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        riddles >> medium >> Celestial Movement
        
(Message started by: Barukh on Aug 26th, 2007, 1:08am)
Title: Celestial Movement
Post by Barukh on Aug 26th, 2007, 1:08am
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.
Title: Re: Celestial Movement
Post by mikedagr8 on Aug 26th, 2007, 1:12am
Because worded problems involving mathematics is my weaknes, I'm going to take a stab at a quarter elipse? Not very confident here.
Title: Re: Celestial Movement
Post by Barukh on Aug 26th, 2007, 1:33am

on 08/26/07 at 01:12:28, 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.

Your intuition deceived you this time.
Title: Re: Celestial Movement
Post by mikedagr8 on Aug 26th, 2007, 1:54am
What, did I fluke it?  ???
Title: Re: Celestial Movement
Post by pex on Aug 26th, 2007, 7:40am
Well, plotting it, it looks a lot like a [hide]limacon (with a cedille under the c, but the server wouldn't let me post it)[/hide], but I haven't yet been able to prove it.
Title: Re: Celestial Movement
Post by Barukh on Aug 26th, 2007, 8:52am

on 08/26/07 at 07:40:48, pex wrote:
Well, plotting it, it looks a lot like a [hide]limacon[/hide]

Nice try, but ... no.  ;)
Title: Re: Celestial Movement
Post by Sameer on Aug 26th, 2007, 1:06pm
Do we need Kepler's laws for this?
Title: Re: Celestial Movement
Post by Barukh on Aug 26th, 2007, 11:22pm

on 08/26/07 at 13:06:54, Sameer wrote:
Do we need Kepler's laws for this?

Not necessarily Kepler's  ;)
Title: Re: Celestial Movement
Post by Grimbal on Aug 27th, 2007, 6:10am
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?
Title: Re: Celestial Movement
Post by Barukh on Aug 27th, 2007, 9:07am

on 08/27/07 at 06:10:09, Grimbal wrote:
I don't know if that curve has a name.

Yes, it certainly does! [hide]...which you may be surprised to hear[/hide]  ;)
Title: Re: Celestial Movement
Post by SMQ on Aug 27th, 2007, 9:32am
I can't prove it yet, but unless I'm very much mistaken, the curve in question is a [hide]circle, regardless of the eccentricity of the orbit[/hide].

--SMQ
Title: Re: Celestial Movement
Post by Barukh on Aug 27th, 2007, 9:39am

on 08/27/07 at 09:32:51, SMQ wrote:
I can't prove it yet, but unless I'm very much mistaken, the curve in question is a [hide]circle, regardless of the eccentricity of the orbit[/hide].

--SMQ

Exactly!  :D :D :D

Now, try to prove it. Remember: I put it in Medium, meaning that there is an elementary argument, and not too difficult.

Hint: [hide]Use one of the fundamental physical laws[/hide].
Title: Re: Celestial Movement
Post by SMQ on Aug 27th, 2007, 10:45am
I'm sure there's a more basic way, but:
[hideb]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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif).

The Wikipedia article on orbital speed (http://en.wikipedia.org/wiki/Orbital_speed) gives the speed at a radius r in an elliptical orbit with semi-major axis of 1 as proportional to http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(2/r - 1).  Substituting the equation for r above gives v = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(2 (1 + e cos http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif) / (1 - e2) - 1) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif((1 + e2 + 2e cos http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif) / (1 - e2)).

Switching to cartesian coordinates, a bit of math gives a tangent vector to the ellipse at point (r cos http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif, r sin http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif) as <-sin http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif, e + cos http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)>.  And normalizing gives t = (1 / http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(1 + e2 + 2e coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif))<-sin http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif, e + cos http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif>.

Finally, multiplying by (scalar) v gives (vector) v = vt = (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif((1 + e2 + 2e cos http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif) / (1 - e2)) / http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(1 + e2 + 2e coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif))<-sin http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif, e + cos http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif>, which, after canceling leaves:

v = (1 / http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(1 - e2))<-sin http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif, e + cos http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif>

which is clearly the equation of a circle centered on (0,e).[/hideb]
--SMQ
Title: Re: Celestial Movement
Post by Aryabhatta on Aug 27th, 2007, 12:32pm
Interesting question Barukh.

I was thinking of [hide] Conservation of Angular Momentum [/hide]
Title: Re: Celestial Movement
Post by Eigenray on Aug 27th, 2007, 3:09pm

on 08/27/07 at 12:32:34, Aryabhatta wrote:
I was thinking of [hide] Conservation of Angular Momentum [/hide]

Ooh, that'll work:

Let the position vector be r=(|r|, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif), and velocity vector v.  By Newton,

v' = - GM/|r|3 r.

By [hide]conservation of angular momentum[/hide],

[hide]L = m (r x v) = m |r|2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif'[/hide]

is constant, and then

v' = [hide]-GMm/L http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif' r/|r| = -GMm/L (coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif, sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif'[/hide],

so

v = [hide]c + GMm/L (-sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif, coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)[/hide]

for some constant vector c.  This works for open orbits too (but of course it will only be a [hide]circular arc[/hide]).

For an elliptical orbit, we should have by symmetry that [hide]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[/hide].
Title: Re: Celestial Movement
Post by Sameer on Aug 27th, 2007, 7:19pm
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!! :)
Title: Re: Celestial Movement
Post by Barukh on Aug 27th, 2007, 11:11pm
Nice work, everybody!  :D

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.
Title: Re: Celestial Movement
Post by Eigenray on Aug 28th, 2007, 2:03am

on 08/27/07 at 23:11:06, 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".

To summarize the argument, he combines

Newton**: dv/dt ~ 1/r2,
Geometry: area ~ r2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cdelta.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif,
Kepler/Momentum*: area ~ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cdelta.gift

to conclude http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cdelta.gifv ~ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cdelta.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif.  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 [hide]circle[/hide].


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 [link=http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Kepler.27s_third_law]here[/link] 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.
Title: Re: Celestial Movement
Post by Barukh on Aug 28th, 2007, 5:49am

on 08/28/07 at 02:03:15, Eigenray wrote:
Newton**: dv/dt ~ 1/r2,
Kepler/Momentum*: area ~ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cdelta.gift

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!  :o
Title: Re: Celestial Movement
Post by SMQ on Aug 28th, 2007, 7:37am
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 [hide]also traces a circle[/hide]!

--SMQ
Title: Re: Celestial Movement
Post by Sameer on Aug 28th, 2007, 9:08am

on 08/28/07 at 07:37:52, 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 [hide]also traces a circle[/hide]!

--SMQ


Was there anything interesting on the tangential component?




on 08/27/07 at 23:11:06, Barukh wrote:
Nice work, everybody!  :D

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.



Is there a online version of the book?
Title: Re: Celestial Movement
Post by Barukh on Aug 28th, 2007, 9:43am

on 08/28/07 at 09:08:32, Sameer wrote:
Is there a online version of the book?

Not that I know... But, according to Eigenray's excerpt, he's probably found something...
Title: Re: Celestial Movement
Post by SMQ on Aug 28th, 2007, 11:06am

on 08/28/07 at 09:08:32, Sameer wrote:
Was there anything interesting on the tangential component?

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
Title: Re: Celestial Movement
Post by Eigenray on Aug 28th, 2007, 1:33pm

on 08/28/07 at 09:08:32, Sameer wrote:
Was there anything interesting on the tangential component?

Now that's a [link=http://en.wikipedia.org/wiki/Lima%C3%A7on]Limacon[/link].

That is, |vt| ~ 1/|r| ~ 1+e cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif), but vt points in the direction http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif+http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/2, so the polar form would be

vt(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif) ~ 1 + e sin(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif).

For e=1 (parabolic orbit) this is a cardioid.


Quote:
Is there a online version of the book?

Not legally  :-/  But if one wanted some, say, "science popularization books" one could always try googling for the title and the word torrent.
Title: Re: Celestial Movement
Post by mikedagr8 on Aug 28th, 2007, 3:24pm
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.
Title: Re: Celestial Movement
Post by ThudanBlunder on Aug 28th, 2007, 4:29pm

on 08/28/07 at 15:24:39, mikedagr8 wrote:
and of course English as it is compulsory.

...unlike punctuation?  
Title: Re: Celestial Movement
Post by Eigenray on Aug 28th, 2007, 8:25pm
I thought I'd attach the following diagram from the book, because it's pretty amazing.

Rotating the velocity diagram by -http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/2, and superimposing the result on the orbit diagram (here S=C), the velocity vector at the instant shown is perpendicular to Op.

Construct a curve as follows: Take a point O (origin of the velocities) inside a circle centered at C (sun/center of velocity circle).  For each point p on the circle, plot the intersection P between Cp and the perpendicular bisector of Op.

Since CP + OP = CP + Pp = Cp is constant, this forms an ellipse.  And, for any point x on line tP, Cx + xO = Cx + xp http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif Cp, with equality iff x=P.  So tP intersects the ellipse only at P, i.e., it is in fact tangent to the curve.

So we've constructed a curve such that the tangent at each point is parallel to the velocity.  So the orbits are ellipses.  No calculus!

Of course, the only reason O was inside the circle is by choice of scale comparing position and velocity.  If we do the same construction with O outside the circle, we get a hyperbola.  As O approaches the boundary of the circle, the orbit "approaches" a parabola.   Can we make the construction work for O on the boundary?
Title: Re: Celestial Movement
Post by Sameer on Aug 28th, 2007, 9:36pm

on 08/28/07 at 13:33:21, Eigenray wrote:
Now that's a [link=http://en.wikipedia.org/wiki/Lima%C3%A7on]Limacon[/link].


My first intuition on the original problem was that the answer was a Limacon. I guess a little flat tire on the left of the intended answer ;)
Title: Re: Celestial Movement
Post by mikedagr8 on Aug 29th, 2007, 12:11am

Quote:
and of course English as it is compulsory.


is now....

and of course English; as it is compulsory.

Are you a happy Pom now?
Title: Re: Celestial Movement
Post by ThudanBlunder on Aug 29th, 2007, 2:08am

on 08/29/07 at 00:11:05, mikedagr8 wrote:
and of course English; as it is compulsory.

Are you a happy Pom now?

Nope, that should have been a comma.
But the whole post looks like a hurried text message, unlike your earlier posts.
Title: Re: Celestial Movement
Post by mikedagr8 on Aug 29th, 2007, 2:23am
Lol, well, that ";" was a typo (Yes I know the name of the symbol). And that particular post I was typing, well, my friend had just woken up and called me away, so I wanted to finish it ASAP, hence the dodgy English. I much prefer Engrish.
Title: Re: Celestial Movement
Post by SMQ on Aug 29th, 2007, 8:22am

on 08/28/07 at 15:24:39, mikedagr8 wrote:
Am I able to get a picture for this.

Sure! 8)

http://www.dwarfrune.com/~smq/wu/orbit.gif

In the graphic above a planet (blue-green) orbits a sun (yellow) in an elliptical orbit.  The point of view stays centered on the planet.  Four velocity vectors and the traces of their endpoints are shown: white is the total velocity, red is the radial component, blue is the tangential component, and green is the tangential component multiplied by the (instantaneous) orbital radius.

--SMQ
Title: Re: Celestial Movement
Post by towr on Aug 29th, 2007, 8:47am
Very nice picture!


on 08/29/07 at 08:22:20, SMQ wrote:
green is the tangential component multiplied by the radius.
Just to be clear; radius is the distance to the sun? (Because typically I think of radius as a fixed parameter, so it'd just give a scaled version of the blue figure)
Title: Re: Celestial Movement
Post by SMQ on Aug 29th, 2007, 9:02am

on 08/29/07 at 08:47:46, towr wrote:
Just to be clear; radius is the distance to the sun?

Yes, it's the instantaneous radius of the orbit.  I included it because by the law of conservation of angular momentum, the instantaneous magnitude of the tangential velocity multiplied by the instantaneous radius of the orbit is a constant.

--SMQ
Title: Re: Celestial Movement
Post by mikedagr8 on Aug 29th, 2007, 5:05pm
That's really cool. I'm going to be sending this to the physics teachers, they'll love it. Nice diagram.   :D


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