# Ground → ship Wi-Fi bandwidth in my fast moving spaceship

I have a hypothetical question that hopefully makes sense considering only the rudimentary amount of knowledge I have on relativity.

Assume I am in a spaceship orbiting around a spherical satellite, at a certain fixed radius. The satellite sends data to the ship via radio (something like satellite internet or other long-range data radio, using electromagnetic signals).

Would the download speed (received data rate seen on-board ship) be faster or slower, compared to when I am stationary with respect to the satellite?

Also consider that attenuation is negligible.

What I initially thought was that time dilation would make "my" clock in the spaceship tick slower, hence the signal from the "satellite" would be received faster. I thought I should add this to clarify my train of thought that led to the question. Please do note that it is a hypothetical question.

(A comment pointed out that a spherically symmetric EM wave is impossible, therefore I have taken down the diagram and the statement regarding so.)

• Comments are not for extended discussion; this conversation has been moved to chat. – Chris Nov 21 at 3:55
• Hi, we've noticed that you have made a large number of minor edits to this post. Please be mindful that every edit bumps the post in the "active" tab of the site and try to make your edits substantial. If you foresee improving this post repeatedly, maybe collect several edits and make them in one go instead of submitting them individually. – Chris Nov 21 at 3:55

If you are in a perfectly circular orbit, then your received signals are affected by the relativistic transverse Doppler effect.

The signals you would receive would be blueshifted by the Lorentz factor of the orbital speed. That is the information transferal rate would (potentially) go up (if you can tune to the new frequency).

If you are in a Keplerian orbit, then (in a Newtonian approximation) $$\frac{mv^2}{r} = G\frac{Mm}{r^2}$$ $$\frac{v}{c} =\sqrt{\frac{GM}{rc^2}} = \sqrt{\frac{r_s}{2r}},$$ where $$r_s = 2GM/c^2$$.

The transverse Dopper blueshift is by a factor of $$f_r = f_t\gamma = f_t\left(1 - \frac{v^2}{c^2}\right)^{-1/2} = f_t\left(1 - \frac{r_s}{2r}\right)^{-1/2},$$ where $$f_r$$ and $$f_t$$ are the received and transmitted frequencies - assuming that the transmitter is at $$r=0$$ or at least stationary with respect to the orbital speed.

However, in General Relativity there is also a gravitational redshift given by $$f_r = f_t \left( \frac{1 - r_s/r}{1 - r_s/r_t}\right)^{1/2},$$ where $$r_t$$ is the position of the transmitter (which has to be at $$r_t > r_s$$) and this may modify the conclusion.

Which will win out depends on the radius of the orbit compared with the radius of any "surface" you put your transmitter on.

There are two effects due to relativity. The motion of the spaceship around the satellite results in a time dilation that can be calculated using special relativity.

If the satellite has significant mass (e.g., enough to keep the spaceship in orbit), then general relativity (gravity) plays a role.

From the spaceship's perspective, the two effects are opposite, but not equal in magnitude.

This Wikipedia article re error correction of GPS signals may help.

Per the suggestion by @ChappoSaysReinstateMonica, the following is included here as an easy way to think about it:

Assuming the satellite has essentially no mass, so the spaceship needs to use its engines to keep going in a circular orbit around the satellite, then only special relativity applies. The spacecraft will perceive its received data rate as being a bit faster than the transmitter's data rate on the satellite. A way to know this is via the fact that a clock taken from the satellite to the spacecraft, then after a while brought back to the satellite, will lose a bit of time relative to a clock left on the satellite, but uploaded data is like clock ticks on the satellite.

• Basically you are sayibg data rate goes up, since curvqtire from a satelite can prob be neglected – lalala Nov 19 at 18:44
• Not sure how to translate "curvqtire" – S. McGrew Nov 19 at 19:46
• Curvature. Typed while walking with a mobile. Sorry – lalala Nov 19 at 20:46
• Assuming the satellite has essentially no mass, so the spaceship needs to use its engines to keep going in a circular orbit around the satellite, then only special relativity applies. The spacecraft will perceive its received data rate as being a bit faster than the transmitter's data rate on the satellite. A way to know this is via the fact that a clock taken from the satellite to the spacecraft, then after a while brought back to the satellite, will lose a bit of time relative to a clock left on the satellite, but uploaded data is like clock ticks on the satellite. – S. McGrew Nov 19 at 20:59
• @S.McGrew the explanation in your comment is excellent and deserves to be included in the answer itself, e.g. introduced by "A simple way of thinking about this is...". :-) – Chappo Says Reinstate Monica Nov 21 at 23:00

Doppler effect (relativistic) of the Electromagnetic Waves will take place. As a result you will receive the emitted Wifi signal (EM wave) as squeezed along the time axis (increases frequency). Naturally the WiFi demodulator available with you would report invalid (unrecognised) signal because digital data is encoded on the carrier wave by some form of modulation like Frequency Shift Keying , Phase shift keying etc.

Let's say it's transmitted by FSK. In FSK, the 0s and 1s are transmitted through distinct changes in Carrier Wave frequency held for a particular duration. First of all the EM wave frequency has changed hence your demodulator won't even recognise the EM wave as Wifi signal.

Assume that you have accounted for Doppler effect EM wave frequency change and distinct change in the changes of the changed frequency (obtained from it's corresponding changed Fourier transform) for zero and one encoding, into your custom demodulator. So the EM wave is now recognised as Wifi signal. But as stated above the distinct changes in EM wave frequency is held for a pre agreed time (between transmitter and receiver) to encode 0 & 1. So due to time dilation this agreement breaks. So your demodulator although recognises Wifi signal presence, reports invalid protocol.

Now you might be tempted to think let's account for the time dilation into the demodulator. But that isn't really a good idea because the above stated pre agreed duration was statistically obtained to ensure noise free transmission. So if you try to account for that in your demodulator, you will occasionally receive corrupted data due to noise in the transmission. But if you are okay with it, then sure download speed increases!!!!

• I'm no RF expert, but I suspect each bit needs a certain number of carrier wave cycles for noise immunity - since the frequency is shifted higher, less time is needed for each bit. – user253751 Nov 20 at 15:12
• Also see Nyquist Bit rate and Shannon Capacity @user253751 – Madhuchhanda Mandal Nov 20 at 17:49
• More mathematically speaking, I expect that the bandwidth will be greater (because the lower and upper frequencies are multiplied by the same factor) so the required SNR will decrease, even though the noise power increases. – user253751 Nov 20 at 17:59

If the satellite's signal is broadcast from the centre of the satellite, so your fixed-radius orbit doesn't alter the distance between you and the antenna, you'll get relativistic doppler shift:

Navigation Technology Satellite 2 carries aboard it two high‐precision cesium beam atomic standards.[...] This observed bias was 4.47 parts in 10^10, which is consistent with a predicted frequency bias arising from the special relativity time dilation effect coupled with a frequency blue shift predicted by general relativity.

I would describe that as a relativity-induced error of 0.447 parts per billion

On the other hand, if the antenna was on the satellite's surface, you'd also see doppler shift as the distance between you and the antenna varies.

The relevant rates of change are usually dominated by Earth rotation, which at the equator amounts to about 460 m/s. As a consequence, two-way Doppler shifts can be as large as ± 440 Hz at 144 MHz, ± 4 kHz at 1296 MHz, and ± 30 kHz at 10 GHz.

So, if your satellite is the size of earth, you might see a doppler shift varying between +3000 parts per billion and -3000 parts per billion over the course of an orbit.

However, communication standards already expect a certain amount of doppler shift - according to "Next Generation Wireless LANs: 802.11n and 802.11ac"

in 802.11n [...] the values for Doppler spread in the 5GHz band is approximately 6Hz

So the makers of the standard already expect a doppler shift of 1.2 parts per billion. Cell phone mobile data standards like 5G worry about doppler even more - allowing them to work even on high-speed trains.

But meanwhile, according to "Clock Solutions for WiFi (IEEE 802.11)" (admittedly a dated source)

The typical performance requirement is ±25ppM all-inclusive frequency stability (includes initial calibration tolerance at 25°C and frequency changes over operating temperature, power and load fluctuations, and aging).

So WiFi hardware makers are already compensating for a 25,000 parts per billion crystal frequency error, which can vary with changing temperature conditions.

I would say, as long as any motion-induced doppler frequency changes are small and slow compared to the thermally-induced crystal frequency changes wifi can already account for, you should get normal performance from a frequency perspective.

Of course, depending on your orbital radius you might encounter problems due to signal latency - and inverse-square-law signal attenuation. And the signal attenuation happens in both directions, so even if the satellite has a megawatt transmitter to give a really long range, you won't get a two-way link unless your wifi hotspot has the same great range.

Basically your question is the same as what Michaelson and Morley asked. Their experiment used the earth traveling around the sun instead of a spaceship. Their results did not show a difference in the speed of light either way.

• Mm didnt look for rotating frame effects. Also see sagnac effect – lalala Nov 19 at 18:42
• @lalala Are you suggesting they missed something because they didn’t account for the rotation of the earth? If you think about it the two situations are still the same. – Bill Alsept Nov 19 at 18:45
• No. They just didnt probe for rotation. Also rotation velocity is very small compared to the assumed linear velocity of earth at that time. What i mean, the begative result of that experiment doesnt tell us anything about rotation – lalala Nov 19 at 18:53
• @lalala Thanks but it’s still the same question. Michaelson and Morley would’ve considered everything. – Bill Alsept Nov 19 at 18:58
• @BillAlsept The written question is: "if I were to download information , would the download speed be faster or slower compared to when I am stationary with respect to the satellite? " In general, download speed is used to mean throughput, not latency. – JiK Nov 20 at 16:10