According to Wikipedia

At a distance of about 125.97 AU (1.884 × 10$^{10}$ km) from the Sun as of September 9, 2012, it is the farthest manmade object from Earth

As of 2013, the probe was moving with a relative velocity to the Sun of about 17 km/s.

How did NASA determine the distance and velocity of Voyager 1 ?


2 Answers 2


Here is a good tutorial on the navigation of deep space vehicles. The two main data types used are two-way Doppler (using an atomic clock reference at the DSN station, with the frequency locked to and sent back to Earth by the spacecraft), which gives the velocity component along the Earth-spacecraft line to better than 0.1 mm/s, and ranging (sending a pseudo-noise signal which is immediately reflected back by the spacecraft radio), which gives the distance from the DSN station to the spacecraft to better than one meter. Though this only gives you one component of position and one component of velocity, you can use many of these high-accuracy measurements over time combined with the constraints of motion in the gravity field of the Sun and planets to solve for the other components of position and velocity.

There is another data type, used less frequently, which is to use two widely separated DSN stations (i.e. at two different complexes on Earth) and a quasar of known position to determine the apparent position of the spacecraft with interferometry. This can determine the angular position of the spacecraft on the projection of the baseline between the two stations to an accuracy of a few nanoradians, letting you solve immediately for one more component. It is more expensive, and so is generally used when higher precision is required over a shorter time, e.g. on approach to Mars to accurately target a landing site.


The distance is quite easy for something so far out there. You send a command to the satellite, and see how long it takes to respond. The time it takes to respond times the speed of light is the distance that it is away from the Earth.

Furthermore, the velocity is likewise easy. If one day it is 17 light hours exactly, and the next day it is 17 hours and 10 seconds, you have a distance, and the amount of time between the two points. Divide one by the other, and you have the speed that the spacecraft is going.

Okay, if you are worried about those measurements being accurate, there's another frequently used method, the Doppler shift. The faster something is moving away from you, the lower it's frequency will appear to be. If the receiver is stable enough, this can be used to determine it's relative speed, which one can use to find the velocity.

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    $\begingroup$ what if the machine takes long time to respond the command so there is a possibility of error in measuring the distance ? $\endgroup$
    – Hash
    Oct 9, 2013 at 12:47
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    $\begingroup$ @Hash: Bottom line is, there's always some commands that always take a fixed amount of time to process. They know what the fixed time is, and thus they can get the measurements accurate. Taking a trend line over several days helps as well. Also, there are other effects I'll add in. $\endgroup$
    – PearsonArtPhoto
    Oct 9, 2013 at 12:50
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    $\begingroup$ Divide one by the other, and you have the speed that the spacecraft is going. — only if it's moving straight away from the source. Plus we need to consider our own velocity (which is well known). But how do we tell the direction where it's going (which is a component of the velocity)? Just from orbital calculations? I suppose we must point very accurately to communicate with the Voyagers. $\endgroup$
    – gerrit
    Oct 9, 2013 at 14:32
  • $\begingroup$ @gerrit: Good point. Hmmm... I'd guess that the later is known. Essentially if you plot it over time for long enough, you'll know how it's going to move via orbital mechanics... $\endgroup$
    – PearsonArtPhoto
    Oct 9, 2013 at 14:39
  • $\begingroup$ Trigonometry would be one factor - we can tell the angle that the return signal is coming from, and with some good math and knowing the location of Earth at the time of sending and receiving we can figure-out the location. $\endgroup$
    – john3103
    Oct 9, 2013 at 15:42

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