# How would you calculate the maximum bandwidth possible in space?

If you'd take Earth and Mars, and at both their closest and furthest position (ignore the Sun being nearly in-between for a moment), what would be the maximum amount of bandwidth possible between these two objects, given an amount of noise and power?

I know in the future (possibly) there will be a colony on Mars, and as bandwidth demand with earth increases (first fairly linearly per colonist and data reporting back to earth, then logarithmically for leisure goals), I'm curious how data transmission would go, exactly.

(For completeness visualisation sake, lets say that after Mars becomes self-sustaining, and sets up infrastructure, there is one - or a set of - radio dishes on Mars' pole, which are able to stay in constant sight to either Earth's low-orbit or worldwide radio dishes (A-la DSN), and maybe 2 satellites orbiting in-between Earth's and Venus' orbits to proxy radio signals when the earth is behind the sun)

(Please note: I am not an expert on electromagnetic signals or similar, I just have a friend who spammed me eons ago about signal-to-noise ratio info, and I'm just echoing what I remember)

Edit: For clarification, I'm asking something along the lines of an output to the Shannon's Channel Capacity equation, with some examples of power inputs, and equivalents to those power inputs (e.g. how much 1KW is in today's world, and how much channel capacity/bandwidth (in bits) that'd create)

• space.stackexchange.com/questions/7776/… Commented Nov 16, 2020 at 17:25
• thank you, though i still would want an answer what specifically the max bandwidth would be between earth and mars, given X power Commented Nov 16, 2020 at 17:43
• This can't be answered. There's a lot more to it apart from power. Antenna sizes, technology, bandwidth... Commented Nov 16, 2020 at 17:51
• Oh, I would still consider it answerable, just that the equations have more than just the sending power. Commented Nov 16, 2020 at 21:33
• Commented Nov 16, 2020 at 21:43

It's probably not what you wanted, but let's ignore engineering and simply calculate the maximum bandwidth from Mars to Earth that physics allows.

So the minimum amount of energy to convey a bit of information is $$kT\ln(2)$$. So if we take the surface of Mars at roughly 250K as giving us $$T$$ (since it will be the background against which we are trying to resolve the data) we get a figure for energy per bit. Now we cover the Earth-facing side of Mars with such transmitters and wonder what limits the power we can use for the transmission. Spurning such plebeian considerations as the total solar power incident on Mars, or the energy we could liberate by reacting Mars with an equal mass of antimatter, let us just consider the density of energy in the communications beam. If that gets too high, the space containing the beam will most likely collapse into a black hole. So if consider a Mars-sized volume of the beam, it cannot contain more energy that the mass of a Mars-sized black hole, which is given by $${\displaystyle r_{Mars}={\frac {2GM}{c^{2}}}}$$ or $$M = \frac{r_{Mars} c^2}{2G}$$

That will represent the signal sent over a period of roughly $$2r_{Mars}/c$$ so if $$B$$ is the bandwidth we see the fundamental limit as given by $$\frac{2 r_{Mars} B k 250 \ln 2}{c} = \frac{r_{Mars} c^2}{2G} c^2$$

Cancelling a few things we get $$B = \frac{c^5}{G k 1000 \ln2}$$

which comes out to be about $$3.7\times 10^{72}$$ bits per second. Probably enough for most entertainment and science purposes.

• One of the many engineering problems with the model is that the beam would mass considerably more than the rest of the solar system, which might be quite badly disrupted by it. Commented Nov 17, 2020 at 14:09
• How would such a flux of information in bits of size $k_B T \ln(2)$ impinging on Earth compare to current rates of global warming?
– uhoh
Commented Nov 17, 2020 at 14:13
• @uhoh it would be thousands of times more energy per second than the total mass energy of the Earth. Capturing all of it in the receivers and efficiently rejecting the waste energy to space would be essential to avoid the planet being instantly blasted to plasma. Furthermore the beam would carry a lot of momentum, so the waste heat would have to be rejected in the direction exactly opposite to Mars or Earth would leave the solar system, probably at relativistic velocities. Commented Nov 17, 2020 at 14:27
• Sounds exciting!
– uhoh
Commented Nov 17, 2020 at 14:30
• This is technically the right answer, but the way you presented is also incredibly amusing to me, I'll mark this as answered, but I'll say here: the other answers gave me insightful information as well, so if you're reading this, please check those out as well. Commented Nov 18, 2020 at 8:00

Communications with lasers between two points in outer space, better known as Free-space optical communication (FSO) has a potential for bit rates similar to fiber-optic cables - up to 1Tb/s. Going through an atmosphere, for instance, from Earth to an orbiting satellite the current effective range drops to about 3.5 km. You would still need radio transmitter to get the signal from a ground station to the satellite, even though we are using lasers to do this today. The biggest drawback would be the communications stream has to be line-of-sight. Once you have a network in place in orbit, you could potentially communicate with Sedna, which is 8 billion miles away and feature a live-feed high density video with a minor 12+ hour delay.

FSO technology is currently being used for earth-based communication. Satellite constellations such as the SpaceX Starlink project using laser to provide global broadband coverage for inter-satellite links between the several hundred to thousand satellites effectively creating a space-based optical mesh network.

The European Space Agency set up a laser-based system called the European Data Relay System (EDRS) on November 28, 2014. The system is operational and is being used on a daily basis.

Imagine being able to watch a live feed of Martian storms. Although I would think a live feed of Jupiter would be far more entertaining.

The problem can't be answered as given as the problem is not adequately specified.

Bandwidth is a function of power and the size of the antennas used. On the transmitting side the bigger the dish the narrower the cone and less energy is wasted on directions that don't lead to the receiver. On the receiving side the bigger the dish the more of the signal it can gather.

Note how dishes meant for deep space use are huge and on deep space craft the high gain antenna can be the biggest thing on the probe.

The spacecraft do not have much power available, the giant dishes on Earth gather more of the very faint signal coming in. The probes can't mount such huge dishes, the Earthbound transmitters compensate by running with vastly more power.

• While it's true that the question might be considered somewhat open-ended technically, they did mention DSN as an attempt to connect to existing technology. Kudos for providing the relatively new user's first question a helpful answer!
– uhoh
Commented Nov 17, 2020 at 1:49