Is gravitational force represented among the orbital elements?

[T.C.]

I'm trying to understand orbital mechanics, and I've become confused by something I read on Wikipedia.

This section says that "In principle once the orbital elements are known for a body, its position can be calculated forward and backwards indefinitely in time". (The article adds appropriate disclaimers for perturbations). I am confused by this because it seems to me that to do orbital predictions, one must know the force of attraction between the bodies, yet the force of attraction doesn't seem to be represented anywhere among the six traditional Keplerian elements.

Am I mistaken? Is the gravitational force represented among the Keplerian elements in a way I don't understand?

Alternatively, is there some unspoken assumption that, whenever one knows the six Keplerian elements, one also knows the masses (or gravitational parameters) of the bodies, effectively making the force an implied seventh orbital parameter?

Or is it possible that the Wikipedia article is wrong, and knowing the orbital elements is

not

sufficient to calculate the position of an orbiting body forward and backward in time?

Note that I asked essentially the same question on the "Orbital elements" discussion page.

-TC

[peadar1987]

I'm trying to understand orbital mechanics, and I've become confused by something I read on Wikipedia.

This section says that "In principle once the orbital elements are known for a body, its position can be calculated forward and backwards indefinitely in time". (The article adds appropriate disclaimers for perturbations). I am confused by this because it seems to me that to do orbital predictions, one must know the force of attraction between the bodies, yet the force of attraction doesn't seem to be represented anywhere among the six traditional Keplerian elements.

Am I mistaken? Is the gravitational force represented among the Keplerian elements in a way I don't understand?

Alternatively, is there some unspoken assumption that, whenever one knows the six Keplerian elements, one also knows the masses (or gravitational parameters) of the bodies, effectively making the force an implied seventh orbital parameter?

Or is it possible that the Wikipedia article is wrong, and knowing the orbital elements is

not

sufficient to calculate the position of an orbiting body forward and backward in time?

Note that I asked essentially the same question on the "Orbital elements" discussion page.

-TC

You don't actually need to know the force, you need to know the acceleration.

A bigger body will have more gravitational force applied to it by virtue of its greater mass, but, again by virtue of its mass, a given amount of force will cause a smaller acceleration. The mass terms on each side of the equation cancel, so acceleration due to gravity is independent of the mass of the smaller object.

[KSK]

From a very quick skim:

Once you've found the semi-major axis, you can calculate the orbital period via Kepler's third law (which includes the gravitational constant and the mass of the larger body in your system). Combine that with Kepler's second law and the shape of the orbit determined from the other orbital elements and you can figure out how the system will evolve with time. The Mean Anomaly at Epoch gives you a reference point to calculate that evolution from.

Admittedly, this is rather hand-wavy but I can sort of see how you could calculate positions given the above information. You can kind of treat the orbiting body as a 'moving bead on a wire' without worrying about what's causing the bead to move.

Caveat - I'm also guessing that this doesn't work well unless you have a very significant difference in mass between your orbiting bodies, such that the barycenter of the system is (to a good approximation) at the centre of the heavier body.

Edited June 18, 2015 by KSK

[Meithan]

Remember that the force of gravity depends only on the distance between the two bodies (it doesn't depend, for instance, on their velocity). That means that if you know, for example, the exact 3D position of a body relative to the Sun (and it's assumed you know the Sun's mass or its gravitational parameter), then the force is implicitly known too. You'd need the mass of the body too to calculate the force, but as peadar1987 pointed out what matters to determine the motion is the acceleration (the force is just an intermediate calculation, if you will, to obtain the acceleration), and gravitational acceleration is independent of the mass of the body.

So you could say that the orbital elements contain in themselves the information that determines the acceleration of the body at any point in time (since all you need is know where it is relative to the other body). This is not an accident: the elements were actually derived using the fact that we can compute the acceleration just by knowing the position of a body.

Also note that if you already know the six orbital elements, you don't need to compute the force or the acceleration anymore to make predictions! The orbital elements are already a solution to the equations that dictate how a body moves knowing its acceleration. So once you have the orbital elements you can forget about force or acceleration altogether: predicting where a body will be or was at any point in time is a matter of geometry.

Edited June 18, 2015 by Meithan

[T.C.]

Thank you for the replies. I believe I have learned the following:

The six Keplerian elements alone are not sufficient for orbital predictions. Additional information is required. I mistakenly believed the additional information is a force, but as peadar1987 explained, it simplifies to acceleration. In any case, this additional information is

not

represented among the orbital elements. However, as peadar1987, KSK, and Meithan explained, the additional information can be inferred from the mass of the central body.

Unfortunately, I am still not clear whether there is an implicit assumption that the additional information is always known.

In other words, is the Wikipedia article correct when it says that "once the orbital elements are known for a body, its position can be calculated forward and backwards indefinitely in time" because we implicitly know the additional information required (e.g. the mass of the central body)?

Or, is the Wikipedia article incorrect because there is no implicit assumption that the additional information is known, and without the additional information, the six orbital elements are insufficient for orbital predictions?

-TC

[ZetaX]

You don't need the mass of the central body. You don't need the forces. As has been said, acceleration is all you need. That acceleration is towards the central body and proportional to 1/distance², not depending on the orbiting body. You can calculate the actual acceleration at a given point (and thus all you need) from a single orbit. For a circular one this is especially easy, as centrifugal acceleration (which is a function in orbital speed and radius) has to match that acceleration towards the center; for general ones you could also calculate it by reducing it to the circular one, or doing it directly (but am now too lazy to find out an actual formula).

Edited June 18, 2015 by ZetaX

[TheShadow1138]

I'm trying to understand orbital mechanics, and I've become confused by something I read on Wikipedia.

This section says that "In principle once the orbital elements are known for a body, its position can be calculated forward and backwards indefinitely in time". (The article adds appropriate disclaimers for perturbations). I am confused by this because it seems to me that to do orbital predictions, one must know the force of attraction between the bodies, yet the force of attraction doesn't seem to be represented anywhere among the six traditional Keplerian elements.

Am I mistaken? Is the gravitational force represented among the Keplerian elements in a way I don't understand?

Alternatively, is there some unspoken assumption that, whenever one knows the six Keplerian elements, one also knows the masses (or gravitational parameters) of the bodies, effectively making the force an implied seventh orbital parameter?

Or is it possible that the Wikipedia article is wrong, and knowing the orbital elements is

not

sufficient to calculate the position of an orbiting body forward and backward in time?

Note that I asked essentially the same question on the "Orbital elements" discussion page.

-TC

I suppose a somewhat short version of the answer is that gravity is tied up in the determination of the semi-major axis, which barring perturbations would remain constant, as would the other five orbital elements. In that case you would not need to explicitly calculate the gravitational forces for all times. This is only true for two-body motion, and I believe the restricted three-body problem (all objects moving in circular orbits among other restrictions), but is not true for n-body systems because of the perturbations caused by all the bodies in the system.

For a true n-body system, such as the Solar System, knowing the orbital elements at a given time will not allow you to know the exact position of the orbiting body at any given time, t because of the other bodies in the system, which the orbiting body itself will affect, which in turn will change the affect that the other bodies have upon the body in question. So, if you have a two-body system, or a restricted three-body system, knowing the orbital elements at a given time will allow you to know the position of all bodies at any time t, but if you have three bodies in elliptical orbits or more than three bodies we cannot analytically solve that system, so it must be simulated and cannot be run in reverse (to my knowledge).

KSP's system is a simplification where everything is simplified to two-body problems, which is what allows for an at least simpler time-warp system.

Edited June 18, 2015 by TheShadow1138

[KSK]

If you don't mind spoilers, TC, the bottom of this page includes a derivation of orbital position as a function of time, starting from Kepler's laws.

[ZetaX]

... I believe the restricted three-body problem (all objects moving in circular orbits among other restrictions),...

If you allow for those unstable solutions, then there are more, e.g. some figure-8 ones.

[T.C.]

Thank you all for the replies, but I think this thread is getting somewhat off-track. I'm not asking what additional information is required to do orbital predictions (e.g. mass, force, acceleration, etc.). I'm asking why Wikipedia claims that the orbital elements alone are sufficient for orbital predictions. Is it a mistake, or is it because everyone with experience in orbital mechanics just takes it for granted that any required additional information is always available, and therefore doesn't need to be mentioned?

-TC

[pincushionman]

If you don't mind spoilers, TC, the bottom of this page includes a derivation of orbital position as a function of time, starting from Kepler's laws.

It does, but part of it is calculating the eccentric anomaly given the mean anomaly. There is no closed-form solution for this:

This equation gives M as a function of E. Determining E for a given M is the inverse problem. Iterative numerical algorithms are commonly used.

So it says "follow these steps," but it doesn't give all the steps.

[ZetaX]

Thank you all for the replies, but I think this thread is getting somewhat off-track. I'm not asking what additional information is required to do orbital predictions (e.g. mass, force, acceleration, etc.). I'm asking why Wikipedia claims that the orbital elements alone are sufficient for orbital predictions. Is it a mistake, or is it because everyone with experience in orbital mechanics just takes it for granted that any required additional information is always available, and therefore doesn't need to be mentioned?

-TC

I answered that. How about reading it instead of claiming that people go off track¿

[T.C.]

I answered that.

I am grateful for your reply, but please forgive me; I am unable to understand your reply as an answer to my question. Can you elaborate? Do you believe the Wikipedia claim that "once the orbital elements are known for a body, its position can be calculated forward and backwards indefinitely in time" is accurate or inaccurate?

-TC

[ZetaX]

I am grateful for your reply, but please forgive me; I am unable to understand your reply as an answer to my question. Can you elaborate? Do you believe the Wikipedia claim that "once the orbital elements are known for a body, its position can be calculated forward and backwards indefinitely in time" is accurate or inaccurate?

-TC

Yes, it's accurate. It's probably a tedium to actually do (unless at least one orbit is circular), but it should be doable.

[TheShadow1138]

Thank you all for the replies, but I think this thread is getting somewhat off-track. I'm not asking what additional information is required to do orbital predictions (e.g. mass, force, acceleration, etc.). I'm asking why Wikipedia claims that the orbital elements alone are sufficient for orbital predictions. Is it a mistake, or is it because everyone with experience in orbital mechanics just takes it for granted that any required additional information is always available, and therefore doesn't need to be mentioned?

-TC

The orbital elements are sufficient for orbital predictions in two-body and restricted three-body systems. You can obtain the mass of the parent body (at the very least), which would allow you to determine the acceleration experienced by the orbiting object.

In real life you will seldom have a two-body or restricted three-body system, so you will need to know the masses of the bodies in the system and the accelerations they all impart upon one another in order to make orbital predictions. These are the perturbations of which the article speaks. The article is correct, since it makes disclaimers about the perturbations. If the system were free of perturbations the orbital parameters would be sufficient, but with the presence of perturbations, the orbital parameters at a given time t are not sufficient to accurately predict the orbit at any later or earlier time t. I hope that helps.

[T.C.]

Yes, it's accurate.

And is it accurate because the orbital elements themselves contain all information required for orbital predictions, or because any information they lack is, by convention, assumed to be known?

-TC

[ZetaX]

And is it accurate because the orbital elements themselves contain all information required for orbital predictions, or because any information they lack is, by convention, assumed to be known?

-TC

The former.

But as TheShadow1138 already pointed out, this is in the simplified system where only the central body's gravity matters. If it doesn't, then the parameters itself already make no sense as they are made to describe ellipses and ellipses only; and no orbit in a real life system would be ellipses.

[T.C.]

The orbital elements are sufficient for orbital predictions in two-body and restricted three-body systems. You can obtain the mass of the parent body (at the very least), which would allow you to determine the acceleration experienced by the orbiting object... I hope that helps.

Yes, that helps! Just to be clear, you're saying that "the orbital elements are sufficient for orbital predictions in two-body and restricted three-body systems" because "you can obtain the mass of the parent body (at the very least)". This corresponds with my hypothesis that there is an implicit assumption that additional information beyond the orbital elements is known (in this case, the mass of the parent body). Thank you.

ZetaX, you've thrown me for a loop. Your claim that the orbital elements themselves contain all information required for orbital predictions seems to contradict other replies I've received here. I don't see how acceleration can be computed from the orbital elements without additional information. I will ponder what you've said.

-TC

[ZetaX]

The mass of the central body is definitely not calculatable unless you assume the gravitational constant to be known. But you simply don't need it to find the acceleration.

As said, acceleration essentially turns down to calculating centrifugal forces (which follow from the orbit and the orbital speed; but hell of a lot of work for non-circular orbits I presume).

[rkman]

The orbital elements define only the size, shape and orientation of an orbit, and the position of the orbiting body at one specific moment in time. That definition says nothing about orbital velocity. It is gravity-agnostic, so to speak.

Or more fundamentally: time-agnostic. To incorporate time (velocity) you'd need to know two instances of mean anomaly at epoch.

Edited June 18, 2015 by rkman

[ZetaX]

The orbital elements define only the size, shape and orientation of an orbit, and the position of the orbiting body at one specific moment in time. That definition says nothing about orbital velocity. It is gravity-agnostic, so to speak.

Or more fundamentally: time-agnostic. To incorporate time (velocity) you'd need to know two instances of mean anomaly at epoch.

Ah, I missed that there is none that gives a flow of time (or is there¿). But then you still know it up to temporal scaling. In other words, you can randomly fix the speed of time. Which is not that problematic: we measure such things in "years", which just happens to be an orbital period...

[Meithan]

Yes, five of the six orbital elements (a,e,i,Ω,É) are completely "time-agnostic", as rkman put it: they contain only geometric information on the size, shape and orientation of the orbit. But the sixth element, the mean anomaly at epoch, does contain a piece of time information. However, it's not enough to make predictions: as ZetaX says, these six elements only determine the motion "up to temporal scaling".

In order to determine this "scaling", you need one additional piece of time-related information: it can be the period (inferred from the mass of the central body in a central force problem, or the masses of the two bodies in the more general two-body problem), or the speed (the magnitude of velocity) at any point along the orbit.

So I guess my answer to you, T.C., would be: you're right, there's an implicit extra quantity apart from the six orbital elements that's required to make predictions. Typically, this is the mass (or gravitational parameter) of the central body, which is a quantity so important in orbital dynamics that I guess it's generally assumed to be known. The "Orbit prediction" section of the article includes an explicit formula of the mean motion n where the dependence on knowing the gravitational parameter of the system is shown; I do agree that there is no explicit and clear mention of this dependence.

Let's discuss it some more, and if all becomes clear and we still feel the Wikipedia article needs to be amended to clarify something, I'd encourage you to do it (or I can do it myself).

Edited June 18, 2015 by Meithan

[K^2]

It does, but part of it is calculating the eccentric anomaly given the mean anomaly. There is no closed-form solution for this:

So it says "follow these steps," but it doesn't give all the steps.

Power method is your best bet.

M = E - e sin(E), we want to solve that using power method, so rearrange it to read E = M + e sin(E).

Start with a guess. In this case, E₀ = M is a good starting guess. So your first approximation is E₁ = M + e sin(E₀) = M + e sin(M). Now take that and plug it back in. E₂ = M + e sin(E₁). For anomalies, power method converges absurdly fast. Typically, 2-3 steps is all you need.

[ZetaX]

Power method is your best bet.

I knew of calculating a fixed point of f by trying \lim_{n\to \infty} f^n(x_0), but I never heard that name. Is that what physicists call it or where did you get it from¿

[YNM]

The six orbital elements, the three kepler's law, can be derrived out of :

- Acceleration at any point

- (Conservation of) Angular momentum

- (Conservation of) Orbital energy

That's why it's precise already, assuming a limited two-body problem. The derrivation is really long (like, truly, really long), needs a hell of vector and vector calculus. I'll not post it here of course.

Edited June 18, 2015 by YNM