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?

  • $\begingroup$ 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. $\endgroup$ – uhoh Nov 17 '20 at 2:54
  • $\begingroup$ Maybe "rational-fractional-synchronous ground track orbits"? hmm.. maybe not ;-) $\endgroup$ – 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.

non-resonant orbital constraints

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

low altitude non-resonant orbit

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

degenerate polar non-resonant orbit

  • $\begingroup$ This is a really nice answer! $\endgroup$ – tfb Nov 16 '20 at 19:17
  • $\begingroup$ 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? $\endgroup$ – user20636 Jan 21 at 0:07

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