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DaveC426913

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The ellipse is the

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My understanding is that Kepler empirically determined that the orbits of planets are ellipses based on Tycho Brahe's observational data. Later, Newton postulated his famous general mechanical force laws and the gravity force law that says gravitational force is directly proportional to the product of the masses and inversely proportional to the square of the distance. From this, he mathematically proved that this generates stable elliptical orbits for a two body system.

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Sorry for all the questions, I appreciate your help...

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Nabeshin

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Sorry for all the questions, I appreciate your help...

Circles are just special cases of ellipses where the eccentricity is zero. The only reason they aren't observed is because the positioning of an astronomical body has to be so perfect that it almost never occurs.

Circles actually represent the lowest eccentricity case for an orbit (e=0), and since eccentricity is always positive, the value can only increase, degrading into an elliptical orbit (0<e<1). Once e=1, the orbit is parabolic (a very special case. One that is again, not usually observed in nature for similar reasons to that of the circle). And if 1<e<infinity, the orbit is hyperbolic. The reason they're all ellipses is because neither parabolas nor hyperbolas are stable, so there's no time to observe these kind of orbits because they quickly leave the host star.

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DaveC426913

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This does not merely apply to orbits.I see, thanks. So, what if the orbit was, for sake of this example, circular? What would be predicted to happen now that Kepler's ideas had been proven mathamatically? How would that change the scenario? Would the force of gravity start to take over where it otherwise wouldn't have? Or could it actually be stable? The way I'm thinking right now, is that the inertia/momentum is what's keeping that orbit stable, would I be right in saying that?

Have you ever seen one of these coin funnels in malls?

http://us.st12.yimg.com/us.st.yimg.com/I/igumballs_2049_5927318

You drop a coin in on its edge and watch it roll round and round until it finally drops out the drain in the middle/bottom.

If it could discount friction, the coin would follow an elliptical path, speeding up as it neared the vortex of the funnel, then slowing down as its speed carried it past the vortex to outer the perimeter again.

OK, well actually, it is still the exact same thing (gravity trading off versus inertia) but I want you to see that this is a

A Youtube video of coins (trying) to follow an elliptical path.

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DaveC426913

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Yes. Without friction, the elliptical path of both coin and planet is stable.Would the mere absence of friction make it go round forever?

Think about throwing a ball in the air. (Let's ignore air friction.)

It starts off with high speed moving upward.

As it rises, it gains potential energy but loses kinetic energy.

As it reaches its zenith, it has max potential energy and zero kinetic energy. Note though, that is has exactly as much energy as it had when it left your hand.

As it begins to fall, it trades off potential energy for kinetic energy.

Once it reaches the ground, it has max kinetic energy and zero potential energy, yet still has exactly as much energy as when it left your hand.

If you manage to somehow have your ball not hit the Earth (say you replace the Earth with a black hole of identical mass but which is only one mile in diameter), the ball will fall all the way to centre, picking up kinetic energy as it goes. It will loop around and come back, shooting past you to its max height again.

And voila! the ball is in

More generally, everytime someone throws a baseball, they are putting the ball into orbit around the Earth's centre - barring air friction and impact with the Earth's surface.

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I see, great explination. I think I get it now, thanks...

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malawi_glenn

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DaveC426913

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I question if this answer is in the spirit of the OP's knowledge level.

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The chances of me calculating anything right now are little to none...

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Chronos

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GR plays a role in this as well, bear in mind how GR solved the precession problem with Mercury.

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malawi_glenn

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I question if this answer is in the spirit of the OP's knowledge level.

OP's answer in post #4 is full of physical quantities that is introduced in courses in Newtonian Mechanics.

One can not prove by intuitive arguments that the paths are elliptical, one must show that angle covered over one period is 2 pi radians.

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atyy

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Would it be too much of a cheat to invoke angular momentum conservation?

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malawi_glenn

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Nabeshin

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If I remember correctly, conservation of angular momentum makes the proof of conic section orbits easier. So, in a sense, I guess you could consider it "cheating" because Newton couldn't invoke such methods.

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malawi_glenn

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If I remember correctly, conservation of angular momentum makes the proof of conic section orbits easier. So, in a sense, I guess you could consider it "cheating" because Newton couldn't invoke such methods.

Then angular momentum is not conserved in GR since then the orbits are not closed, if we turn the argument.

If you invoke Angular Momentum conservation, you must prove it as well.

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Nabeshin

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If you invoke Angular Momentum conservation, you must prove it as well.

Well the proof is pretty trivial. Call some value r x v angular momentum, and it's fairly straightforward to show from the gravitation equation that d/dt (r x v) = 0, so angular momentum is conserved. Then if you use this value it makes the derivation flow pretty nicely. Not that much extra work.

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malawi_glenn

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Well the proof is pretty trivial. Call some value r x v angular momentum, and it's fairly straightforward to show from the gravitation equation that d/dt (r x v) = 0, so angular momentum is conserved. Then if you use this value it makes the derivation flow pretty nicely. Not that much extra work.

Of course, everything that is spherical symmetric angular momentum must be conserved, i meant that angular momentum is conserved in GR as well (as far as I know, but orbits are not ellipses)

I looked up how to 'derive' ellipses from angular momentum conservation, but that derivation is as long as the one I meant so...

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atyy

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I was thinking of Kepler's law of equal areas. But more as a handwavy thing - the only true way I'm familiar with is to solve the equations as you said.

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atyy

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I looked up how to 'derive' ellipses from angular momentum conservation, but that derivation is as long as the one I meant so...

Hmm, interesting - it's possible in a non-handwavy way?

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atyy

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I haven't followed the details, but just for reference: "Conservation of the direction of angular momentum means that the particle will move in a plane. .......... The precession of perihelia reflects the fact that noncircular orbits are not closed ellipses; to a good approximation they are ellipses which precess, describing a flower pattern. http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll7.html

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Nabeshin

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Of course, everything that is spherical symmetric angular momentum must be conserved, i meant that angular momentum is conserved in GR as well (as far as I know, but orbits are not ellipses)

I looked up how to 'derive' ellipses from angular momentum conservation, but that derivation is as long as the one I meant so...

Ah, sorry I thought you were just using GR as a counter-point. But again, like someone previously said, using GR is likely above what the OP wanted to ask anyways (and besides the point, because the main interest seems to be the trade-off between inertia and the gravitational force).

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