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In a science fiction movie some aliens come to earth and when asked why, they say they received our radio transmission and came to investigate, the radio transmission being the first television broadcast (the Berlin Olympiad in 1936). So, the idea is that it took decades for the signal to reach them.

I am wondering how attenuated such a signal would be after traveling for 50 years in space, assuming the transmitter was about 100,000 Watts?

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What you are looking for is called free-space path loss.

Let's assume the signal is broadcast from a spherical source (e.g. the signal isn't sent with a directional antenna). Neglecting all other sources of loss (e.g. diffraction, reflection), the simplest way to calculate this loss is $\frac{1}{4\pi r^2}$.

The signal must be received by another antenna as well; therefore we must also take into account the aperture efficiency. This describes the ability of an antenna to pick up a signal, which is dependent on wavelength, and given by the equation $A_{eff} = \frac{\lambda^2}{4\pi}$. A wonderful derivation can be found elsewhere on this site.

All together, this gives us the free-space path loss (FSPL):

$$ FSPL = \left(\dfrac{4\pi r}{\lambda}\right)^2 $$

Where $r$ is the distance between transmitter and receiver and $\lambda$ is the wavelength of the signal.

Useful resource: http://www.radio-electronics.com/info/propagation/path-loss/free-space-formula-equation.php

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If we ignore dissipative losses, then the intensity of the signal will go as: $$ I = \frac{P}{4 \ \pi \ r^{2}} \tag{1} $$ where $I$ is the intensity [e.g., W m-2], $P$ is the power [e.g., W] at the source and $r$ is the distance [e.g., m] between the source and the observation point. If we also ignore Faraday rotation and assume the radio waves propagate at the speed of light in vacuum, then the result is relatively trivial to calculate.

Let's ignore leap years for the moment, thus the time between the initial signal creation and measurement by the aliens is 50 years = 18,250 days = 438,000 hours = 1,576,800,000 seconds. In that time, light could travel ~4.727 x 1017 m (i.e., ~49.96578 light-years, where the difference from 50 arises from differences in the length of a year used).

So we have $P$ = 100,000 W and $r$ = 4.727 x 1017 m, then $I$ ~ 3.56 x 10-32 W m-2. As a reference, the threshold of hearing is roughly 10-12 W m-2 or ~20 orders of magnitude higher.

For another comparsion, the sun's total output is on the order of 1026 W, or 21 orders of magnitude larger than your assumed signal strength. Thus, the sun's intensity at our value of $r$ would be ~3.6 x 10-11 W m-2, which is much more reasonable/measureable.

I am wondering how attenuated such a signal would be after traveling for 50 years in space, assuming the transmitter was about 100,000 Watts?

Definition: There is a unit of measure called the jansky (Jy), defined as 1 Jy = 10-26 W m-2 Hz-1. For example, 1 Jy at 1 MHz would be 10-20 W m-2.

I do not think attenuation would matter much, though it is certainly present. Even with all my assumptions that ignore several factors (many of which would make the intensity even smaller), the intensity 10-32 W m-2 is an absurdly small value. The value could be even smaller due to Doppler effects and some of the other complications I mentioned above.

Reference: For reference, the standard reference values, $F_{o}$, for apparent magnitudes in astronomy range from 670–4810 Jy or ~10-24–10-23 W m-2 Hz-1. Note, however, that apparent magnitude does not apply to radio or microwave frequencies.

If we know the apparent magnitude, $m_{x}$, of an object and the spectral band in which we measure it, then we can estimate its spectral flux, $F_{x}$, using: $$ F_{x} = F_{o} \ 10^{-m_{x}/2.5} \tag{2} $$

The faintest objects observed with Hubble have $m$ = 31.5, which corresponds to ~10-37–10-36 W m-2 Hz-1. Thus, the minimum spectral intensities in this range would be ~10-26 W m-2 at ~300 GHz.

Answer: The intensity dependence on distance as r-2 alone reduces the received value enough to make indistinguishable from noise for our capabilities. The capacities of a hypothetical extra terrestrial civilization are purely speculative.

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  • $\begingroup$ If the answer is -32EE Watts after attenuation and the Hubble can hear -37EE, then it should be able to hear it, shouldn't it? $\endgroup$ Commented Jul 17, 2017 at 20:11
  • $\begingroup$ Be careful, look at the units... the $10^{-37}$ values have a $Hz^{-1}$ dependence and Hubble's longest(highest) wavelength(frequency) is in the infrared, thus why I chose 300 GHz. Things are slightly different in radio astronomy than optical too, so that will affect things as well. $\endgroup$ Commented Jul 17, 2017 at 20:25

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