Are planetary orbits really ellipses?

In a universe with just two objects, A, a large body (a star) and B a much smaller body (a planet). If B is in orbit around A, is that orbit truly an ellipse with A at one focus?

Intuitively, it seems like it should be more egg shaped.

Is there a simple proof of this (one that my rusty brain might be able to follow)?

Thanks

 

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What you are describing is Keplers laws of planetary motion, which he came up with by looking at tons of observational data. The proof of Keplers laws is done by studying a system of two massive objects interacting gravitationally using Newton's law of gravitation. It's commonly taught in the first year of a physics degree, although it's much more simple if you know a bit about Lagrangian mechanics, which normally comes later.

 

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Any proof should also encompass the parabolic and hyperbolic orbits which also have the central mass (in the limit of small test masses) at a focus.
There's a demonstration that this follows from Newton's Universal Gravitation on Wikipedia.
http://en.wikipedia.org/wiki/Kepler_problem#Solution_of_the_Kepler_problem

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Newton's physics is about motion, so we have to have a relationship between time and angle if we hope to talk about shape.

Let A sit at the origin of the coordinate system and be the source of a central inverse-square attractive force. Such a central force necessarily does not change the angular momentum of the planet B about the origin. Neither does it cause it to stray from the initial plane of A, B and the direction of B's movement.

Let us assume that B has some movement not in a line with A, or else it would fall directly in. So we can write the position of B in the x-y plane as (x,y) = (r cos θ, r sin θ). r and θ should be thought of as functions of time.

So the motion of the planet in polar coordinates about the star can be expressed in terms of r and θ. Our goal is to show that r and θ have the same relation as an conic section with focus at the origin. Thus we expect to find \(r = \frac{a}{1 \, + \, b \cos (\theta - \theta_0)}\) for some length a and some number b. If b is positive then \(\theta = \theta_0\) happens at time of closest approach.

Since angular momentum is constant and non-zero, we write it as \(L = L_0 = m r^2 \frac{d \theta}{d t}\).
Then since F = ma we have in polar coordinates: \(-\frac{G M m}{r^2} = m \frac{d^2}{dt^2} \left( r, \theta \right) = m \frac{d^2 r}{dt^2} - m r \left( \frac{d \theta}{dt} \right)^2 = m \frac{d^2 r}{dt^2} - \frac{L_0^2}{m r^3}\)
http://en.wikipedia.org/wiki/Centripetal_force

From \(-\frac{G M m}{r^2} = m \frac{d^2 r}{dt^2} - \frac{L_0^2}{m r^3}\) we use \(\frac{d \quad}{dt} = \frac{L_0}{mr^2}\frac{d \quad}{d \theta}\) to first express this as: \(-\frac{G M m}{r^2} = m \frac{L_0}{mr^2}\frac{d \quad}{d \theta} \left( \frac{L_0}{mr^2}\frac{d r}{d \theta} \right) - \frac{L_0^2}{m r^3} = \frac{L_0^2}{m r^2} \left( \frac{1}{r^2} \frac{d r^2 \quad}{d \theta^2} - \frac{2}{r^3} \left( \frac{d r \quad}{d \theta} \right)^2 \right) - \frac{L_0^2}{m r^3}\) and then as \(-\frac{G M m^2}{L_0^2} = \frac{1}{r^2} \frac{d r^2 \quad}{d \theta^2} - \frac{2}{r^3} \left( \frac{d r \quad}{d \theta} \right)^2 - \frac{1}{r}\).

If \(u(\theta) = \frac{1}{r}\) then \(\frac{du}{d\theta} = - \frac{1}{r^2} \frac{dr}{d\theta} \) and \(-\frac{d^2u}{d\theta^2} = \frac{1}{r^2}\frac{d^2r}{d\theta^2} -\frac{2}{r^3} \left(\frac{dr}{d\theta} \right)^2\) so we may substitute:
\(-\frac{G M m^2}{L_0^2} = -\frac{d^2u}{d\theta^2} - u\)
Or:
\(u + \frac{d^2u}{d\theta^2} = \frac{G M m^2}{L_0^2}\)

Now \(u + \frac{d^2u}{d\theta^2} = 0\) is a famous differential equation with families of solution \(u = C_0 \cos ( \theta - \theta_0 )\).
It follows that \(u = \frac{G M m^2}{L_0^2} + C_0 \cos ( \theta - \theta_0 )\) is a solution to Newton's equations of motion for some constants \(C_0\) and \(\theta_0\).
But since \(u = 1/r\) this is equivalent to writing:
\(r = \frac{ \frac{L_0^2}{GMm^2} }{1 + \frac{L_0^2 C_0}{GMm^2} \cos ( \theta - \theta_0 )} = \frac{a}{1 + b \cos ( \theta - \theta_0 )}\)

Thus conic sections are solutions to motions in Universal Gravitation.

If I did the math right, a circular orbit has \(C_0 = 0\) and so
\(r = \frac{L_0^2}{GMm^2} = \frac{m^2 r^4 \left( \frac{d\theta}{dt} \right) ^2}{G Mm^2}\) or \(\frac{dt}{d\theta} = \sqrt{\frac{r^3}{GM}}\) for a period of \(P = 2 \pi \sqrt{\frac{r^3}{GM}}\)

Not a bad approximation for Earth.

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If we assume some physics and some vector math, we can get this result a little faster:
http://en.wikipedia.org/wiki/Laplace–Runge–Lenz_vector#Derivation_of_the_Kepler_orbits

 

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Quite impressive rpenner, and I'm sure that is exactly what Jennifer was looking for.

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The simple answer is, yes, it's an ellipse. The more complex answer is that precession effects can muddle things up, but in no scenario does it look like an egg.

 

to form the planet in the first place means the Solar System would have had angular momentum. Even though the planet might end up with the major part of it the Star will still be spinning. In our own solar system 95% of the angular momentum is in the planets etc. and just the 5% left in the Sun, which takes about a month to spin once on its axis.
The fact that the star is spinning will be the main reason it is oblate (egg shaped on cross section).
I do believe the planet might add to this to some minimal way. But planets circling the star will cause a tidal wave as well.

 

Oblate does not mean egg shaped. It means "squashed"

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What does (egg shaped in cross section) mean then?
Did you agree that any star with a planet would necessarily be spinning?

Do you agree with any of the following?

It would always rotate faster than the orbiting planet orbits the star so the tidal bulge would cause tidal acceleration.
The acceleration would cause a precession even without other planets in the system.
The other factors causing precession of an elliptical orbit are:
1. the continual loss of mass of the star.
2. Frame dragging of space time.
3. Relativity

One factor that would set the planet into an elliptical orbit, would be the sudden change of star mass at the time of turning on (Protostar - t tauri transition.)

 

I would say egg shaped is where one end is squashed a bit compared to the other on, so an ellipse would have an extra axis of reflection symmetry. See a more technical discussion here

No it's not required, although I imagine it would be very hard to find a physical counter example.

All of these effects are tiny compared to the other effects - Newtonian gravity very accurate for most gravitational calculations despite it playing second fiddle to GR. The largest effect you mention that can't be explained by Newtonian gravity (the oblateness of the sun can be) you can illustrate with the following. Mercury's orbit was observed to precess at ~570 seconds of arc per century, while the Newtonian prediction was ~530. The GR contribution is ~40 seconds of arc per century. Now go an look up what a second of arc is (Clue - very small angle).

 

What they said, plus:

(1) Kepler's law says that everywhere along the orbital, the object sweeps out an equal area in an equal time. This is a pretty good illustration:

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(2) The geometry of an ellipse:

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(3) The formula for an ellipse:

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You will notice that when a=b you get a circle, a special case of an ellipse.

Odds are, nature won't find the special case for a=b, so the planetary orbital tends to be elliptic. But it's not impossible - just very unlikely - to have a circular orbit.

 

There is no such thing as a stable orbit. An object will not stay in the same orbit. Over time, the orbit changes. An example is that earth's orbit is increasing, so, the path the earth takes around the sun is increasing in distance as time goes by. We are in fact getting further away from the sun.

As the sun loses mass the earth's orbit increases. The only reason it appears to us that the earth is in a stable orbit is because our timeline is a drop in the bucket compared to the timeline of the earth orbiting the sun. The change of orbit is very gradual and it takes a long long time for us to observe a change.

 

That's an interesting observation. Are there theoretical orbiting masses which are 100% stable?

 

I believe there are planets around protostars which are 100% stable but you can't see them as the the star is not on yet. With the sudden loss of mass from the "turning on" there is a kink in the orbital path and from then on it is a continuous oscillation of the orbital radius trying to restore equilibrium.
I am toying with the idea of trying to calculate the amount a planets radius would increase at the T- Tauri start point. You just need to balance centripetal force with Gravitational force and then find out what period the loss of mass took, so you get an idea of the momentum the G force had to over come (causing the degree of eccentricity, the oscillation) which makes the orbit elliptical.

 

I'm sorry for my sloppy wording, Robittybob1, but I meant is there a massive structure which is itself 100% theoretically stable such that it is not prone to decay, etc?

 

There are some stable end stage stars, neutron stars and black dwarfs just from the top of my head, but they are still cooling which in the end increases their stability. They may (virtually)never cool entirely.

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In condition, as envisaged in the question, body B will fall straight into body A. In order to orbit around body A, body B has to have an initial velocity in its orbital path in right direction. Parameters of its motion should be precise for its current orbital path.

There are no static bodies, except stable galaxies, in any universe. Hence, if body A is moving, orbital path of body B will not trace any type of closed geometrical figure. Sun is a moving body and hence planetary orbital paths are wavy about sun's path. Orbital paths of satellites are wavy about orbital paths of their parent planets.

Kepler's 'laws on planetary motion' are derived from observations of relative positions of few planets in solar system. Therefore, they are true to find relative positions of bodies in any planetary system. It is not right to use them to derive other parameters of orbital motion. Reference for detailed description will be provided on personal request.

 

Please explain where Jennifer said the two bodies were at rest with respect to one another.

 

The orbit will be an eclipse, even thought in a 1 star 1 planet universe it could just about be a perfect circle. (Practically circular is a possibility but circular is the extreme.)
If the planet is big enough and if the star is spinning the Star will have a oblateness. (which I called "egg shaped" in error.)

 

jennifer,
nevermind rpenner, he's the egghead of the bunch.
anyway, yes, the fact that at least the earths orbit is an ellipse is easily provable by measuring certain aspect of its travel around the sun.

the math provided by rpenner proves it for all orbits.

 

No. In any real situation involving stars they spew out material and you get the effect MD refers to. Even if you had an object which didn't emit any material, ie a classical black hole or just hypothetical inert blob of material, general relativity says that things moving in orbits, like the Earth going around the Sun, bleed energy via gravitational radiation. It's the gravity version of electrons classically emitted EM radiation when they go in circles.