I guess that everyone is familiar with the problem of aligning a TV satellite dish: some degrees off and the signal you desire is gone.

How sensitive is communication with deep space probes to this problem? Is it addressed with special solutions or the radio emissions are not focused enough for this to be a problem?

I understand also that sometimes the same antenna is used for 2-way communication with a spacecraft. Is this possible for the above-mentioned reason?


2 Answers 2


You can calculate this using the diffraction limit:

$$\sin\theta\approx{\lambda\over D}$$

Where $\lambda$ is the wavelength being used for the radio communication, $D$ is the diameter of the dish (either on the spacecraft or on the ground), and $\theta$ is the beam width. You need to point the antenna accurately enough to keep the target in the beam. The ratio is usually very small, so you can just drop the $\sin$, where $\theta\approx{\lambda\over D}$ is then in radians.

  • 1
    $\begingroup$ This gives (approximately) the first minimum of the Airy disk for the antenna, if you treat it as a telescope (a reasonable assumption for a parabolic dish). The actual precision required varies: for example, a weak transmitter such as Voyager might only be detectable closer to the first maximum, while the ISEE-3 Reboot project made initial contact using a sidelobe (a secondary maximum) of the Arecibo dish. $\endgroup$
    – Mark
    Commented Jul 7, 2015 at 21:32
  • $\begingroup$ Yes, this just gives an approximate result (hence the $\approx$). You generally like to do better. (Not worse though. ISEE-3 was a bit of an oddball case, since it is difficult to change which way the Arecibo antenna is pointed.) The formula above gives about 7° for MER. The antenna pointing requirement for MER was 0.5°, and the overall system performance was about 2°. $\endgroup$
    – Mark Adler
    Commented Jul 7, 2015 at 22:53
  • 1
    $\begingroup$ This is exactly at the level asked in the question. +1 $\endgroup$ Commented Jul 8, 2015 at 3:07
  • $\begingroup$ Do I understand correctly then that the formula for the size of the Airy disk is the same as the one for the size of the transmission lobes? $\endgroup$
    – Federico
    Commented Jul 8, 2015 at 6:54
  • $\begingroup$ For the central lobe it's not bad. Real antennas and feeds however can have very different side lobes than what you see in the Airy function. $\endgroup$
    – Mark Adler
    Commented Jul 8, 2015 at 13:54

It really depends on the gain of the antenna used. In antenna design you can either go for gain (directing all the energy into a narrow cone) or wide response (sensitivity in all directions).

The further away the space craft you want to communicate with, the more important gain becomes. The gain of an antenna is "almost free" - up to a limit it does not add a lot of noise. Once you get to the electronic amplifier, any amplification includes amplification of he (input) noise of the amplifier. So you want to collect as much as possible of the weak signal. That means high gain, large dish, and precise alignment.

Of course astronomers are used to this kind of thing - they know exactly where their spacecraft is, and where their antenna is pointing.

For example, the Mars Orbiter has a 3 m dish that is used in the X band (8-12 GHz) - a wavelength of approximately 3 cm. that makes the dish 100 wavelengths across. This means that it has to be pointed to a small fraction of a degree - for which it is equipped with a precision gimbal. The antenna on earth is much larger - 34 m across - for even greater gain (and requiring pointing to better than a milliradian). See for example http://mars.nasa.gov/mer/mission/comm_size.html http://marsmobile.jpl.nasa.gov/mro/mission/spacecraft/parts/antennas/ and other links on the NASA site. In fact the angular position is determined to tens of nano radians- meaning that knowing where to point the antenna is trivial by comparison.


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.