# How is the altitude of a satellite defined, given that the Earth is not spherical?

The altitude of a satellite is the distance between the Earth's surface and the satellite, but the Earth itself is not spherical. At the equator the Earth's radius is 21 km more than at the poles, and in fact the shape of the Earth is not even a perfect oblate spheroid.

So how is the altitude of a satellite actually defined?

For a satellite with a circular orbit which is constant - the distance between the Earth's surface to the satellite or the distance between the Earth's center and the satellite?

update: 6378.137 km is what I use now.

By convention the altitude of a spacecraft is the distance to the center of the Earth minus roughly 6378 kilometers, or some reference radius that is representative of the equatorial radius of the Earth. Spacecraft altitude is not really used as a precise description of a satellite's position, since its only a scalar and it requires a definition, but if you have to state some value for an altitude, use the distance to the Earth's center of mass minus some reference radius which is generally an average equatorial radius. For example I see the following equation a lot (e.g. 1, 2, 3, 4, 5, 6 and in this answer):

$$\text{altitude} = |\mathbf{r}| - 6378 \ \text{ or } \ 6378.1\text{ km}$$

Another way you can confirm this is to simply search this SE site for "6378". (it turns out most of the time those are from me, so it doesn't count). A more accurate value for the equatorial radius of the Earth could be used, such as 6378.137, but the difference is not meaningful because a spacecraft's altitude per se is not really a precisely defined quantity.

You can read more about the nadir and the question of precisely what point on Earth's surface is actually "directly beneath" the satellite at a given moment, and the difference between the geodetic subsatellite point rather than the geocentric subsatellite point in this excellent answer.

Now the last part of the question:

For the satellite with circular orbit which is constant - the distance between earth surface to satellite or the distance between earth center and satellite?

The short answer is "neither", but the answer that's likely to be closest to correct is that the distance from the satellite to the center of mass of the Earth is likely to be more constant than the distance to the surface, assuming you figure out whether you mean the geodetic or geocentric subsatellite point. This is because the the higher order gravitational terms beyond the monopole are weak and have only a small effect.

A quick way you can confirm that is to note that the Earth's radius varies by more than 20 kilometers from equatorial to polar, and yet satellites in circular orbits can have periapsis and apoapsis within a kilometer of each other. For example I chose two of the satellites in the "A-Train" constellation, which has a nearly polar orbit. Aqua and Aura and looked up their info in Wikipedia. Both of them have a difference between periapsis and apoapsis of only 2 kilometers or less.

A useful convention:

Using this soft definition, it is common to see orbits described using the altitudes at periapsis and apoapsis. For example, this discussion of a 625 x 625 km orbit of an Iridium satellite means that it's circular with a radius of 625 + 6378 km, and a Geostationary Transfer Orbit of 185 km x 35,786 km would have its apses also at about those values plus 6378 km added to each.

• This is a great answer! – ChrisR Nov 28 '17 at 14:06
• A note that, e.g., 6378.137 km is certainly a theoretical or some average value. Mean sea level, or so, I guess. It goes down to a precision of a centimetre, so it does not account for mountains. Or even for dust piling up on the ground :-) – Rolazaro Azeveires Nov 29 '17 at 15:45
• @uhoh Right. :-) I missed it. Sorry, and thanks. – Rolazaro Azeveires Nov 29 '17 at 18:03
• @uhoh Thanks! I was plotting orbit of one student satellite (sun-synchronous LEO satellite) using SGP4. The fact that for that satellite the eccentricity was 0.0087 which was higher than Aqua's eccentricity 0.0000979. Due to high eccentricity the difference between semi-major and semi-minor axis was around 110 km. This 110 km is small compared to radius of Earth but 1/6th compared to 600 km (in general LEO satellite altitude). I had a notion that sun-synchronous orbit is nearly circular but this seems highly elliptical (from Earth's surface point of view) even with 10^-2 order of eccentricity. – Sumit Agrawal Nov 29 '17 at 20:41
• @uhoh I guess as it was a student satellite so due to convenience of launching, the eccentricity was a little higher. – Sumit Agrawal Nov 30 '17 at 2:01

In orbital mechanics, the position of a satellite can be defined using orbital elements (which have some advantages over other coordinate systems, for typical orbits). Orbital elements describe a position in relation to the centre of mass of the Earth or any other body. For example, one of the orbital elements is the semimajor axis, which for a circular orbit is identical to the distance between the satellite and the Earth centre of mass — not the Earth surface.

As you say, the distance to the Earth surface changes over even a circular orbit. One can calculate this based on the geoid, and it's useful for many purposes, but it's rather a diagnostic than a prognostic variable.

• I would say "can be defined using OEs." All one needs is six components, and these could be position and velocity, or some kind of orbital elements. Most OE definition go singular at specific points requiring a change in their definition, whereas vectors don't. (Sorry I'm a big fan of the vector notation.) – ChrisR Nov 28 '17 at 10:36
• @ChrisR Fair point. Edited. – gerrit Nov 28 '17 at 10:41

Earth is approximated as a point-mass when doing this calculation. The oblatness of the planet and different gravity potentials can then be computed using spherical harmonics or "mascon" (short for mass concentration).

When designing special orbits, such as those around the Lagrange points, it isn't uncommon to define a new reference frame whose center is at barycenter of the two celestial bodies considered for the problem.

To expand on the other answers, it really depends on what you want to use the altitude for.

In the Shuttle Mission Simulator, we calculated the orbital altitude several different ways.

• Altitude above the Fischer Ellipsoid
• Altitude above Mean Sea Level
• Altitude above the selected landing site
• This is a really important point! Not all spacecraft are in long-term stable orbit. Occasionally one is full of people trying to get home, in which case altitude becomes incredibly relevant. – uhoh Nov 30 '17 at 1:01