_{1},d_2$ are the diameters of the receiver/transmitter antennas, $k$ is the Boltzmann constant, $\lambda$ the wavelength of the signal, $W$ is the power of the transmitter per 1 Hertz, $\Delta f$ is the frequency window of the receiver.
They argue, that in the cm range, the only antenna temperature is due to internal electronic noise, so $T=50$ degrees C. They also give as an example $\Delta f=10^4$ and $\tau=100$ s.
