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Keplerian orbits, those around spherically symmetric mass distributions (Newton's Shell theorem collapses them to a point) have analytical solutions in that you can write $t(\theta)$ as a simple equation. Of course the reverse $\theta(t)$ can't be written and still has to be solved numerically.

But my guess is that for anything other than Keplerian orbits proper, there is no simple equation for the orbit. There are plenty of equations out there, the result of various perturbation calculations or other approximations.

For example, see

These refer to Kelperian-like orbital parameterization of orbits around an oblate/prolate spheroid characterized by $J_2$. For example the equation for nodal period including effects of eccentricity and $J_2$ was typed out in @Chris' answer:

$$T = T_0\left[1 - \frac{3J_2(4-5\sin^2 i)}{4\left(\frac{a}{R}\right)^2\sqrt{1-e^2}(1+e\cos\omega)^2} - \frac{3J_2(1-e\cos\omega)^3}{2\left(\frac{a}{R}\right)^2(1-e^2)^3}\right]$$

Here, according to the answer, $\omega$ is the argument of perigee, and $e$ and $i$ are eccentricity and inclination.

We can suspect this is might be an approximation because a real orbit around a body with nonzero $J_2$ won't have easily recognized Keplerian elements (the orbit won't even be planar!), they'd have to be redefined. But maybe in the context of this equation the non-elliptical orbit can still have a well-defined eccentricity somehow.

Question: When all is said and done, Any exact analytical solutions for non-Keplerian orbits; those around non-radially symmetric mass distributions (e.g. J₂≠0)? Or once $J_2$ deviates from zero or the central mass deviates in any way from spherically symmetric do the equations of motion always become approximations?

For the purposes of this question, "exact analytical solutions" could include infinite series, as long as they can be and have been written as such.

note: This is a complicated question to write; I'm open to comments recommending adjustments to the wording or the scope.

uhoh
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  • Very good question, I have been cogitating on this too. But "orbits around an oblate/prolate spheroid characterized by $J_2$" seems to contain a misinterpretation: IMO, the $J_n$ can only model oblateness. BTW, as a first step, I would limit the scope to gravity potential involving only a $J_2$ > 0 (I don't think it can be negative). – Ng Ph Nov 04 '21 at 15:39
  • @NgPh there's no such mathematical sign restriction. $J_2$ and $J_{22}$ are simply the two lowest order terms in an infinite multipole expansion of a Geopotential model. It's true that for a large but slowly rotating planet in hydrostatic equilibrium it will be oblate, things get complicated and it can become triaxial if it rotates faster. On the other hand a small rocky body does not have to be in hydrostatic equilibrium and can have any $J_2$ it wants to, positive or negative. – uhoh Nov 04 '21 at 17:20
  • @NgPh If an asteroid happens to be rotating around its long axis, it will likely have a negative $J_2$, whereas if the same asteroid happens to be rotating around its short axis (axis of largest moment of inertia), its $J_2$ will now (likely) be positive. It simply depends on which direction you decide to call the object's $z$-axis, and traditionally that's the object's current axis of rotation. – uhoh Nov 04 '21 at 17:52
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    can you point me to the definition of $J_{22}$ in the Wikipedia geopotential model? – Ng Ph Nov 04 '21 at 20:57
  • @NgPh looks like they don't go that far, but $J_{22}$ would be the zonal harmonic coefficient for the $P_2^2$ term ($J_2$ is associated with $P_2^0$) in the stuff between Equations 9 and 10. This answer links to The Earth's Longitude Gravity Field as Sensed by the Drift of Three Synchronous Satellites – uhoh Nov 04 '21 at 22:46
  • From what I have read, the term "zonal" seems to be reserved for the case m=0, that is no variation longitudinally (Phi angle in Wiki notation). In addition, I can't understand your explanations of negative $J_2$ with inertia moments. I would think that the spherical harmonics are related to the geometrical form only (homogeneous distribution of mass within this form). – Ng Ph Nov 05 '21 at 10:23
  • @NgPh rather than try to have a multi-threaded discussion in comments, ask a new question, perhaps "Can $J_2$ have eithwr sign?" – uhoh Nov 05 '21 at 10:53
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    Does this question include non-physical solutions? That is, shapes that are impossible or highly unlikely to occur in nature. I think I can come up with some examples originating in electric fields, the cousin of gravity. But you won't find any asteroids with a funny enough shape. – SE - stop firing the good guys Nov 09 '21 at 11:21
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    @SE-stopfiringthegoodguys I'm torn, because no, except you will probably come up with something really interesting so yes... :-) There's this: Could you stably orbit around a square (cubic) body? Would the orbit destabilize automatically if not corrected by input? for example – uhoh Nov 09 '21 at 11:23
  • You should be able to use Lagrangian/Hamiltonian mechanics to any mostly sane gravitational field, and come up with equations of motion, which will be differential equations, usually 2nd degree. Now whether these equations have analytic solutions, that's a different matter; certainly some will, but you might need particular, specific layouts of the masses. – SF. Nov 10 '21 at 13:05
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    The motion of a satellite under the $J_2$ perturbation is described by a dynamical system that is not integrable, which implies that there exist no exact, closed-form analytical solutions. Source: 1, p. 5, which cites proofs 2, 3. You could still write the solution as an infinite Taylor series about the initial conditions, but I'm not sure if there might be some convergence issues. – LeWavite Nov 14 '21 at 18:31
  • @LeWavite that's so informative and constraining that I'd really welcome it as a partial answer, please feel free to do so! – uhoh Nov 14 '21 at 23:48
  • @LeWavite answered the question of integrability. You used the terms exact analytical. While it seems obvious that an "integrable" ODE results in an "exact analytical" solution, is the converse true? Is it your belief that it is not necessarily true, that is the two terms ("integrable" and "having exact analytical solution" are not synonymous)? – Ng Ph Nov 18 '21 at 22:51

2 Answers2

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The motion of a satellite under the 2 perturbation is described by a dynamical system that is not integrable, which means that there exist no exact, closed-form analytical solutions. Source: [1, p. 5], which cites proofs [2], [3].

You could still write the solution as an infinite Taylor series about the initial conditions, but I'm not sure if there might be some convergence issues.

LeWavite
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  • This is great! While not a complete answer yet, this offers helpful insight and constraints. Sounds like that if there are any solutions they are likely curiosities; special cases. – uhoh Nov 15 '21 at 19:32
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    I agree. In any case, even if one could write down an analytical solution not in closed form (such as an infinite Taylor series), the computational effort for such solution would probably be higher than that of a high-accuracy, numerical solution... – LeWavite Nov 16 '21 at 09:12
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    @LeWavite, nevertheless, a solution based on "brute-force" integration will not provide the insight that an infinite series may give, such as by truncating the series and analyze the closed-form approximation. – Ng Ph Nov 16 '21 at 18:40
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This is to address OP's note:

This is a complicated question to write; I'm open to comments recommending adjustments to the wording or the scope.

When the force potential field of a 2-body system is perturbed, the Newtonian motion of the non-central body is no longer an ellipse around the central body. However, the motion's trajectory, limited in time, can be viewed as a distorted ellipse whose deformation varies with time. The position and velocity of the satellite at a given instant, also called the state vectors, can be used to uniquely define an ellipse, in the Keplerian sense, that would be the satellite trajectory if all perturbations were removed. This uniquely defined elliptical orbit has correspondingly Keplerian orbital elements: a, e, I, Ω, ω, and τ (semi-major axis, eccentricity, inclination, RAAN, argument of perigee, true anomaly). It is usually designated as the (Keplerian) osculating orbit, at the particular instant.

In a perturbed motion, in most practical cases (e.g. when the perturbating forces are small), the osculating elements a(t), e(t), I(t), Ω(t), ω(t), and τ(t) can be assumed to vary slowly (wrt to the "period" of any osculating orbital elements).

I will digress at this point and address one implicit question in the OP's Question: How can we define what is an "exact analytical solution"?

Let's assume that we are at the center of a circle and we observe a moving car whose path follows the circle. Assume that very far away we have a tree that we can use as a reference point.

  • The question to us is: when do we have an "exact analytical solution" for the position of the car?

We can resolve the "loose-end" of the definition of "exact analytical" by re-phrasing: Knowing that the car is in front of the tree at time T0, can you give the position of the car at any time T, so that the accuracy of your prediction is independent of how far T is from T0 and without knowledge of the positions at intermediate times?

Obviously, if the angular speed of the car is constant and known, we can close our eyes and say "give me any T, I can predict the position exactly". That is the accuracy of our prediction is dependent only on the accuracy of our knowledge of the angular speed (and the accuracy of our time keeping function, of course).

Now, this is equally true even when the car is not moving at constant angular speed. For example, it can move at one constant speed for half the circle, then at twice that speed for the 2nd half, etc.... As long as we know how the angular speed changes with time, analytically, we can predict the position without computing the intermediate positions. And this would be our definition of "exact analytical solution".

Would that mean that, mathematically speaking, the motion of our car is an integrable solution of the equation of motion? This is far beyond my mathematical reasoning capability. And furthermore, I would say it may not be of much interest in practical applications to know the answer to that theoretical question.

Ng Ph
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