There distance between Earth and Mars varies between 50 - 400 million kilometers because of their orbits. Does this affect data transmission noticeably? I'd naively expect you could transmit 8x faster at the nearest point than the furthest because that's how the distance varies.

Does two-way communication with a Mars rover work better when we are closer?

I have one wireless base station in my house. When I take my laptop to the far end of the house, my signal strength goes down and/or errors may occur. Obviously home wifi has much different concerns such as obstructions and short range design. Distance to Mars varies much more than my laptop, but within a known range. Communication works because certain design considerations were made such as batching commands.

So I'm wondering how that affects communication with the Rover, as that is known today, but also thinking ahead to when/if we have a Mars colony. That would mean communication between a larger number of devices with more network applications that would have to account for the difference.

Would the connection bandwidth or latency change significantly between near and far end?

It is like a large scale cellular tower design where the distance to mobile antenna is variable within a range. Except that in this case, all mobile antennas are always together at the same varying distance and therefore communicate could be optimized for that known distance/latency.

  • $\begingroup$ What do you mean by "better"? Do you think about latency, bandwidth, or are you interested in some particular application? $\endgroup$ Commented Mar 8, 2018 at 10:12
  • 1
    $\begingroup$ hang on, I'll try to post some DSN data... $\endgroup$
    – uhoh
    Commented Mar 8, 2018 at 12:06
  • $\begingroup$ I added clarification so you can see where I was going with the question. It is an interesting computer network design concern...but the question may not be clear enough yet to work on this site. Based on the comments posted, I may need to close this question and re-ask smaller questions to built up my understanding of the issue. $\endgroup$ Commented Mar 8, 2018 at 13:56
  • $\begingroup$ FWIW, here's a plot of the Earth-Mars distance, using JPL Horizons data, for 2015-2025. The distance is between the planet centres, not surface to surface. $\endgroup$
    – PM 2Ring
    Commented Sep 22, 2021 at 16:59

3 Answers 3


If we apply the Shannon Hartley theorem to this problem and use the aproximation for a signal to noise ratio much smaller than 1 we get the formula

C ~= 1.44 * B * S/N

C is the channel capacity in bits per second
B is the bandwidth in Hertz
S is the received signal power in watts
N is the noise power in watts

Doubling the distance reduces the signal power S by 1/4. If noise power is constant the signal to noise ratio S/N and also the channel capacity C is reduced by the same factor.

Unfortunately increasing the distance by a factor of 8 will reduce the channel capacity C by a factor of 1/64.

If you have a video transmission with 60 frames per second at the minimal distance, you get less than one frame per second at the maximal distance.

  • $\begingroup$ Signal strength does vary by a factor of 64, yes. Which would be of paramount importance if signal strength were a limiting factor in the communication. Which it is not. $\endgroup$ Commented Sep 22, 2021 at 17:23
  • 1
    $\begingroup$ @PcMan The limiting factor in the communication is the signal to noise ratio. $\endgroup$
    – Uwe
    Commented Sep 22, 2021 at 21:05

Adressing the "Does two-way communication with a Mars rover work better when we are closer?" part of the question:

In theory the communication would work better when Mars is closer, especially follow-on commands that have to wait for the results of a preceding command. In practice, however, as I'm informed, by someone who was actually assigned to the Mars Science Laboratory (Curiosity rover) mission, typically a whole Sols (a Martian day) worth of programming is prepared ahead of time and uploaded to the rover for it to complete over the next day, allowing for analysis of results and planning the next days programming.

The rest is speculation on my part, but as you're already introducing a whole days latency between communications, I don't think the distance between the planets have much practical impact on the communication.

  • 1
    $\begingroup$ The question is not talking about latency, but about bandwidth. $\endgroup$ Commented Mar 8, 2018 at 9:08
  • $\begingroup$ @NathanTuggy The question is unclear in this respect at the moment. But I like this answer since it explains that there is a latency of 1 day independent of the distance, so one has to think in packages anyway. $\endgroup$ Commented Mar 8, 2018 at 10:15
  • 2
    $\begingroup$ @derwodamaso: The question is not actually unclear: "data transmission rate" cannot and does not mean latency. It means, well, the rate at which data is transmitted, which is exactly the same as bandwidth. (For a longer discussion, see Wikipedia.) $\endgroup$ Commented Mar 8, 2018 at 10:21
  • $\begingroup$ @NathanTuggy, the question is about two way communication, which does indeed relate to the transition rate but also to the latency. The fact that the MSL takes the latency out of the equation by sending a days worth of commands is in my opinion an important part of the answer. (Although not the complete answer) $\endgroup$
    – Martini
    Commented Mar 8, 2018 at 10:27
  • $\begingroup$ @NathanTuggy Yes, there's that part about the rate. But then there is also the general question about when communication works "better". Also, "8x faster" indicates he is thinking about latency, because 8 is the ratio between the greatest and the smallest distance. $\endgroup$ Commented Mar 8, 2018 at 12:35

In the question you state that the distance varies by a ratio of 8:1. Let us assume that the signal transmitted by a spacecraft at Mars has a constant signal level and pointing accuracy from the transmitting antenna, i.e. the signal strength pointing back at earth is a constant. If this is the case then the signal energy received at earth will vary with a ratio of 64:1 due to the inverse square law.

Now, for a signal to be detected then there is a minimum energy per bit that needs to be received by the antenna on Earth. As the signal level from the planet at its furthest distance is 1/64th the signal level at the closest approach the maximum data rate will also be reduced by this factor of 64.

The communications system would probably be designed with a suitable link budget to allow for correct operation with the greatest expected separation between transmitter and receiver, resulting in a much lower error rate in the data transferred when the signal path length is shorter.

Edit following the clarification in the question

The time taken for the messages to get to Mars will increase by a factor of 8 as the signal is travelling at a constant velocity but the data that can be carried by that signal will reduce to 1/64th of the amount that is possible when Mars is at its closest.

As an additional aside cellular power management gets even more complicated with 3G CDMA systems with IIRC the ideal being that the power level of the signal received from each handset at the cell mast being equal.

  • $\begingroup$ Ah! Good point. I was thinking 8x difference, not 64x. $\endgroup$ Commented Mar 8, 2018 at 14:10
  • $\begingroup$ So the signal strength theoretically varies by 64x. But doesn't latency of a constant frequency vary as well? (Because of the varying distance) $\endgroup$ Commented Mar 8, 2018 at 14:12
  • $\begingroup$ The HF signal travelles within the vacuum of space with the speed of light. Double distance needs double time, 8 times the distance increases the delay by a factor of 8. $\endgroup$
    – Uwe
    Commented Mar 8, 2018 at 14:17
  • 1
    $\begingroup$ The time taken to travel is proportional to the distance so the latency will double as the distance doubles, but the power received varies as the square of the distance so the power received at double the distance is reduced to a quarter. This reduced power means that the receiver either has to be 4 times more sensitive or has to listen to each data bit for 4 times as long, hence the reduced data rate that is possible. $\endgroup$
    – uɐɪ
    Commented Mar 8, 2018 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.