As far as I can tell when the Voyagers listen to Earth they have the Sun in the same beam. I never did calculate how bright a blackbody Sun would be because I have no idea if that's representative of the Sun's output in S-band

I also don't know how wide of a slice in frequency the front end is exposed to, or if the Sun would be strong enough to saturate it.Frequency responses of Voyager's S-band high gain antenna's feed-horn and receiver front end & IF? remains unanswered.

This is related Why is the operating temperature for the Voyagers' receiver noise calculation about 1550K? and answers there are informative.

I assume the designers were optimistic that Voyager might survive this long and designed the system to be able to pick out transmissions from Earth from the Sun's output, but I'm not quite able to put the whole puzzle together yet.

So I'd like to ask:

Question: How bright is the Sun in S-band? Is there a measured value for power per unit frequency? How close is it to what a blackbody would produce at say 5800 K? At this frequency is there a big difference between when the sun is quiet and when it is active?

See also Do stars have "radio photospheres"? Are they different from their optical photospheres?

note: The linked questions and answers have a lot of Voyager-specific information which is relevant to Voyager's operation in S-band, but I have nothing on the Sun's behavior at 2.1 GHz Actually the first link gives the downlink frequency as

2113.312500 MHz for Voyager 2


1 Answer 1


Well, I did some digging and found a helpful chart here. The image itself is located at this link.

Wikipedia defines the S band as the section of the electromagnetic spectrum from 2 to 4 GHz. To interpret the respective values the graph displays at these frequencies, we need to convert these two figures to wavelengths:

$\lambda_1 = \dfrac{c}{\nu} = \dfrac{300\:000\:000\:m/s}{2\:000\:000\:000\:Hz} = 0.15\:m$

$\lambda_2 = \dfrac{c}{\nu} = \dfrac{300\:000\:000\:m/s}{4\:000\:000\:000\:Hz} = 0.075\:m$

And, we might as well find the wavelength at 3 GHz, since it's at the middle of the S band and might help us (hint: since the graph is on a log scale, and this calculation comes out to a nice power of 10, it definitely will help us):

$\lambda_m = \dfrac{c}{\nu} = \dfrac{300\:000\:000\:m/s}{3\:000\:000\:000\:Hz} = 0.1\:m$

So, we can take a look at the graph at 0.1 m. The units seem to be in $W\: m^{-2} \: (c/s)^{-1}$, and I'm assuming that $(c/s)$ is just $Hz$, which would make these units have the same units as Janskys, though, as Connor Garcia mentions in his comment, Janskys are $10^{-26}\: \rm{W\: m^{-2}\: Hz^{-1}}$. So, the quiet sun in the S band lies around $10^{6}\: \rm{Jy}$, and the "disturbed" or "active" sun lies around $10^{7}\: \rm{Jy}$.

And as for the second part of your question, the graph also shows some blackbody curves. You asked for 5800 K, but the closest one on the graph is at 6000 K - hopefully that's okay. As expected, the Sun seems to closely match that line for a good portion of the wavelengths on the graph, but seems to deviate in the S band, which is interesting, to say the least. So, from the blackbody curves in the graph, the sun is brighter than a typical blackbody at ~6000 K in the S band, both in its quiet and disturbed states. This difference is around one order of magnitude in the quiet state, and around two orders of magnitude in the disturbed state.

And lastly, you asked about the difference between the "brightness" of the sun in its active and disturbed states in the S band. From visual inspection of the graph, it seems like the difference in the S band is about one or two orders of magnitude.

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    $\begingroup$ Excellent, thank you!! I'll review later today but it looks like you've nailed it. $\endgroup$
    – uhoh
    Nov 16, 2020 at 2:16
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    $\begingroup$ @Calc-You-Later +1 vote, outstanding answer, but since a Jansky is 1e-26 W/(m^3Hz), shouldn't your post read 10^6 Jy instead of 10^-20? $\endgroup$
    – Connor Garcia
    Nov 16, 2020 at 19:36
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    $\begingroup$ @ConnorGarcia ahh, you're right! Thank you for telling me, I'll edit this straightaway. $\endgroup$ Nov 16, 2020 at 19:57
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    $\begingroup$ @Calc-You-Later yep, m^3 is a typo. $\endgroup$
    – Connor Garcia
    Nov 16, 2020 at 20:02
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    $\begingroup$ That question looks really interesting! I'll do some more digging along with looking at the resources you've send and see if I can come up with a substantive answer :). $\endgroup$ Nov 17, 2020 at 1:42

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