I saw a series of half hour documentaries on various satellites that had been sent out to explore various aspects of our solar system and found myself wondering how we aim our telemetry signals to hit (typically) tiny satellites that are (typically) millions of kilometers away and actually reach them.

The signal sent from the satellite back to Earth naturally has a similar issue. The Earth is obviously much bigger than the satellite but given the immensity of the universe, it still strikes me as exceedingly difficult to hit from great distances away.

Only the very best snipers in history can hit a target more than a mile or two away so how do we send a radio signal so accurately that it hits a satellite so very far away?

I know that space is a vacuum so there are no currents to blow a spacecraft off course but the gravity of the various bodies passed by the spacecraft can change its course, even if the body is relatively distant. Although we've mapped many planets and moons, we're still discovering moons regularly. I don't expect we've mapped nearly every asteroid or comet yet but every one of those bodies could affect a passing spacecraft so the calculation of where a spacecraft will be at a given time must be extremely challenging, perhaps enough to make it difficult to find.

Although I've heard of satellites couldn't transmit or receive any more, I can't recall ever hearing of one that had been "lost" in the sense of not being able to find it to get a signal to it. Have we ever lost satellites in that sense?

FWIW, I read the Wikipedia article on telemetry but it was very general and didn't explain how we locate satellites precisely enough to send them signals. I am NOT a mathematician or engineer so I doubt I would follow explanations in engineering textbooks. I'm just a space enthusiast who is looking for a layman's understanding of how telemetry finds spacecraft.

  • $\begingroup$ Telemetry ===> the ground, commands=====> the spacecraft $\endgroup$ Apr 22 at 0:36

Firstly, because we know exactly where the target is.

The location of a Satellite, or planetary probe, is known down to literally a small number of meters. Yes, even when this probe is moving at hundreds or thousands of meters per second relative to Earth.

So we know where it is, and more importantly where it will be when our message beam gets there.

Secondly, because we are not shooting a "sniper bullet" at it, but more like a "shotgun blast".

Due to not only technical limitations but fundamental laws of physics, there is no such thing as a perfectly focused radio transmission. To achieve a perfectly focused beam, your transmitter would need to be infinitely wide. We approximate this by having huge transmitting and receiving dishes, and by synchronizing multiple dishes to electronically "fake" having even bigger dishes, effectively kilometers wide.

This spreading of the beam means that the signal gets weaker with distance, of course. This, too, is a fundamental issue with longrange transmissions that we have to work with. We mostly solve this by having huge transmitter dishes, and pumping a lot of power into the transmissions.

The same problem occurs in reverse when the probe has to talk back.
Probes have tiny antennae, and meager power systems, so the returning signal is both weak and much less focused than our outgoing signals.
This is handled on the ground by having a huge antenna, very sensitive receivers, and having our antenna point at the exact correct point in space at a predetermined time for receiving the probe' feeble signals.
We only have one, or a very few, receivers of suitable capability. And we have many probes out there that we want to talk to. So the communication schedule for use of this facility is very rigorously and jealously scheduled. A big part of the budget for a distant probe is the cost and time of the DSN to communicate with it.

  • $\begingroup$ Thank you! It seems I overestimated the difficulty of pinpointing the location of a spacecraft at a given time. $\endgroup$
    – Henry
    Apr 22 at 21:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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