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Topic: Celestial Movement (Read 1136 times) |
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Barukh
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Celestial Movement
« on: Aug 26th, 2007, 1:08am » |
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Consider a stationary orbit of a celestial body moving in a gravitational field of a single star. At every point in the orbit, the instantaneous velocity of the object has different direction and magnitude. Take all these “velocity vectors” and plot them on a single diagram and originating from the same point (say, O).
What is the shape traced by the endpoints of these vectors?
If there is a need, I will add a drawing later.
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mikedagr8
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Re: Celestial Movement
« Reply #1 on: Aug 26th, 2007, 1:12am » |
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Because worded problems involving mathematics is my weaknes, I'm going to take a stab at a quarter elipse? Not very confident here.
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Barukh
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Re: Celestial Movement
« Reply #2 on: Aug 26th, 2007, 1:33am » |
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on Aug 26th, 2007, 1:12am, mikedagr8 wrote:| Because worded problems involving mathematics is my weaknes, I'm going to take a stab at a quarter elipse? Not very confident here. |
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Your intuition deceived you this time.
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mikedagr8
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Re: Celestial Movement
« Reply #3 on: Aug 26th, 2007, 1:54am » |
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What, did I fluke it? 
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| « Last Edit: Aug 26th, 2007, 2:12am by mikedagr8 » |
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pex
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Re: Celestial Movement
« Reply #4 on: Aug 26th, 2007, 7:40am » |
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Well, plotting it, it looks a lot like a limacon (with a cedille under the c, but the server wouldn't let me post it), but I haven't yet been able to prove it.
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| « Last Edit: Aug 26th, 2007, 7:41am by pex » |
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Barukh
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Re: Celestial Movement
« Reply #5 on: Aug 26th, 2007, 8:52am » |
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on Aug 26th, 2007, 7:40am, pex wrote:| Well, plotting it, it looks a lot like a limacon |
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Nice try, but ... no. 
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Sameer
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Re: Celestial Movement
« Reply #6 on: Aug 26th, 2007, 1:06pm » |
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Do we need Kepler's laws for this?
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Barukh
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Re: Celestial Movement
« Reply #7 on: Aug 26th, 2007, 11:22pm » |
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on Aug 26th, 2007, 1:06pm, Sameer wrote:| Do we need Kepler's laws for this? |
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Not necessarily Kepler's
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Grimbal
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Re: Celestial Movement
« Reply #8 on: Aug 27th, 2007, 6:10am » |
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At least if the orbit is circular, the plot of the speed vector over time is a circle.
If the orbit is elliptical, I would have guessed an ellipse as well. Except that when I think of it, near the body, the speed is high and the direction changes quickly. An extremely flat orbit would result in something that looks like a half disk, i.e. The speed moves along the diameter when it is far away, and goes around a half-circle when zooming near the celestial body it orbits.
I don't know if that curve has a name. Maybe the inverse of the speed vector is an ellipse?
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Barukh
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Re: Celestial Movement
« Reply #9 on: Aug 27th, 2007, 9:07am » |
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on Aug 27th, 2007, 6:10am, Grimbal wrote:| I don't know if that curve has a name. |
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Yes, it certainly does! ...which you may be surprised to hear
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SMQ
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Re: Celestial Movement
« Reply #10 on: Aug 27th, 2007, 9:32am » |
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I can't prove it yet, but unless I'm very much mistaken, the curve in question is a circle, regardless of the eccentricity of the orbit.
--SMQ
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--SMQ
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Barukh
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Re: Celestial Movement
« Reply #11 on: Aug 27th, 2007, 9:39am » |
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on Aug 27th, 2007, 9:32am, SMQ wrote:I can't prove it yet, but unless I'm very much mistaken, the curve in question is a circle, regardless of the eccentricity of the orbit.
--SMQ |
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Exactly! 
Now, try to prove it. Remember: I put it in Medium, meaning that there is an elementary argument, and not too difficult.
Hint: Use one of the fundamental physical laws.
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| « Last Edit: Aug 27th, 2007, 9:40am by Barukh » |
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SMQ
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Re: Celestial Movement
« Reply #12 on: Aug 27th, 2007, 10:45am » |
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I'm sure there's a more basic way, but:
| hidden: | WLOG except for scaling, consider an elliptical orbit about point (0,0) with semi-major radius of 1 and eccentricity e. The polar equation of the ellipse is r = (1 - e2) / (1 + e cos  ).
The Wikipedia article on orbital speed gives the speed at a radius r in an elliptical orbit with semi-major axis of 1 as proportional to  (2/r - 1). Substituting the equation for r above gives v = (2 (1 + e cos ) / (1 - e2) - 1) = ((1 + e2 + 2e cos ) / (1 - e2)).
Switching to cartesian coordinates, a bit of math gives a tangent vector to the ellipse at point (r cos , r sin ) as <-sin , e + cos )>. And normalizing gives t = (1 / (1 + e2 + 2e cos))<-sin , e + cos >.
Finally, multiplying by (scalar) v gives (vector) v = vt = (((1 + e2 + 2e cos ) / (1 - e2)) / (1 + e2 + 2e cos))<-sin , e + cos >, which, after canceling leaves:
v = (1 / (1 - e2))<-sin , e + cos >
which is clearly the equation of a circle centered on (0,e). |
--SMQ
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Aryabhatta
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Re: Celestial Movement
« Reply #13 on: Aug 27th, 2007, 12:32pm » |
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Interesting question Barukh.
I was thinking of Conservation of Angular Momentum
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| « Last Edit: Aug 27th, 2007, 12:32pm by Aryabhatta » |
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Eigenray
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Re: Celestial Movement
« Reply #14 on: Aug 27th, 2007, 3:09pm » |
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on Aug 27th, 2007, 12:32pm, Aryabhatta wrote:| I was thinking of Conservation of Angular Momentum |
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Ooh, that'll work:
Let the position vector be r=(|r|, ), and velocity vector v. By Newton,
v' = - GM/|r|3 r.
By conservation of angular momentum,
L = m (r x v) = m |r|2 '
is constant, and then
v' = -GMm/L ' r/|r| = -GMm/L (cos, sin) ',
so
v = c + GMm/L (-sin, cos)
for some constant vector c. This works for open orbits too (but of course it will only be a circular arc).
For an elliptical orbit, we should have by symmetry that the radius GMm/L = (|v1| + |v|2|)/2, where v1, v2 are the max and min velocities at perihelion and aphelion, and then the center c = (v1 + v2)/2.
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| « Last Edit: Aug 27th, 2007, 3:21pm by Eigenray » |
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Sameer
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Re: Celestial Movement
« Reply #15 on: Aug 27th, 2007, 7:19pm » |
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Actually wikipedia has a nice interconnection of Kepler's law and Newton's law.. I started with the same equation SMQ used as mentioned in the wiki and then thought I would rather ask!! 
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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Barukh
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Re: Celestial Movement
« Reply #16 on: Aug 27th, 2007, 11:11pm » |
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Nice work, everybody!
I am glad you liked this problem.
The soluton I had in mind is inspired by Aryabhatta's comment. Unlike Eigenray's, it's almost purely geometrical, and is attributed to R. Feynman. It is nicely presented in a fascinating book I've read recently (pages 111-122):
D.L.Goodstein & J.R.Goodstein, "Feynman's Lost Lecture".
I strongly recommend this book to everybody.
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| « Last Edit: Aug 27th, 2007, 11:12pm by Barukh » |
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Eigenray
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Re: Celestial Movement
« Reply #17 on: Aug 28th, 2007, 2:03am » |
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on Aug 27th, 2007, 11:11pm, Barukh wrote:It is nicely presented in a fascinating book I've read recently (pages 111-122):
D.L.Goodstein & J.R.Goodstein, "Feynman's Lost Lecture". |
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To summarize the argument, he combines
Newton**: dv/dt ~ 1/r2,
Geometry: area ~ r2  ,
Kepler/Momentum*: area ~ t
to conclude v ~ . Thus:
Quote:Now, what are the changes in the velocity? The point is that in the first motion, this is the velocity. However, there is an impulse toward the Sun, and so there is a change in velocity, indicated by the green line that produces the second velocity, vK. Likewise, there's another impulse toward the Sun again, but this time the Sun is at a different angle, which produces the next change in the velocity, vL, and so on. Now, the proposition that the changes in the velocities were equal -- for equal angles, which is the one that we deduced -- means that the lengths of these succession of segments are all the same. That's what it means.
And what about their mutual angles? Since this is in the direction of the Sun at this radius, since this is at the direction of the Sun at that radius, and since this is the direction of the Sun at that radius, and so on, and since these radii each successively have a common angle to one another -- so it is likewise true that these little changes in the velocity have, mutually to one another, equal angles. In short, we are constructing a regular polygon. A succession of equal steps, each turn through an equal angle, will product a series of points on the surface underlying a circle. It will product a circle. Therefore, the end of the velocity vector -- if they call it that, the ends of these velocity points; you're not supposed to know what a vector is in this elementary description -- will lie on a circle. |
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He then goes on to use this fact to prove that the orbit is an ellipse!
*He gives an argument for this too, which he attributes to Newton. To paraphrase, suppose the particle starts at point A, and moves to point B in some unit of time. In the next unit of time, without the Sun, the particle would continue on in a straight line to some point c, with AB=Bc. But the particle is deflected towards the Sun, ending instead at a point C, such that the vector Cc is parallel to the line from the Sun S to B ("the middle instant"). Now, because triangles SAB and SBc share the altitude SB, and have equal bases AB=Bc, they have equal area. And because SB is parallel to Cc, triangles SBC and SBc have equal area. So in the limit, equal areas are swept out in equal times.
**In fact, Feynman explains how Newton's law follows from Kepler's laws 2&3. The fact that equal area = equal time is equivalent to the fact that gravitational force points towards the Sun, and then the argument here shows that period ~ r3/2 is equivalent to force ~ 1/r2. So really, his lecture shows that Kepler's first law is a consequence of the other two.
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Barukh
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Re: Celestial Movement
« Reply #18 on: Aug 28th, 2007, 5:49am » |
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on Aug 28th, 2007, 2:03am, Eigenray wrote:
Newton**: dv/dt ~ 1/r2,
Kepler/Momentum*: area ~ t |
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An interesting thing about this Kepler's law is it doesn't require Newton's law, but is a consequence of the fact that the gravitational force is directed towards the Sun.
The whole thing is amazing! 
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| « Last Edit: Aug 28th, 2007, 5:49am by Barukh » |
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SMQ
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Re: Celestial Movement
« Reply #19 on: Aug 28th, 2007, 7:37am » |
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I found it interesting from my exploration that as well as the total planetary velocity vector, the (sun-centered) radial component of the velocity vector also traces a circle!
--SMQ
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Sameer
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Re: Celestial Movement
« Reply #20 on: Aug 28th, 2007, 9:08am » |
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on Aug 28th, 2007, 7:37am, SMQ wrote:I found it interesting from my exploration that as well as the total planetary velocity vector, the (sun-centered) radial component of the velocity vector also traces a circle!
--SMQ |
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Was there anything interesting on the tangential component?
on Aug 27th, 2007, 11:11pm, Barukh wrote:Nice work, everybody!
I am glad you liked this problem.
The soluton I had in mind is inspired by Aryabhatta's comment. Unlike Eigenray's, it's almost purely geometrical, and is attributed to R. Feynman. It is nicely presented in a fascinating book I've read recently (pages 111-122):
D.L.Goodstein & J.R.Goodstein, "Feynman's Lost Lecture".
I strongly recommend this book to everybody. |
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Is there a online version of the book?
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"Obvious" is the most dangerous word in mathematics.
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Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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Barukh
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Re: Celestial Movement
« Reply #21 on: Aug 28th, 2007, 9:43am » |
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on Aug 28th, 2007, 9:08am, Sameer wrote:| Is there a online version of the book? |
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Not that I know... But, according to Eigenray's excerpt, he's probably found something...
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SMQ
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Re: Celestial Movement
« Reply #22 on: Aug 28th, 2007, 11:06am » |
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on Aug 28th, 2007, 9:08am, Sameer wrote:| Was there anything interesting on the tangential component? |
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The trace of the tangential component "looks like" a cartoid, but I didn't investigate beyond that. However, the magnitude of the tangential component multiplied by the radius is a constant (by conservation of angular momentum), so the scaled tangential component would obviously trace a circle.
--SMQ
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Eigenray
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Re: Celestial Movement
« Reply #23 on: Aug 28th, 2007, 1:33pm » |
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on Aug 28th, 2007, 9:08am, Sameer wrote:| Was there anything interesting on the tangential component? |
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Now that's a Limacon.
That is, |vt| ~ 1/|r| ~ 1+e cos(), but vt points in the direction  =+ /2, so the polar form would be
vt() ~ 1 + e sin().
For e=1 (parabolic orbit) this is a cardioid.
Quote:| Is there a online version of the book? |
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Not legally  But if one wanted some, say, "science popularization books" one could always try googling for the title and the word torrent.
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| « Last Edit: Aug 28th, 2007, 1:49pm by Eigenray » |
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mikedagr8
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Re: Celestial Movement
« Reply #24 on: Aug 28th, 2007, 3:24pm » |
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Am I able to get a picture for this. That's really interesting I found, never knew that. It's a good start for next year for when I do my subjects. 2*Maths, Chemistry, Physics, Accounting and of course English, as it is compulsory.
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| « Last Edit: Aug 29th, 2007, 2:24am by mikedagr8 » |
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