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Imagine a satellite in an elliptical orbit around Earth. As the earth travels around the sun, does the elliptical orbit of the satellite swing around Earth, as pictured in A, or does it maintain a consistent orientation as pictured in B? Are both of these options wrong?

Option A

diagram of elliptical orbit changing orientation

Option B

diagram of elliptical orbit not changing orientation

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I would suspect option B, since it would be weird if the motion of an object would both have an angular motion around its planet and the sun (with different frequencies). Also did a quick test in Algodoo (as a numerical simulation) and I can confirm option B with it. I do not have any mathematical or physical proof. –  fibonatic May 22 '13 at 20:18
@fibonatic If the problem is truly hierarchical, then the sun's tidal forces are nothing and I think B should be provable. Because you can apply any acceleration function over time, and if its done uniformly then the geodesic is a valid inertial reference frame. So the moon maintains its orbital orientation as it would normally. –  AlanSE May 22 '13 at 20:31
See also Bertrand's theorem –  Ikiperu May 22 '13 at 23:18
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2 Answers

up vote 9 down vote accepted

Actually, both are quite possible. In the general case, for an arbitrary elliptical orbit, what you'll tend to find is that B is true (granted there will be some precession, but not usually in line with the Sun). However, it is possible to set up an orbit (such as a Sun-synchronous one) in which A is true. But orbits such as this require planning and precise positioning.

To understand why A does not have to be true (and in fact, why A is not the general case), you should bear in mind that when orbiting a body, your orientation is irrelevant. The Sun ensures both Earth and the satellite will orbit around it, but it has very little influence over the specific way they are pointing.

Imagine that Earth did not rotate on its axis; you'd see the same stars in the same positions every night. However, you'd still experience the Sun rise and set (once per year). The sun doesn't drag the orientation of orbiters (at least, not in a significantly mention-able way). All that is important from the satellite's perspective is Earth, which has the dominant sphere of influence.

As I mentioned earlier, however, it is possible to establish orbits that follow the Sun. A sun-synchronous orbit can be very useful as it allows a satellite to always be in view of the Sun and thus continuously collect solar power. This is possible due to a combination of several effects including the small amount of sway the Sun's gravity has over the satellite. However, to establish this type of orbit requires the use of a specific angle of incidence (which varies based on altitude). In Low Earth Orbit, the inclination is around $98^\circ$.

An important thing to note is that while most orbits do not generally follow the Sun, they technically do precess (meaning they usually do not look exactly like B but it's a good approximation/generalization). This precession happens at different rates depending mostly on altitude, inclination, and eccentricity of the orbit. But that's just extra info in case you were curious.

<3 rocket science

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Thank you for the answer. That is actually great news. I'm writing a simulation and the orbits typically turn out like B, so I assumed I had a bug somewhere in the orbit code. Thanks! –  Andrew Garrison May 22 '13 at 20:38
@Jim A small correction: the inclination $i=63.5^\circ$ corresponds with a Molniya orbit instead of a Sun-synchronous orbit. –  Pulsar May 22 '13 at 23:41
@Pulsar right. I keep switching up my inclinations. Thanks –  Jim May 23 '13 at 16:12
"The Sun ensures both Earth and the satellite will orbit around it" - So that means satellites in the earth's orbit don't take into account their motion around the sun? It happens automatically? –  capex Sep 30 '13 at 1:47
@humbleware that means the satellites in Earth's orbit are in free-fall around Earth and around the Sun. Earth's sphere of influence allows us to essentially ignore the Sun's presence most of the time. But the context of the statement indicates that the orientation of the objects is not as relevant as the fact that it is orbiting. –  Jim Sep 30 '13 at 15:50
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The orbit of a satellite is determined by 6 parameters, the orbital elements:

  • The eccentricity $e$
  • The semimajor axis ($a$)
  • The inclination ($i$)
  • The longitude of the ascending node ($\Omega$)
  • The argument of periapsis ($\omega$)
  • The mean anomaly at a specific time ($M_0$)

enter image description here

The plane of reference is the equator, and the reference direction is the Vernal Point.

If the reference frame of the Earth were an inertial frame, then these parameters would be constants, and your Figure B would be valid. However, the orbits of satellites are perturbed due to several effects:

  • The non-sphericity and non-homogeneous mass distribution of the Earth
  • The gravitational field of sun and moon
  • Solar radiation pressure
  • Atmospheric drag (for low-orbit satellites)
  • General-Relativistic corrections

The first effect is the largest. The non-sphericity of the Earth can be approximated by a quantity $J_2\approx 1.083\times 10^{-3}$, and it has two consequences:

  1. a rotation of the line of apsides: the position of the perigee varies as $$ \dot{\omega} = \frac{3\pi}{T}\frac{J_2\,R_\text{e}^2}{a^2(1-e^2)^2}\left(2 - \frac{5}{2}\sin^2 i\right), $$ where $T$ is the orbital period of the satellite, and $R_\text{e}$ is the equatorial radius of the Earth. Interestingly, one can choose the inclination $i$ in such a way that $\dot{\omega}=0$, namely by setting $i = \sin^{-1}\sqrt{4/5}$, so that $i=63.43^\circ$ or $i=116.57^\circ$. Such orbits are called critically inclined orbits or Molniya orbits.
  2. a nodal precession: $$ \dot{\Omega} = -\frac{3\pi}{T}\frac{J_2\,R_\text{e}^2}{a^2(1-e^2)^2}\cos i. $$ One can choose the orbital parameters in such a way that the period of this nodal precession is equal to the orbital period of the Earth around the Sun, which implies $\dot{\Omega} =0.9855^\circ/\text{day}$. The result is a Sun-synchronous orbit, which means that the orbital plane of the satellite rotates as in your Figure A.

For more details, see also this link and this link.

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+1 But I noticed you didn't actually answer the question. This is mostly just background info. Perhaps a more explicit answer of the question would help gain you more votes? –  Jim May 23 '13 at 16:30
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