Which is stronger per Hz, radio signals from a Mars spacecraft, or Mars' blackbody radiation?

Question

It just occurred to me that while Mars is cold from a human's perspective, from a radio engineer's perspective it's still kind-of hot.

When even the largest DSN antennas are pointed at Mars to receive signals from spacecraft in orbit around Mars, or even potentially on the surface, they can't really resolve or separate thermal radio noise from the planet from radio transmissions of the same frequency by a spacecraft.

Using 32 GHz (9 mm wavelength, from here) and 70 meters, I get a beam width much larger than the planet even at closest approach.

Even on a per-Hz basis, wouldn't the thermal radiation from the whole planet be much stronger than radio transmissions from spacecraft there?

note: I've just found out that the first 70m dish was built specifically to "talk to Mars' or actually Mariners 3 and 4, who's missions flew past Mars. While Mariner 3 didn't make it, Mariner 4 did. Why does the word “Mars” shows up in google maps when viewing the Goldstone DSN complex?

Answer 1

First of all, there is no fundamental problem if the amplitude of the signal is lower than the background noise. As long as the shape of the signal sent is known, one can receive the information even if the amplitude is orders of magnitude below noise level. A very prominent example is GPS that sends signals using 20 Watt transmitters that emit the to almost half of Earths surface.

Nevertheless, let's calculate the relative amounts of radio signal and black body radiation received from a Mars orbiter.

Assumptions:

• Transmission takes place in the $K_a$ band at 30 GHz or 9 mm wavelength.
• The antenna diameter is 3 meter with a gain of 60 dB and a transmitter power of 35 W (like on MRO)
• Transmission bandwidth is 500 MHz
• Mars (6700 km diameter) has a temperature of 250 K and is an ideal black body

Note that we do not need to include the distance to Earth or the size of the DSN antenna as both cancel out when we look at the ratio of powers.

According to Planck's law, the radiance of a black body is $$B(\lambda, T) =\frac{2hc^2}{\lambda^5}\frac{1}{ e^{\frac{hc}{\lambda k_\rm B T}} - 1}$$ $$B(\lambda=9~\rm{mm},T=250~K) = 3.14 \frac{\rm W}{\rm {sr~m^2~m}}$$ To get the total power output, we have to multiply by the apparent surface of Mars $A = \pi \cdot (6.7\cdot 10^6)^2\rm m^2$ and the bandwidth (in terms of wavelength) of $b = \frac{30~\rm GHz}{500~\rm MHz} \cdot 9~\rm mm = 0.14~\rm mm$ as well as the total surface of a sphere of $4\pi$ $$P_{\rm Mars} = 4\pi \cdot B \cdot A \cdot b = 7.8\cdot 10^7 ~\rm W$$ This is the power Mars emits in the spectral band our receiver is working at.

The antenna, on the other hand emits an apparent power of $$P_{\rm Antenna} = 35~\rm W \cdot 10^{6} = 3.5\cdot10^7~\rm W$$

In summary, both are emitting roughly the same amount of power, resulting in a signal-to-noise ratio of about 0.5. Please note that many additional sources of noise are not included here, like amplifiers, atmosphere and so on.