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All satellites have the same inclination.

At an altitude of 400 km, the circumference of their orbit is $2\pi \cdot (6371+400) = 42\ 543$ km.

If, for example, you have only 2 satellites separated by half the distance (i.e. 21271 km), will they be able to communicate with each other? I guess not because Earth would be in the way.

If we have 3 satellites separated by a third of the distance (i.e, 14000 km), will they still be able to communicate with each other?

If the maximum distance is X, how many satellites are needed to cover the whole orbit? enter image description here

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    $\begingroup$ No answer considered the wavelength so far. IMHO your satellites are close to the ionosphere, so short wave radio signal communication may not be possible due to reflection or refraction. $\endgroup$ Apr 25 at 6:20
  • $\begingroup$ Also wondering about any potential Fresnel zone considerations. $\endgroup$
    – Dan
    Apr 25 at 11:03
  • $\begingroup$ See astronomy.stackexchange.com/q/49172/31410, nearly the same question $\endgroup$ Apr 25 at 12:37
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    $\begingroup$ If you're interested in the pure line-of-sight calculation, Math.SE might give a more straightforward answer on how to turn the geometry into a formula. Here you'll get lots of answers about the technicalities :D $\endgroup$
    – Nick T
    Apr 25 at 16:40
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    $\begingroup$ 400 km is inside the upper ionosphere. Ground radio transmitters/receivers can achieve beyond-the-horizon communications by bouncing radio waves between the ionosphere and the ground. I'm not sure if something similar is possible for satellites inside the ionosphere, but it might be worth investigating. $\endgroup$
    – 8bittree
    Apr 25 at 17:40

5 Answers 5

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If, for example, you have only 2 satellites separated by half the distance (i.e., 21, 271 km), will they be able to communicate with each other? I guess not because Earth would be in the way.

Technically, like @Christopher James Huff said in his great answer, bouncing signals off the moon or using other relay satellites to send information to the other spacecraft would be possible, but direct communication would not be possible.

If we have 3 satellites separated by a third of the distance (i.e, 14000 km), will they still be able to communicate with each other?

No, see this from my recent answer to a similar question.

The absolute minimum number of satellites needed is 3, like I said in a comment, if the orbital distance is greater than or equal to the Earth's radius (at least 6371 km). However, if the orbital distance is less than 6371 km, you need more. For instance, if the radius is more than $6371(\sqrt2-1)\approx 2639$ km, then you will only need 4. Continuing, using the formula for the incircle and excircle of bicentric, regular polygons, where the incircle is the Earth and the excircle is the satellite network orbit, (see the Wikipedia article), the orbital radius is equal to $R=\frac{6371}{\cos{\frac{\pi}{n}}} - 6371$, where $n$ is the number of satellites in orbit. Then given some orbital distance, $R$, from the Earth (relative to the surface), the minimum number of satellites needed is $n=\bigg\lceil\frac{\pi}{\cos^{-1}\big(\frac{6371}{R+6371}\big)}\bigg\rceil$, where $\lceil x \rceil$ is the smallest integer greater than or equal to a number, $x$.

This implies that you cannot have three equidistant satellites orbiting as the Earth will still block signals.

If the maximum distance is X, how many satellites are needed to cover the whole orbit?

Just apply the formula in the above quote for 400 km and get $9.09$ (which rounds up to 10) satellites. If you don't want a perfectly symmetrical network, the minimum angular separation would be $\alpha = 39.5852^\circ$, corresponding to a distance separation of $4677.98$ km, which requires 9 satellites if you are willing to have a temporary loss of signal between the outlying satellite which is about 0.4 degrees behind its neighbors. If you want the network to be symmetrical around the Earth, you need 10 satellites, separated by 36 degrees and 4254.3 km.

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Imagine a circle (the circumference of the Earth) and a triangle with the angle points being the locations of the satellites and the centre of the Earth. (the centre of the circle)
You can divide that triangle into two equal rectangular triangles with a line from the centre of the circle to the middle of the opposing line that connects the two satellites.

Let's suppose the angular distance between the two satellites is $2\alpha$, measured from the centre of the Earth, and also that the midpoint of the straight line between them with length m would barely touch the surface.

Then:

$ \cos\alpha = R_{Earth}/(R_{Earth} + 400)$ = 0.94092453, .. ----> . $\alpha$ = 19.7926⁰

$ \tan\alpha = m/2R_{Earth}$, ---> . m = $2R_{Earth}\tan\alpha$ = 12742 x 0.359876 = 4586 km.

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  • $\begingroup$ I just wanted to calculate the arc of the orbit, not the the distance to the horizon for each one of the two satellites. I will edit my question shortly to include a sketch showing what I need. I'm sorry for not having been clear enough with my question. $\endgroup$ Apr 24 at 22:45
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    $\begingroup$ @SarahWilliams the difference is only a couple percent: 2π * (R_Earth + 400km) * 2α = 4683 km. $\endgroup$
    – hobbs
    Apr 25 at 3:21
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Probably not quite what you're looking for, but two satellites on opposite sides of Earth, 13542 km apart, can still stay in almost-uninterrupted contact via EME communications (bouncing the signal off the moon).

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  • $\begingroup$ But then they'd need a moon-synchronous orbit to always see the moon? $\endgroup$ Apr 24 at 23:29
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    $\begingroup$ @PaŭloEbermann I was thinking of a non-synchronous polar orbit, which would only experience eclipses twice a month. A moon-synchronous orbit that ensures the moon is always visible is possible, but probably not with a 400 km altitude. $\endgroup$ Apr 25 at 12:18
  • $\begingroup$ Most low earth orbits will have the moon invisible roughly half of each orbit (when they are on the other side of Earth), or am I missing something here? (Polar or not doesn't really matter here.) $\endgroup$ Apr 25 at 23:26
  • $\begingroup$ With this kind of answers, you can always bounce signals off terrestrial relay stations or other satellites. How is that an answer? $\endgroup$ Apr 26 at 11:43
  • $\begingroup$ @PaŭloEbermann the polar orbit does matter: for large parts of the moon's orbit, all parts of a polar orbit are in view. The ideal orbit would be inclined at 90 degrees to the moon's orbit. $\endgroup$ Apr 26 at 18:12
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For line-of-sight communication it's not longer than $2\sqrt{ (6371+400)^2 - (6371)^2} = 4585$. Or $2\cos^{-1}\left(\frac{6371}{6371+400}\right)$ = 40 degrees, so 9 satellites at least.

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  • $\begingroup$ Thank you. Can you please elaborate on the formulas you used? Where did they come from? $\endgroup$ Apr 24 at 13:31
  • $\begingroup$ @SarahWilliams the first equation is based on en.wikipedia.org/wiki/Pythagorean_theorem $\endgroup$
    – Uwe
    Apr 24 at 18:26
  • $\begingroup$ @SarahWilliams The other is finding the angle of a right triangle by using two sides. In this case the triangle is half of the triangle between the two satellites from the centre of the Earth where the base of the triangle is tangent to the Earth's surface at the midpoint (ie: the triangles in the picture in your question). Both equations are usually taught around year 8 (11-13 years old) mathematics when geometry is introduced. $\endgroup$
    – J...
    Apr 26 at 13:18
  • $\begingroup$ @J... Im aware of Pythagoras theorem. I just wanted to calculate the arc of the orbit, not the distance to horizon for each satellite. $\endgroup$ Apr 26 at 19:04
  • $\begingroup$ @SarahWilliams It is trivially calculated from the angle produced in the result above. See : Circular arc $\endgroup$
    – J...
    Apr 26 at 19:09
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This depends on the wavelength. While short waves spread more like a light, long waves bend over horizon and can cover distances in excess of 17,000 km even from the surface of the Earth. This is more than diameter of the Earth (12,742 km).

These waves require very long antennas but it may be easier to build them when where there is no weight.

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