# Elliptical Orbital Mechanics

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Okay,

I wanted to try and find a way to apply circular motion and gravitational field calculations to elliptical orbits to prove why all orbits must have opposite periapsis and aopsis and only one of each (as oppose to for example two opposite aopsis and two opposite periapsis) but I couldn't do it.

Anyone with more experience give me a hand, it would be handy if the derivation tried to use relatively simple mechanics because I'm second year A level in the UK.

Cheers

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4 hours ago, Whyx said:

Okay,

I wanted to try and find a way to apply circular motion and gravitational field calculations to elliptical orbits to prove why all orbits must have opposite periapsis and aopsis and only one of each (as oppose to for example two opposite aopsis and two opposite periapsis) but I couldn't do it.

Anyone with more experience give me a hand, it would be handy if the derivation tried to use relatively simple mechanics because I'm second year A level in the UK.

Cheers

To explain, difficult.

We need to do a little history, Issac Newton basically described a mysterious force that acted at a distance, and although he did not particularly like the idea he did not have a better explanation and the concept of gravity has stuck since then until Einstein. The concept of general relativity, and so gravity as we are familiar with is a faux force, there is no force acting at a distance. Another faux force we are all familiar with is centripedal acceleration which is not acceleration at all, but comes from the fact that objects are in a non-intertial reference frame.

Our first example exemplifies the first. You have a world that has zero rotation (like venus) and you are standing on the surface. In a moment all the mass becomes a quantum singularity at the center. As you are standing you begin to fall and you are absorbed into the singularity. While you are falling you were in a inertial reference frame, since there was no forces acting on you. Therefore the force you were feeling on the ground was the ground and the electrostatic interactions holding the ground up, all the way to the center of planet, and cooperate as a force acting against your inertia. The reason you feel is because you had potential energy but no kinetic energy. The exact metric is the specific potential energy (SPE it means the energy per unit of mass). Lets say you were standing on a planet that had a radius of 10,000km and standard gravimetric parameter of 1E16 (=Mass * universal gravity constant= mu, written like u but is greek m). The equation actually comes from the integral of MGH which at the microscale determines the moment of energy exchanged at each radius.

SPE=uo/r-u1/r

Between any two radius we can know the specific potential energy gained from a movement to the other. For example if you move from (2-body problem) deep space to 10,000 km SPE is equal 1,000,000 joules per kilogram.
So lets say the you fall from 10,000km to the 'new' singularity, which happens to be 0.01M you would be approaching c.Your apoapsis on this orbit would be the point from which you fell and if that point was infinitesimally small your random motions would have created a periapsis at that point and retruned you to your apoapsis, a perfect line from your starting position to the center of the planet and back. The reason is that the space-time next to the point mass is highly warped by the nearby energies as you approach you need to increase your velocity (inertia) but in fact the only way for inertia to increase in a free fall is our faux friend gravity. You have to think of it like this, the energy in space creates space-time (this is a combination of mass and other forms of energy). If you lack kinetic energy then it draws you in, but if you have enough kinetic energy there is an energy level by which you can leave. As you fall you convert potential energy to kinetic energy allowing you to escape. The kinetic energy you gain is Specific Kinetic energy (SKE). Specific potential and kinetic energy compose the specific mechanical energy of an orbit (as in orbital mechanics).

dSKE=-(uo/r-u1/r) Total SKE = starting V^2/2 + dSKE  resulting in a velocity of (2*tSKE)^0.5

If we however pushed your body 1 meter tangential to the radius you would not hit the surface of the point mass, but instead the second faux force would kick in, centripedal acceleration. So there are a couple of logics here. In our reference frame (think stars and galaxies far away) we can define a state which is not in rotation. If you fell from a dead stop circled the point mass and returned back to a different point from which you fell relative to the point mass, then it means the point mass has imparted information . . . energy would not have been conserved.  That explains why you return to the same position and altitude but why does the periapsis need to be 180' opposed. The reason is that as you fall through each radius you impart potential energy and gain kinetic energy and as you rise on the other side of the point mass in an exact reversal of the process you gain potential energy at the expense of kinetic energy. Consequently both sides look the same, in practice however this is seldomly true but approximate. This is because in our universe, there are no true 2-body problems. The other issue is that relativity applies, so what appears to be Mercury's starting position to us, is not the starting position when viewed from mercury.

The second logic is an empirical one. Before Newton had applied gravity to the planetary motions Kepler had already noted the elliptical motion of the planets. He also noted that 'massive' bodies were typically close to a foci. IF the central mass (say the sun) is at one of the foci it means that every planets orbit has two symmetrical parts, one that approaches the sun and the other part that moves away.

The equation is this r =   l / 1 + e cos (angular position). l is the semilatus rectum, e is the eccentricity. As you know cos 0 = 1 , 0 is the periapsis, the reference position for the ellipse. As you move in either direction from zero, so for example π and -π radians place a satellite exactly at the same distance from the point mass that it orbits (roughly). In an ellipse there are only two unique radius, those that intersect the semi-major axis, all other radii mirror the other half of the ellipse.

How do we connect Kepler and Einstein. [See general relativity]. As you move about an orbit there is an inter-conversion of energy, you fall into a warping of space time but at some point a (the semi-major axis) you have enough energy to remain at a static potential energy level. If by chance SKE has a tangential velocity vector where speed = Vtan and Vtan = SQRT(u/r) then you have a circular orbit. If however you have at a some radial velocity then the satellite will 'cycle' from pE to Apo. There are two ways we could have orbits. if you took a cylinder and sliced it along a diagonal you would have a elliptical shaped object with two equivilant foci. In such a state we assume the mass is at the elliptical center and the satellite would approach the central body twice per orbit.

You should note in the first example that change in PE from deep space (Infinity) to 10,000M is much, much less than the change of PE from 10,000 M to 0.1 M. This 'gravity' acts as a radiative force F=k/r^2. This implies that the warping of space-time is not linear with respect to radius. The decrease in the warping of space slows as one measures further and further away.  Consequently the type of ellipse that is more suited is a conic section where the center of the cone is the major foci (central body) and the minor foci is at some distance from the central body. So because the rate of warping relative to distance is not uniform we need a conic section to explain orbital motion, and with a conic section the major body allows only one pass at close distance per orbit.

I should point out the Kepler's laws of planetary motion are a conservative guess. In fact with 100s of years of study since there are lots of n-body corrections. There are a great many places in our solar system where planetoids and other things could reside in stable orbits, but the problem is that massive bodies have a tendency to tweek the orbits of smaller objects, potentially catastrophically, but often throwing them out of our system. The la grange points are examples of places where we can find 'trojan' asteriods, and for jupiter and saturn there are a fair number. But earth and venus have few of these objects. So even n-body solutions are not perfect explanation.

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The form of Kepler's Law that we know can be derrived from Newton's Law of Motion and Gravitation.

The key ideas are :

- Prescribe the forces in derivatives of distance.

- Turn into spherical coordinates.

- Do calculus.

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7 hours ago, YNM said:

The form of Kepler's Law that we know can be derrived from Newton's Law of Motion and Gravitation.

The key ideas are :

- Prescribe the forces in derivatives of distance.

- Turn into spherical coordinates.

- Do calculus.

That's not too bad.  I can't really translate UK educational levels, but I remember writing down the derivation for *circular* orbits during a second year university physics exam (had to be done in less than 5-10 minutes just for that question).  This can be done with polar coordinates and doesn't require calculus (and thus doesn't explain ellipses nor the area law, but handles the degenerate case of ellipse==circle well).  [this may have been deriving gravitation from circular motion, it has been a long time].

Unfortunately polar coordinates are unlikely to remain your friends with ellipses, although it appears that again solving for the degenerate case of ellipse==line works well here, and then showing that orbiting in two orthogonal "lines" forms an ellipse is the method used above (maybe not explicitly, but certainly that is what x and y are doing).

The spherical coordinates might be a bit of an issue.  Working from Keplar we can show that everything stays on a plane, and I can see how you can show that Kepler's orbits are stable, but I'd hate to show that every [large/small] body will follow conic sections (and remain on a single plane).

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4 hours ago, YNM said:

The form of Kepler's Law that we know can be derrived from Newton's Law of Motion and Gravitation.

Safe to say, it is very, very involved and intensive to derive so.

The key ideas are :

- Prescribe the forces in derivatives of distance.

- Turn into spherical coordinates.

- Do calculus.

Newton's application works for slow moving bodies around distant stars, it tends to break down as things speed up.

The page needs an awful lot of work to be complete. First GM = u. Assuming that one is a infinitesimally small satellite around a massive central (and far enough away such that relativity does not need to be taken into account) and get rid of m, temporarily.

r = 1 / (u/L) (1 + e Cos O)) versus  r = l / (1 + e Cos O) where l is the semi-latus rectum or r = a * (1-e2) / (1 + e Cos O) where l = a * (1-e2) or l = b2/a
for keplerian mechanics a = is the arithmetic mean of Pe and Apo as measured from the center of massive bodies, b is the geometric mean. Both represent the axis of an ellipse. Therefore the semi-latus rectum goes undefined as L/u.

To truly bridge the gap between Kepler and Newton (although neither formulated their math to the above stated description we assume that their understanding was equivalent) you need to convert one understanding to the other.

Here it is, the X axis is the axis that runs from Pe though CB to Apo and also runs along O = 0(2π)- π. It is the axis where Y = 0. The Y axis runs through the systems center of mass therefore it runs through π/2 and 3π/2 each radial to the radius of the semi-latus rectum; it is the axis were X = 0. When Cos (O = π/2) then e Cos O = 0 thus r = 1/(u/L) = L/u (where L is the angular momentum of the orbit, mass reduced in our example). The problem between Kepler and Newton is a discrepancy in the central body that is related to the planets mass, for example Jupiter would pull on the sun gives the sun orbital moment, so that the semi-latus rectum is not at right angles to Jupiter's periapsis through the sun but through the system center. Thus if you are looking at binary stars you cannot reduce the mass.

Therefore there are three (x2) other orbital landmarks. There are a, the radii in which V is equal (u/r)1/2 [Note that for a circular orbit all radii = a]. In terms of the angular distance from the periapsis a is variable. The limit of Oa as e-->0 is Ol this means that as eccentricity increases a is almost always closer the minor focus than the central body. This has to be the case, because for a to have a velocity of (u/r)1/2the and also be eccentric means that the absolute value of the radial velocity vector is large and thus the tangential component of velocity must be small and similar to the velocity at the apoapsis. There is the l radii, where (reduced angular momentum)/u. The third points are r = b, which is more or less trivial since b is a measure from the elliptical center to along the minor axis.

I should point out also the link is incomplete in one regard for the stated purpose. There is no reason that upon leaving O = 0 after completion of an orbital cycle O since the conversion from polar coordinate system makes the assumption that the process is a cycle (IOW its a circular definition), which of course it is, but the Newtonian systems assumes it is a static cycle, in order to do that you need to know whether that stasis has an energy dependence or not; it does. No total energy (information) change is what causes the cycle to be 'a standing wave'.

Here is the reason using the Newtonian system.

*Note: For a Newtonian system (a benefit) you can choose a coordinate reference that is not, say, going to be effected by change, say the position of three far off galaxies that appear at right angles to each others position and whose positions evolve slowly. From that coordinate system you would need two O to define every position in the orbit, one in X-Y and other in X-Z (or Y-Z or all three). This is actually the most thorough way of testing the system, the problem is that it is the most complicated particularly so if you don't place the systems barycenter at X=0, Y =0, Z=0.

Here are the flaws in the Newtonian system.
Information is always leaving and entering the system, the system is never in stasis. Also, the Newtonian system only assumes that matter can contribute to gravity and that energy can change the state. However energy and matter both contribute to gravity. Energy is always entering and leaving the system. For example, the moon's orbit is affected by the water on the surface of the earth (The earth is not  a point mass, who knew). The sun is loosing mass via d(hv)/dtc2 . If the sun were to stop converting mass into hv then the water on Earth would freeze, and the moon's orbit would change more slowly (but still change due to perturbations in elevations on Earth). This is an example of the complexity of the system. Another example, a comet behave by Newtonian orbital mechanics until is reaches a certain radius, but strictly Newton does not reduce mass, and yet the comets mass is reduced by solar winds and hv. As a consequence the comets orbit is subject to change or alternatively disappear. In this 2-body problem the larger body is converting mass into both hv and energetic ejecta. The comet in response is converting its mass into the circumsolar atmosphere (compositely know as the wind). A third example, a satellite orbiting the sun is taking pictures of deep space, in doing so most of its reaction wheels and can no longer maintain attitude, by using the remaining reaction wheels and solar panels it braces itself again photon pressure therefore able to take pictures at specific positions in space as the satellite rotates around the sun, by bracing itself against photon pressure it is changing dr/dt. A forth example, a super nova is an object whose fusion pressure is no longer sufficient to keep heavier atoms from accumulating in its deepest core, the core collapses, in that immediate moment the core converts tremendous amounts of a matter to energy. In the exact moment before the star explodes a pulse of space-time energy leaves the core of the star and as energy is converted to light in the subsequent explosion another larger pulse of space-time density  radiates from the star. Any object orbiting that star would observe a change of its position relative to the systems center of mass. Finally as the mass ejecta fly by this superman planet the gravity that once held the planet in orbit diffuses into the deepness of space, if it was in a circular orbit it would be free to orbit as an ellipse or a hyperbola and wander the galaxy as an orphan planet. In each of these scenarios the information of the system, but the kinetic energy in the objects has not changed in complete response.

IN the first the moving  tides are providing information from the sun transmitted to the  earth revealing differences in the viscosity  of matter on the surface that rigid matter would not provide.  In the second the objects mass is dividing itself and while some of the mass remains in predicted motion, much of the mass is taking a different motion, it is doing so as a consequence of non-gravimetric information the central body is transmitting. In the third example information is provided from humans to adjust the random position of the spacecraft such that different forces balance each other. In the forth example simple newtonian function is not static, the warping of space-time changes several times before the central tenant of the function, that the system center acts as a point mass, becomes invalid and after a time, becomes valid again but with a different mass.

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@PB666 I just linked that site because it seems to be the closest one to what I know. Derivations are usually shown in astronomy books; Fundamentals of Astronomy is one. I have some lecture slides but they're not easy to show here.

Kepler's original formulation have no math form at all AFAIK. It's only through Newton's gravity and motion that one can get those differential form.

Also, Newton's law doesn't work with relativity, so ofc you can't get the relativistic corrections. Haven't seen one derrived through EFE pr tensors or something.

@wumpus Was it through centripetal ? Centripetal is the easiest.

The newtonian derrivation indeed also proves the planar and conic-section nature of the trajectories. It's because of the way the forces works - I could imagine that if the other body were to be assumed moving, you can get a spiral model instead.

In any case, I've always seen the derrivation as "huh" and take it for granted that it is possible to show such, but I take no interest in exploring it any further.

Edited by YNM
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5 hours ago, YNM said:

@PB666 I just linked that site because it seems to be the closest one to what I know. Derivations are usually shown in astronomy books; Fundamentals of Astronomy is one. I have some lecture slides but they're not easy to show here.

Kepler's original formulation have no math form at all AFAIK. It's only through Newton's gravity and motion that one can get those differential form.

Also, Newton's law doesn't work with relativity, so ofc you can't get the relativistic corrections. Haven't seen one derrived through EFE pr tensors or something.

The elliptical mark ups as we use them today were not developed in Keplers time, so the equations used today are sort of projections of the elliptical math that Kepler knew of or expanded up[on.

The required elements of a conic section are a a = semi-major axis, e= eccentricity from which one can draw an ellipse.

Major axis = the longest diameter of the orbit. The specific problem here is a reference point. . . how do you know.

Eccentricity: Angular width of it face at full moon from its closest and furthest points, we would need to know the radius of the earth.

The part astonomers work with such as inclination (i) require the choice of a reference plane, then you need to know the inclination line in that plane and create a cross-reference measure of distance from that line.
From that you need to measure the argument of Periapsis (w). Once you get this far fabrication of an orbital model would be pretty strait forward.

The constant sweep rate is pretty good intuition. For Newtonian orbital mechanics you have the modern vectors and v. If the barycenter of the system is xyz = 0,0,0 then magnitude r is dot product (x,y,z) of its cartesian coordinates and v is the direction of motion perpendicular  direction of r, r x v = h (a vector)  which is invariant (dot products and cross products are from the 1890s). The sweep rate is roughly 1/2 h. In this system if the ellipse is in X,Y then h projects into Z (its own dimension). Thus the two systems are congruent, in the Newtonian system its more refined however because specific (mass-reduced) angular momentum has a vector. This means any addition of energy to the orbit therefore would change that vector h2/u is however the semi-latus rectum (l)  of the orbit with reference to the systems barycenter, this particular component is not derivable from Kepler, but is derivable from Newtonian celestial mechanics. l is important since b is a trivial aspect of the orbital ellipse, and it is possible to predict the orbits using r = l / (1 + e Cosine O). Since both the system center and its satellites would obey this it is possible to  map orbits and predict future positions.

In the Newtonian 2-body problem since h can also be made a function of angle r2dO/dt (where V = wr and w = dO/dt) and r = (h2/u) / (1 + e Cosine O): specifically the periapsis is the angular reference point and h2/u = L/m, there is the immediate consequence that since h is constant and can be related to r that the ellipse cannot evolve in any direction without the change in energy.

To see this, for the inclination of an orbit to change h (which is also a direction vector) would have to change, this is a constant. The periapsis then could change in the inclination plane. There are two ways to do this, the periapsis could change its magnitude, but that changes v without changing r (at the change point) this means magnitude of h changes, you would have to a another change at 180' to change v again and r to maintain h' this however would change eccentricity (making it circular). You could then go to a third point and burn out to the original rmax, then go to rmax and draw in the rmin to the original pe.  h would not have remained but would have been restored but no r = (h2/u) / (1 + e Cosine O) would be invalid because O = 0 is no longer the periapsis.

To be clear apsidal precession occurs in orbits, its one of the factors that contributes to the ice ages. And it occurs without any energy being put into the system, but the system in n-body, which means that energy is being transferred in the system from one orbit to the next and back again. Basically every year Ope, new = Ope, old + .000056 radians. This precession occurs as a consequence of the effects of the major planets in the solar System, namely Jupiter and Saturn. So this goes to a frequent disclaimer about h (or L) that h is local and not infinite. If we forget about relativistic differences, the two major problems are, for satellites traveling in LEO around the Earth, that the earth is not a point mass nor is it of uniform density, this alters u/r2, that external, other non-centric celestials, perturb orbits. Finally for LEO orbits as orbital speed increases so does the effect of drag, both increase in proximity to planetary surfaces, so that loses of h can be permanent.

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