How do we track the exact location of the spacecraft which is millions or billions of miles away from us? What things do we study for tracking purpose? How exactly do we predict?

  • 1
    $\begingroup$ This answer is a good start but it is not a complete answer to your question. Tracking precise locations of spacecraft in deep space in 3D is much more than sending and receiving radio signals to get distances, and you may want a location at a certain point in the present or future when a spacecraft passes near a planet while the data you receive might be 12 or 18 hours old! A good answer will address how this is done. $\endgroup$ – uhoh Apr 26 '17 at 8:02
  • $\begingroup$ Thank you, can you please share how it is done? Really appreciate your time here. $\endgroup$ – Pras4 Apr 26 '17 at 8:08
  • 1
    $\begingroup$ I think someone will leave a good answer. I just wanted to make sure nobody closed this question as a duplicate, so I left a comment to point out why your question is particularly interesting, and different. $\endgroup$ – uhoh Apr 26 '17 at 8:16
  • 1
    $\begingroup$ Really appreciate that. Hope, somebody would answer. $\endgroup$ – Pras4 Apr 26 '17 at 8:20
  • 1
    $\begingroup$ Re How do we track the exact location of the spacecraft? We don't. There are always errors in measurements, in the behaviors of the spacecraft when it makes maneuvers, and in our models of the solar system. Perfection is unattainable. $\endgroup$ – David Hammen Apr 27 '17 at 22:50

How do we track the exact location of the spacecraft which is millions or billions of miles away from us?

We don't track the exact location of spacecraft. There are always errors in measurements, errors in the behaviors of the spacecraft when it makes maneuvers, and errors in our models of the solar system. Exactness (perfection) is unattainable. It is far better to model those errors than it is to pretend they don't exist.

What things do we study for tracking purpose?

This depends very much on the object of interest. A key distinction is whether the object of interest is a cooperative or uncooperative target. A cooperative target somehow actively cooperates in determining the object's trajectory. The Moon, the Sun, the planets, asteroids, etc., are obviously uncooperative. All objects of interest in space were uncooperative until the middle of the 20th century, and the only available measurements were the object's angular position (azimuth and elevation) as seen by an observer on the surface of the Earth.

This changed drastically in the latter half of the 20th century when humanity developed radar and when humanity started sending cooperative targets into space. Radar enables the measurement of the range to somewhat nearby objects in space, where "somewhat nearby" means a few astronomical units or so. These range measurements alone are quite precise, enough so that direction measurements (azimuth and elevation) are of lesser importance.

Cooperative targets that relay a signal sent from the Earth back to the Earth drastically improve the range of this already very precise range measurement and enable an even more precise measurement, range rate (the rate at which the distance to the target changes with respect to time).

With one exception, measurements of where a cooperative target is in the sky are so imprecise compared to range and range rate that those measurements are essentially useless. That one exception is Delta-Differential One Way Ranging, or ΔDOR, for short. This involves simultaneous observations of a deep space probe by two or more ground stations separated by thousands of kilometers. Interferometric techniques yields nanoradian level precision on the angular position of the observed object.

How exactly do we predict?

Even with range, range rate, and ΔDOR measurements, a single set of measurements provides information on only four measurements of an object's state. Moreover, because ΔDOR requires using two ground stations simultaneously, it is rather expensive. In many cases, there are only one (range) or two observations (range and range rate) at any point in time. Whether it's one, two, or four measurements, that's does not suffice because an object in space has six translational degrees of freedom. Inferring orbital state from a single set of measurements is an underdetermined problem.

What's done is combine multiple sets of measurements gathered over time. This is an overdetermined problem. Each of those measurements has an error associated with it, the propagation from one measurement to another induces additional errors (process noise), and imprecise behaviors of a probe when it performs a correction burn or adjusts its attitude induce yet other errors (plant noise).

Describing the techniques needed to combine those multiple measurements and to account for measurement noise, process noise, and plant noise would require multiple books. If you want to study this on your own, you'll need to understand statistical filtering techniques. One set of keywords of interest is "precision orbit determination", or POD for short.

  • $\begingroup$ Is triangulation used too by simultaneous distance measurements from several ground stations? Unfortunately the diameter of the earth is very small when compared with the enormous distance of these spacecrafts. Even the diameter of the earth orbit around sun is small in comparison. $\endgroup$ – Uwe Apr 28 '17 at 10:18
  • $\begingroup$ @Uwe you can consider ΔDOR to be fancy triangulation. It's even got a triangle in the name. But seriously, it's a time-difference-of-arrival thing, which is basically the same concept. $\endgroup$ – hobbs Apr 29 '17 at 0:54
  • $\begingroup$ "an object in space has six translational degrees of freedom"? I think of three degrees of freedom for location and another three for the attitude. But is attitude translational, I would call it rotational? $\endgroup$ – Uwe Apr 29 '17 at 8:33
  • $\begingroup$ @Uwe - Position and velocity. Attitude maintenance is the spacecraft's job. With regard to your previous question, triangulation is pretty much useless. The Deep Space Network antennae with the smallest beamwidth are the 34 meter dishes when used in the Ka band. These have a half power beamwidth of 0.016 degrees. As an example, for a spacecraft 5 astronomical units away, that corresponds to a circular patch of sky 200000 km in diameter (50% larger than Jupiter). All other antenna/receiver combinations involve significantly larger beamwidths, making that already huge patch even larger. $\endgroup$ – David Hammen Apr 29 '17 at 10:12

While this site does not give a very detailed answer, it gives a relatively good idea on how this is achieved:

The JPL has five groups, that handle the navigation together.

  • Ephemerides Group

    Calculates the positions of astronomical objects at predicted times.

  • Orbit Determination Analysts

    Study radio transmissions and images from cameras to determine the spacecraft’s current location.

  • Maneuver Designers

    Plan propulsive maneuvers to keep the spacecraft on the correct flight path.

  • Radiometric Tracking Team

    Assesses techniques for acquiring and improving the accuracy of radio-based measurements of the spacecraft’s velocity, location, and angle relative to Earth.

  • Trajectory Designers

    Plot the most efficient path for the spacecraft.

So it really is a team effort involving large groups of experts employing several different methods to locate a satellite.

  • $\begingroup$ This is an interesting breakdown. In which group or category would the actual computation reside? For example, the Ephemeris group might supply the state vectors and masses of all gravitational bodies that could affect a spacecraft, but who combines those gravity fields with the radio-based measurements and actually does the mathematical integration and iterative fitting to compute the complete trajectory solution? When I look up a spacecraft's trajectory in the JPL Horizons database as I did here who does that final calculation? $\endgroup$ – uhoh Apr 27 '17 at 0:28
  • $\begingroup$ Also, note that the question is how, not who. While your first sentence contains the word "how", the rest of the answer is really just a list of groups, not an explanation of a process $\endgroup$ – uhoh Apr 27 '17 at 0:28

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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