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
    Commented Nov 17, 2020 at 2:54
  • $\begingroup$ Maybe "rational-fractional-synchronous ground track orbits"? hmm.. maybe not ;-) $\endgroup$
    – uhoh
    Commented Nov 17, 2020 at 3:10

1 Answer 1


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! $\endgroup$
    – user21103
    Commented Nov 16, 2020 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
    Commented Jan 21, 2021 at 0:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.