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cj

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**Why are orbits elliptical?**

Any ideas?

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- Thread starter cj
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- #1

cj

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Any ideas?

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- #2

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What orbits...?

Daniel.

Daniel.

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tony873004

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Round is a state of perfection, and nothing is perfect.

But, most orbits in our solar system are close to circular.

Pertabutions from the other planets is one of the main reasons that planets can never achieve a perfectly circular orbit.

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Daniel.

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neurocomp2003

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pick up a standard astronomy text and find out =] eg Carroll and Ostlie

- #6

cj

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Again, why are some (all in our solar system) planet's

orbits elliptical?

dextercioby said:

Daniel.

- #7

cj

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neurocomp2003 said:pick up a standard astronomy text and find out =] eg Carroll and Ostlie

I guess if I had access to an astronomy text, I'd do just that.

I guess this is one reason why discussion boards exist.

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tony873004

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dextercioby said:

Daniel.

I don't understand the "nope" part . But I agree with the rest.

I imagine you're responding to my post.

It's kinda the point I was trying to make. Everything is either hyperbolic or elliptical because circular or parabolic are perfect conditions that only exist on paper. If you think you have a perfectly circluar orbit, try expressing its eccentricity accurate to 15 digits.

- #9

cj

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cause this deviation from perfection?

tony873004 said:

I don't understand the "nope" part . But I agree with the rest.

I imagine you're responding to my post.

It's kinda the point I was trying to make. Everything is either hyperbolic or elliptical because circular or parabolic are perfect conditions that only exist on paper. If you think you have a perfectly circluar orbit, try expressing its eccentricity accurate to 15 digits.

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[tex] \epsilon=\sqrt{1+\frac{2EM^2}{m\alpha^{2}}} [/tex]

,where

[tex] U_{C}=-\frac{\alpha}{r} [/tex]

Daniel.

P.S.U judge from that formula what the eccentricity may be.

- #11

neurocomp2003

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look up kepler's proof for the eccentricity of planetary orbit online. Its like a 1-2page proof.

Some things to consider.

[1] Gravitation of other planets(i think this is minimal if my memory serves me right)

[2] When the Object started orbiting a particular point.

If the point(axis) of rotation is not directly in the center than the orbit will not be perfectly circular...the sun is not located in the center of Earth's rotation.

Some things to consider.

[1] Gravitation of other planets(i think this is minimal if my memory serves me right)

[2] When the Object started orbiting a particular point.

If the point(axis) of rotation is not directly in the center than the orbit will not be perfectly circular...the sun is not located in the center of Earth's rotation.

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- #12

Andrew Mason

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The general solution to the differential equation of motion of a mass in a gravitational field is a conic section. See, for example:cj said:Any ideas?

http://en.wikipedia.org/wiki/Keplerian_harmonic_law#Kepler.27s_first_law

Newton provided a complicated geometric proof. Feynman provided a similar geometric proof (see: Feynman's Lost Lecture)

AM

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selfAdjoint

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DaveC426913

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No. They would be elliptical even without other planets.cj said:So it's the gravitational effect from other planets that

cause this deviation from perfection?

An orbit has two components: orbital speed and radial distance. The only way you would get a circular orbit is of these two (otherwise independent) values were just right.

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SpaceTiger

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It really just depends on how precise you need to be. For many cases, it's perfectly alright to approximate a planet's orbit as circular. In others, you may want to approximate it as an ellipse (as Kepler did), and in others you'll need to go to higher order (as Einstein did with Mercury).

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amt

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There is still no clear and concise answer to the original question though.

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selfAdjoint

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amt said:There is still no clear and concise answer to the original question though.

I thought my answer in post #13 was concise. Perhaps it wasn't clear? The conic section/focus property falls out of simultaneously satisfying the conservation of angular momentum and the inverse square law of gravity. This is essentially what Newton showed in the first 13 propositions of his

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Chronos

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The sun provides a certain centripital force with its gravity, which changes depending on your distance. If a planet comes along and it is moving with a certain velocity and has a certain mass, you could do the math and figure out the radius of the circle it would move in...if it did move in a circle that is. For a certain specific velocity and mass of a planet, there is only ONE radius that has the exact centripital force to keep it moving in a circle around the sun. However what if as it approaches the sun, the radius is slightly larger than what it should be for a circular orbit? Imagine the planet moving in a straight line. As is passes the sun, it gets pulled sideways towards the sun slightly, but not enough to hold it in a circular path. So now it is slowly picking up velocity in the direction perpendicular to its original path - in addition to its original velocity. Right as the planet passed the sun, the pull of gravity was perpendicular to its direction, so it couldn't slow the planet down. As it moves away, the angle becomes smaller, and the sun beings to slow the planet down and eventually stop it and reverse direction. Now it is being accelerated towards the sun, and its sideways velocity is slowing. Now that I've tried, it's a little hard to put into words, but it makes perfect sense to me. Draw it out on paper, draw the sun and draw a circular orbit path. Now start with a radius a little larger and same velocity, now just imagine the forces that the sun is applying on the planet.

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Locrian

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Thanks spacetiger, that's an interesting perspective, and not one I've really heard before.

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Chronos

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- #22

εllipse

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Sounds interesting. Do you happen to have a link or know of a text that goes into detail on this?Chronos said:

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Chronos

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- #24

russ_watters

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...With one caveat: if an elliptical orbit were unstable and somehow "decayed" into a circular one, then circular ones would exist/be prevalent. But elliptical orbits are stable.

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Symbreak

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The reason why the planets have a slight eccentricity is due to minor effects in the gravity of other planets (as said by others) but also due to the rectilinear movement of a planet at its formation. Also, large impacts of comets can cause the planet to deviate in its orbit, causing more eccentricity in its orbit round the sun.

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selfAdjoint

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Trying once more. The general quadratic in two variables, such as x and y, is a conic section. As Symbreak says, these consist of ellipses and hyperbolas with circles and intersecting lines as special cases. Given a massive body like the sun, regarded as at rest, and a planet with any instantaneous velocity relative to it, the future orbit will be in the plane of that velocity and the COG of the bodies (since there's no force acting normal to that plane), Given any x-y coordinate system in that plane, a quadratic equation will result from the law of gravity and conservation of angular momentum. The solution, path of the planet, will be some conic section, but there is no reason to suppose it will be the special case, a circle. That will only happen if the initial velocity vector happens to be exactly at right angles to the radius vector from the sun to the planet, and its speed is exactly the circular speed for the distance of the planet from the COG.

So the orbit will be an ellipse or a hyperbola, but if it is a hyperbola the planet will fly off on it into outer space and never return. So the only class of stable solutions are the ellipses. Note that the influence of other planets has not entered at all. In fact they do not serve to form ellipses but to destroy their stability by causing rotation of the axes and off-plane forces.

So the orbit will be an ellipse or a hyperbola, but if it is a hyperbola the planet will fly off on it into outer space and never return. So the only class of stable solutions are the ellipses. Note that the influence of other planets has not entered at all. In fact they do not serve to form ellipses but to destroy their stability by causing rotation of the axes and off-plane forces.

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- #27

LeonhardEuler

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This is not the reason for eccentricity. I'll try to prove this more formally. Not axiomatically or anything, but just a good convincing argument.Symbreak said:

The reason why the planets have a slight eccentricity is due to minor effects in the gravity of other planets (as said by others) but also due to the rectilinear movement of a planet at its formation. Also, large impacts of comets can cause the planet to deviate in its orbit, causing more eccentricity in its orbit round the sun.

Assuming the sun is much larger than the planets so that it remains stationary, and that the force of gravity varies inversly as the square of the didtance and directly as the product of the mass of the sun and the planet, and that the sun is the only body exerting a force on the planet, the orbit will be an elipse.

Proof:

Put the sun at the center of a coordinate system. The sun exerts a force directly toward itself, so the torque is zero, implying that the angular momentum of the planet is constant. Introduce the position vector of the planet, [itex]\vec{r}[/itex], and the veloctity vector, [itex]\vec{v}=\frac{d\vec{r}}{dt}[/itex]. Introduce the unit vector, [itex]\vec{i}[/itex], pointing towards the planets initial position. Defeine [itex]\vec{j}[/itex] perpendicular to [itex]\vec{i}[/itex] and lying in the plane spanned by the position and velocity vectors. Define [itex]\vec{k}[/itex] by the right hand rule. Then the angular momentum is equal to [itex]\vec{r} \times m\vec{v} = mh \vec{k}[/itex] for some constant h.

Now, niether the force exerted by the sun nor the initial velocity vector has a component outside of the plane spanned by the intitial position and velocity vectors. Therefore, the orbit lies in this plane. Now consider the area swept out by the position vector as the planet moves over a small interval of time. This area is approximately the area of the triangle between the vectors [itex]\vec{r}[/itex], [itex]\vec{r}+\Delta \vec{r}[/itex] and [itex]\Delta \vec{r}[/itex]. This error in this approximation approaches 0 as [itex]\Delta \vec{r}[/itex] approaches 0. This is exactly half the area of the parralelogram spanned by [itex]\vec{r}[/itex] and [itex]\vec{r}+\Delta \vec{r}[/itex], which is equal to [itex]||\vec{r} \times \Delta \vec{r}||[/itex] The amount of area added over a short period of time, per unit time is then [itex]\frac{1}{2}||\vec{r} \times \frac{\Delta \vec{r}}{\Delta t}||[/itex]. In the limit of [itex]\Delta t\rightarrow 0[/itex], this is [itex]\frac{1}{2}||\vec{r} \times \frac{d\vec{r}}{dt}||=\frac{1}{2}||\vec{r} \times \vec{v}||[/itex], which is the angular momentum. The total area swept out by the postion vector is

[tex]A=\frac{1}{2} \int_{t_0}^{t} r^2 d\theta = \frac{1}{2} \int_{t_0}^{t} r^2 \frac{d\theta}{dt} dt[/tex]

From the fundamental theorum of calculus, the rate at which the vector sweeps ot area is

[tex]\frac{1}{2} r^2 \frac{d\theta}{dt} [/tex]

Equating the two rates for area sweeping, you get:

[tex]\frac{1}{2} r^2 \frac{d\theta}{dt}=\frac{1}{2}||\vec{r} \times \vec{v}||[/tex]

[tex]\rightarrow r^2 \frac{d\theta}{dt}=||\vec{r} \times \vec{v}||[/tex]

The magnitude of the crossproduct is now known. Combining this with our knowledge of its direction, you get:

[tex]\vec{r} \times \vec{v}=r^2 \frac{d\theta}{dt} \vec{k}[/tex]

Now define the unit position vector, [itex]\vec{u}=\frac{\vec{r}}{||\vec{r}||}[/itex]. Since the length of u is constant, it is perpendicular to [itex]\frac{d\vec{u}}{dt}[/itex]. Now:

[tex]\vec{r} \times \vec{v}=r\vec{u} \times \frac{d(r\vec{u})}{dt}[/tex]

[tex]=r\vec{u} \times (r \frac{d\vec{u}}{dt} +\vec{u}\frac{dr}{dt})[/tex]

[tex]=r\vec{u} \times r \frac{d\vec{u}}{dt} +r \vec{u} \times \vec{u}\frac{dr}{dt})[/tex]

Since [itex]r\vec{u} [/itex] and [itex]\frac{dr}{dt}\vec{u}[/itex] are parrallel, this reduces to:

[tex]r\vec{u} \times r \frac{d\vec{u}}{dt}=r^2\vec{u} \times \frac{d\vec{u}}{dt}[/tex]

We also have

[tex]\vec{F}=m \vec{a}= -\frac{MmG\vec{u}}{r^2}[/tex]

Taking the cross product of the acceleration and angular momentum, you get:

[tex]\vec{a} \times (r^2\vec{u} \times \frac{d\vec{u}}{dt})[/tex]

From the formula for the vector triple product:

[tex]=r^2[(\vec{a} \cdot \frac{d\vec{u}}{dt})\vec{u} - (\vec{a} \cdot \vec{u})\frac{d\vec{u}}{dt}][/tex]

[itex] \vec{a}[/itex] and [itex]\frac{d \vec{u}}{dt}[/itex] are perpendicular becuse the the accelerationis parallel to the force, the force is antiparalel to the position vector, and the postion vector is perpendicular to [itex]\frac{d \vec{u}}{dt}[/itex]. Therefore this reduces to:

[tex]=-r^2(\vec{a} \cdot \vec{u})\frac{d\vec{u}}{dt}[/tex]

[tex]=-r^2((-\frac{GM\vec{u}}{r^2}) \cdot \vec{u})\frac{d\vec{u}}{dt}[/tex]

[tex]=GM\frac{d\vec{u}}{dt}[/tex]

Now consider [itex]\vec{v} \times h\vec{k}[/itex] and [itex]GM\vec{u}[/itex]:

[tex]\frac{d(\vec{v} \times h\vec{k})}{dt} = \vec{a} \times h\vec{k} + \vec{v} \times \frac{dh\vec{k}}{dt} = \vec{a} \times h\vec{k}[/tex]

[tex]\frac{d(GM\vec{u}}{dt}=GM\frac{d\vec{u}}{dt}[/tex]

[tex] \vec{a} \times h\vec{k} = GM\frac{d\vec{u}}{dt}, So: \vec{v} \times h\vec{k} = GM\vec{u} + \vec{C}[/tex]

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- #28

LeonhardEuler

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[tex](\vec{r} \times \vec{v}) \cdot h\vec{k} = (h\vec{k}) \cdot h\vec{k} = h^2[/tex]

[tex]\vec{r} \cdot (\vec{v} \times h\vec{k}) = \vec{r} \cdot (q\vec{u} + \vec{C}) = rq + \vec{r} \cdot \vec{C}[/tex]

Therefore, from the vector identity [itex](\vec{A} \times \vec{B}) \cdot \vec{C} = \vec{A} \cdot (\vec{B} \times \vec{C})[/itex]:

[tex]h^2 = rGM + \vec{r} \cdot \vec{C} = rGM + rc \cos{\theta}; c=||\vec{C}||[/tex]

Solving for [itex]\vec{r}(\theta)[/itex]:

[tex]r = \frac{h^2}{GM + c \cos{\theta}}[/tex]

Which we recognize as the equation of a conic section in polar coordinates. Requiring the conic section to be a circle is seen to be equivalent to requiring r to be independent of theta. This will happen if and only if (assuming h is not zero and q is finite) c is zero. c will be zero if and only if [itex]\vec{C}[/itex] is zero. This implies:

[tex]\vec{v} \times h\vec{k} = GM\vec{u} + \vec{C} = GM\vec{u}[/tex]

The cross product is a vector perpendicular to the velocity. Thus, requiring the directions of these vectors to be the same means that the initial velocity must be exactly perpendicular to the the initial postion vector(which makes sense for a circle). [itex]\vec{k}[/itex] was defined to be perpendicular to the velocity, so requiring the magnitudes to be the same gives:

[tex]vhk \sin{90} = vh = GM[/tex]

h was defined as [itex] \vec{r} \times \vec{v} = h\vec{k}[/itex]. Since we must have position and velocity perpendicular for the first condition, this means that h=rv. So:

[tex]v^2r=GM[/tex]

Dividing by [itex]r^2[/itex] gives the fammiliar result for acceleration in circular orbit:

[tex]\frac{v^2}{r}=\frac{GM}{r^2}=\frac{F}{m}=a[/tex]

This is the reason that orbits are eliptical. They must be conic sections due to the 1/r^2 nature of the force. They could not be hyperbolas or parabolas because these are unbounded, while orbits are bounded. And finally, they are not circular because this would require two independant initial conditions to take on certain exact values, making elipses the general case. The beginning of this argument was taken with some modifications from "Calculus and Analytic Geometry" by Stein and Barcellos.

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