# Under what conditions can a non-resonant satellite stay hidden from an observer on the ground?

From the perspective of someone on the ground, a satellite is usually not above the horizon all the time. But can a satellite have such an orbit that it is never above the horizon?

Such orbits clearly exist. The easiest example is a satellite in geostationary orbit, which will never be visible from the antipode of the stationary foot point.

However, geostationary and geosynchronous orbits are resonant orbits with an orbital period with a 1:1 resonance to the rotation of the Earth. In the grand scheme of things, such simple fractions are a special case.

Theoretical motivation: Orbital resonance requires the orbital period to be some rational numbers. The rational numbers form a countable set, while the real numbers do not, so "almost all" orbits are non-resonant.
Practical motivation: Satellites are subject to various perturbations, requiring active station keeping to stay in a resonant orbit.

An alternate and equivalent formulation is to find orbits such that a satellite can be located anywhere in it and still stay below the horizon all the time.

What are the constraints of such orbits?

• While it's convenient to think of these as resonant, we shouldn't, unless you really mean that there is a regular exchange of energy "locking" their motion. Is Dawn's upcoming low periapsis orbit for XMO7 “resonant”? Do you mean instead repeat-ground track orbits? Resonance is "a thing" in physics and only happens in coupled dynamical systems, and I don't think that is what you are after here.
– uhoh
Nov 17 '20 at 2:54
• Maybe "rational-fractional-synchronous ground track orbits"? hmm.. maybe not ;-)
– uhoh
Nov 17 '20 at 3:10

This is a case of an answer originating before a question. Originally a spin-off problem from this question asking about satellite footprints, it doesn't fit very well as an answer there. Hence this separate question and answer.

As it turns out, this has a straightforward geometric solution from observer latitude $$\phi$$ and inclination $$i$$:

$$r_P < \frac{r_{earth}}{\cos(\phi - i)}$$

$$r_A < \frac{r_{earth}}{\cos(\phi + i)}$$

From which a few corollaries follow:

• $$i < \phi$$

• $$r_A$$ is unbounded iff $$i + \phi \geq \frac{\pi}{2}$$

• $$r_P$$ is only unbounded iff $$\phi = \frac{\pi}{2}$$ and $$i = 0$$

• There are no such orbits iff $$\phi = 0$$

Some practical consequences:

• For a hidden non-resonant orbit to have a perigee altitude above 200km, the observer must have a latitude greater than 14 degrees.

• The Moon is visible from the poles.

Alternate view, showing the tight squeeze when the observer is at low latitude:

Alternate view, showing the degenerate polar case with unbounded apogee:

• This is a really nice answer!
– user21103
Nov 16 '20 at 19:17
• perhaps this would have been better starting with the (hopefully) verily easily understood "something in equatorial orbit will never be visible from the poles" , and working outward from there?
– user20636
Jan 21 at 0:07