Is it possible to extend high speed data transmission with lasers to the distance Earth to Mars?

Data transmission using an optical laser between ground station and a GEO satellite may offer very high data rates, up to 1.8 GBit/s for instance.

But what about a transmission from Earth to Mars without any repeater between the planets? Will the huge distance decrease the maximum data rate to MBit/s or even kBit/s? Using a much more powerful laser and larger aperture sizes for transmitter and receiver may increase the possible data rate, but to what extent? At least some photons per data bit are needed for an acceptable error rate.

• This is a great question. By asking about photons per bit, it avoids things like aperture size and instead addresses the underlying principles and limitations.
– uhoh
Jan 12, 2018 at 23:18
• Actually, that can be misleading. You can have more than one bit per photon. Jan 13, 2018 at 16:57

No, you don't need "at least some photons per data bit". 13 bits per photon has been demonstrated with laser communications. You calculate the data rate capability the same way you do with any other wavelength, which is using the power, range, transmit and receive apertures, noise, modulation scheme, and coding gain.

This paper summarizes detailed analyses of the power and mass comparisons of spacecraft radios for the same data rate at Ka-band and laser comm at $1.55\,\mathrm{\mu m}$. The ground stations were equivalenced based on the cost of construction, which ended up being an array of radio dishes with an aperture equivalent to an $80.5\,\mathrm m$ antenna, and an optical telescope with a $10\,\mathrm m$ aperture.

Also for apples-to-apples comparison, both systems assumed the same pointing accuracy requirement for the spacecraft, with the laser terminal responsible for fine-tuning its telescope pointing with the additional accuracy needed for the smaller beam width.

The benefit is much more dramatic farther than Mars, but at maximum Mars distance, the mass of a 1 Gbps RF system would be more than twice that of a laser system, and the power required for that RF system would be 13 times the laser system. There is no question that at 1 Gbps, even at Mars, you would use a laser system.

• That's incredible! While the IEEE paper seems paywalled, clicking view/open in this link trs.jpl.nasa.gov/handle/2014/44268?show=full shows some related slides. Does the sentence "In PPM, a single laser pulse in one of $M$ symbol slots encodes $\log_2(M)$ information bits" (where PPM is Pulse Position Modulation) concisely summarize the way this works? Since $h \nu$ is so much larger than $k_B T$, individual photons can be detected, and timed to sub-nanosecond accuracy?
– uhoh
Jan 14, 2018 at 3:38
• Yes. Simply consider the arrival time of the single photon as carrying the information. Jan 14, 2018 at 3:38
• @uhoh Fundmental physics will allow you to time a photon to an accuracy roughly the same as it's inverse frequency. For 1.5 $\mu m$ that is about $5 fs$. Engineering is likely to limit you well short of this, but picosecond accuracy should be attainable. Especially as you only need to meaure relative pulse arrival times, not absolute times. Apr 22, 2019 at 15:27
• @uhoh at single photon brightness levels? JUst asking because I'm interested. Apr 22, 2019 at 20:09
• @uhoh A non-paywalled version is available at sci-hub.tw/10.1117/12.2007000. Aug 15, 2019 at 23:51

It's generally true that the error rate in communication is proportional to the energy per bit. Rigorously you can see this through things like $$E_b/N_0$$ and the Shannon Hartley theorem. To compensate for the greater distance to Mars, one could either increase the transmitter power or antenna gain to maintain the same power, or transmit each bit for a longer time thus accumulating more energy in each bit.

GEO is a distance of approximately 36e3 km, while Mars is between 55e6 km and 401e6 km, depending on the relative orbits of Mars and Earth.

Applying the inverse square law, this means the power will be reduced by some factor of at least

$$\left(36 \times 10^3 \over 55 \times 10^6 \right)^2 = 3.6 \times 10^{-7}$$

but not more than

$$\left(36 \times 10^3 \over 401 \times 10^6 \right)^2 = 6.7 \times 10^{-9}$$

Reducing the bitrate by the same factors means the communication system given in the example would operate between 648 and 12 bits per second, all else equal.

That's not to say that laser communication to Mars couldn't work, just that the greater distance requires equipment engineered for the purpose. A Mars communication system will necessarily have higher powers, bigger apertures, and more expensive infrastructure which simply wasn't necessary for a GEO system.