# Is an Earth synchronous (?) polar orbit possible?

I don't even know how to exactly ask this question. A coworker is doing some experiments on a data visualization software and he has some test data showing fake telemetry from a simulated satellite. This data shows a polar orbit, but once displayed on a globe it looks weird: the satellite always follows the same path around the globe, like if it was rotating at the same speed as the earth. It looks like this (Sorry, I can't show you the original but only this very accurate Paint 3D reproduction):

Our question is: is this kind of orbit possible ?

• That is what a polar orbit would (roughly) look like, if the earth was spinning under it, but at that altitude it would be spinning much faster than the Earth.
– user20636
Commented Jan 10, 2020 at 11:56
• observablehq.com/@jake-low/satellite-ground-track-visualizer shows a visualisation of a satellite orbiting the Earth, while it rotates under it.
– user20636
Commented Jan 10, 2020 at 12:27
• This reminds me of a Sun-synchronous orbit. Not the answer to your question, but weird in that it always track the same spot on Earth with regard to the Sun, e.g. always tracking the sunrise/sunset-line. It works because the Earth is slightly oblate, and a nearly polar orbit at the right altitude will slightly precess (precessing exactly once per year) Commented Jan 10, 2020 at 23:51

A polar orbit can be geosynchronous and always follow the same path, but that path cannot be straight along a meridian. If the orbit passes on top of both poles, then the orbit lies on a plane containing both poles, which cuts the earth longitudinally like two opposite meridians do. However, as the satellite travels along that orbit, the Earth moves so its path superimposed on the Earth will not be straight along a meridian.

The simulation linked by JCRM shows how the path of a polar orbit looks like on Earth: https://observablehq.com/@jake-low/satellite-ground-track-visualizer

Set orbit inclination to -90° for a polar orbit.
You can also set orbital altitude: see how by rising the orbit up to 35000km the multiple paths gradually merge into one.
(Geosynchronous should be at 35,786km, which is at the limit of that simulation parameters and unfortunately only a part of the path is shown)

• It's simple to change the limits and parameters of this, it's an online script that can b edited by anyone: observablehq.com/@asdfex1/satellite-ground-track-visualizer This shows a 10 times longer track, allows super-synchronous orbits and higher simulation speed. Commented Jan 11, 2020 at 12:13
• What a great tool! Commented Jan 11, 2020 at 16:12
• In what sense could this orbit be "geosynchronous"? Its position over the Earth is constantly moving, correct? Commented Jan 12, 2020 at 7:02
• @CarlLeth It's synchronous to the Earth rotation, it gets back to the same point every day. Only geostationary orbits are not moving at all. These are geosynchronous orbits with 0° inclination, at all other inclinations the orbit is geosynchronous, but not geostationary. Commented Jan 12, 2020 at 10:52
• @asdfex I see; you are pointing out the distinction between "geostationary" and "geosynchronous". I suspect "geostationary" is what people usually mean when they say the latter. Doesn't an orbit with a 12-hour period (or 8, or 6...) also return to the same point every 24-hour period? But your description does seem to match OP's question. Commented Jan 12, 2020 at 19:20

No, such an orbit is not possible (or: it's not possible unless the satellite is accelerated most of the time, which is not plausible in terms of fuel). In particular it's not possible for the path of the orbit to trace out a great circle over surface the Earth (in a coordinate system rotating with the Earth). Your picture shows a great circle.

For a circular polar orbit with period $$p$$, then for successive passes over the equator, the point where the satellite is directly overhead will move west by a distance of about

$$\frac{2\pi R p}{d}$$

where

• $$p$$ is period of the orbit;
• $$R$$ is the radius of the Earth;
• $$d$$ is the length of the sidereal day (which is slightly less than the solar day).

In real life there will be various perturbations which make this not quite accurate.

From this it's easy to see that the track of the orbit can never be a great circle: it must necessarily pass over both poles, but the angle at which it cuts the equator can never be $$\pi/2$$, and any great circle which passes through the poles must cut the equator at that angle.

There is, I am sure, software which will compute & plot orbit tracks: I'm afraid I don't know of any to recommend however: if someone else does feel free to edit this answer.