I am researching into how coordinate systems of solar systems objects are created by reading some of the reports written by the Working Group on Cartographic Coordinates and Rotational Elements (e.g. 2009). However, I am finding it difficult to completley understand the role of time in defining reference systems.

When observing at a planet from Earth, for example Jupiter, there are a variety of factors that make it difficult to construct a reference system (including no solid surface and planetary precession), so we use geometry to define a reference system. However, because our perspective is dynamic, meaning Jupiter's surface changes and the planet's move, we say that at time J2000 we know the precise orientation and position of Earth and therefore can say from position defined at J2000 this is the coordinate reference system for Jupiter.

So, does incorporating time (e.g. J2000) mean we can say a coordinate reference system is based on the situation of an object, Jupiter in this example, at a given moment?

  • $\begingroup$ I'm not sure I understand your question exactly. In order to define a reference system for a planet's latitude and longitude, we need a fixed frame that doesn't depend on the Earth's own precession. For this, we choose the J2000 frame which is based on the Earth's equatorial reference frame at a given instant. In other words, we set the J2000 frame so Earth's own frame changes don't affect the latitude and longitude of positions on another planet. $\endgroup$
    – user21
    Jan 26, 2016 at 15:01
  • $\begingroup$ So, I've edited the question to try and clarify what I'm on about. However, I think you've answered my question. Choosing J2000 means we can define a coordinate system for a planet or object without having to compensate for changes in observation because of Earth's precession. $\endgroup$
    – spk578
    Jan 27, 2016 at 12:07
  • $\begingroup$ But how do we compensate for another planets precession? How do we account for the object's precession affecting the reference system we defined at J2000 for the object? $\endgroup$
    – spk578
    Jan 27, 2016 at 12:11
  • $\begingroup$ The formulas given in astropedia.astrogeology.usgs.gov/download/Docs/WGCCRE/… are dependent on the date, so they do compensate for precession. $\endgroup$
    – user21
    Jan 27, 2016 at 15:52
  • $\begingroup$ Just wait till you get to Saturn, where it's even more axisymmetric than Jupiter.... ;-) $\endgroup$
    – jvriesem
    Aug 13, 2019 at 20:21

2 Answers 2


Look at J2000 Solar System Barycentric (spatial reference) and Barycentric Dynamical Time (TDB). Together they are good space and time coordinate system. NASA/JPL has some good info and data on these.

In terms of time, TDB is rescaled so that from the Earth, it appears to be close to the same as TT (roughly ~UTC). The rescaling is due to the fact we are in a gravitational well, as well as moving at 30 km/s with respect to the solar system barycentre, due to relativity. The original unscaled is called Barycentric Coordinate Time (TCB) and differs by ~0.5seconds / year.


We need a system to describe "where things are in the sky". Even a cursory glance at the sky will find that "things move across the sky daily". So instead of describing where something is directly, we will describe where it is relative to the stars.

However some of the stars move (due to their actual motion relative to the sun) and appear to wobble (due the motion of the Earth around the sun). So let us consider those objects that are so far that any such motion is undetectable. For example, quasars. Other distant stars are also suitable as they don't have a measurable motion. I'll call these sources "fixed stars". The goal is to describe a system of coordinates in which the fixed stars don't move.

For our coordinate system we will use the plane of the Earth's equator on the March Equinox (chosen in part so the plane passes through the sun). Declension is defined as the angle relative to this plane. Right Ascension is then the angle between the line through the Earth and Sun, and the line formed by projecting the object onto the plane. For very distant objects, it doesn't matter if we use the sun or the Earth as the centre as the angle will be the same to any reasonable level of accuracy.

However, choosing the March equinox in this way causes a problem, because the plane of the Earth's equator is slowly changing, and this means that the position relative to this plane will also slowly change. The RA and Dec of a quasar will slowly change due to this precession.

The solution to this issue is to define the coordinate system on a particular date "Jan 1st 2000". With this convention we can assign a position to the quasar, and it won't change. This is a coordinate system that can describe the position of any object relative to the fixed stars.

We can now define the position of any object in the same coordinates. For nearby stars we can describe their proper and apparent motion relative to this coordinate system. For planets the position relative to the fixed stars varies from day to day, due to the relative motion of the planets. It also depends on the location of the observer. So I can talk about the location of Jupiter at midnight on 28 June 2018, from Perth, WA, using the J2000.0 coordinates.

The 2000.0 defines exactly which fixed coordinate system we are using. But to describe the location of Jupiter in the sky we also need to use an observation time and date and location.

  • 2
    $\begingroup$ I think the question is about the features on the surfaces of the planets, not their positions in the sky. $\endgroup$
    – Mike G
    Jun 28, 2018 at 15:26
  • $\begingroup$ This is a great answer to some question, but it does look like @MikeG has a point about this question. $\endgroup$
    – uhoh
    Jun 28, 2018 at 23:04
  • $\begingroup$ I read it as asking if the J2000.0 is an observation date for (say) Jupiter, which has it "surface changes and the planet's move" $\endgroup$
    – James K
    Jun 29, 2018 at 4:23

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