What causes orbits ?
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dbrya ...@optusnet.com.au (D Bryan)
If gravitation was to suddenly disappear, would planets continue to orbit their Suns in order to conserve angular momentum ?
db
"Robert J. Kolker" robert_kol...@hotmail.com
No. Spacetime would flatten out and the planets would move in a geodisical fashion with respect to the new curvature. They would move in what we call straight lines.
Bob Kolker
Ian Stirling r...@mauve.demon.co.uk
No.
If gravity disappeared, the planets would continue on in straight lines.
The sun would start to expand at some 200Km, increasing in size by 10% in 5 minutes or so.
I suspect the dynamics of a star expanding without gravity will be interesting.
The angular momentum is conserved through the interaction of gravity and inertia.
If you remove gravity, then you break conservation of angular momentum.
Hmm, it'd be an interesting weapon if you could kill gravity in a star.
"Dirk Van de moortel" dirkvandemoor...@ThankS-NO-SperM.hotmail.com
The planets would stop orbiting *and* nicely conserve angular momentum.
But of course the Suns would explode and stop burning hydrogen, and the planets would fall apart or maybe even explode as well.
Dirk Vdm
Sam Wormley sworml...@mchsi.com
Kepler discovered that planetary orbit of Mars was best described by an ellipse. He discovered this AFTER he discovered that Mars swept out equal areas in equal times (Kepler's second law).
Newton was able to show that under the incluence of a central force, in this case gravity, that the orbits of objects follow conic sections.
These include hyperbola, parabola and ellipse. The closed paths of planets are ellipses, sweeping out equal areas in equal times (conservation of angular momentum).
Total Angular Momentum is the vector sum of the spin angular momentum S and orbital angular momentum L http://scienceworld.wolfram.com/physics/TotalAngularMomentum.html Conservation of Angular Momentum http://scienceworld.wolfram.com/physics/ConservationofAngularMomentum... Angular Momentum can be expressed as the cross product of r and p http://scienceworld.wolfram.com/physics/AngularMomentum.html it is the L that is conserved. The r and p are changing because of the force of gravity (in the case of a planet orbiting a star).
Magically take away the Sun's gravity, then o r is meaningless o r cross p has zero magnitude o the planet continues to move in a straight line with momentum p
"Dirk Van de moortel" dirkvandemoor...@ThankS-NO-SperM.hotmail.com
Look at http://users.pandora.be/vdmoortel/dirk/Stuff/AngularMomentum.jpg Before (with gravitation): R1 x p = |R1| |p| sin(pi/2) = |R1| |p| After (no gravitation): R2 x p = |R2| |p| sin(pi/2-a) = |R1/cos(a)| |p| cos(a) = |R1| |p| Angular momentum would be conserved, wouldn't you agree?
Dirk Vdm
"Dirk Van de moortel" dirkvandemoor...@ThankS-NO-SperM.hotmail.com
Nitpick: I forgot to write the unit vectors in the up-direction on the RHS.
Dirk Vdm
Sam Wormley sworml...@mchsi.com
Dirk Van de moortel wrote: How about that! Angular momentum would be conserved... but does it have physical meaning?
"Dirk Van de moortel" dirkvandemoor...@ThankS-NO-SperM.hotmail.com
Angular momentum can be calculated "around" any point in space...
When calculated around the center of attraction, does it have physical meaning in the first place?
An arrow pointing up from the plane of motion?
We already have made a non-physical choice by taking r x p in stead of p x r ...
Dirk Vdm
Sam Wormley sworml...@mchsi.com
Dirk Van de moortel wrote: Good point! (no pun intended)
bri ...@encompasserve.org
Just because the vector is meaningless doesn't mean it doesn't exist.
r cross p has the same magnitude it had before you took away the object at one end of the r vector.
Of course. And it has the same angular momentum it started with.
John Briggs
Sam Wormley sworml...@mchsi.com
Yes--You are correct!
-Sam
"Greg Neill" gneill...@OVE.netcom.ca
About the same meaning as the angular momentum for a system taken about an arbitrary point, I suspect.
One wonders, if one had the technology to "kill" gravitation in a body, would its inertia necessarily have to disappear too? In other words, is killing gravity equivalent to breaking linear momentum conservation while preserving angular momentum conservation?
af ...@FreeNet.Carleton.CA (John Park)
_Closed orbits_ are maintained through the interaction of gravity with inertia; a planet moving off in a straight line still has its angular momentum about the star. (L=rxp --it doesn't say anything about curved paths.) --John Park
"Keith Stein" ks012a2...@blueyonder.co.uk
Good question.
The total angular momentum would be conserved at ZERO !
Remember that the sun is also rotating about the center of m***, so the total angular momentum about the center of m*** is zero, both before and after gravity disappears.
keith stein
Sam Wormley sworml...@mchsi.com
Careful here Keith--Are you saying that the total angular momentum of the solar system is zero?
pcardin ...@volcanomail.com (Paul Cardinale)
If you take a weight attached to a rope and swing it around over your head, and the rope breaks, would the weight keep going in a circle over your head in order to conserve angular momentum?
Note that conservation of angular momentum does not require any particular path shape.
Paul Cardinale
"Keith Stein" ks012a2...@blueyonder.co.uk
Well i was saying that, but i was wrong eh!
On reconsidering the total angular momentun is clearly NOT zero, either before or after gravity disappears, although the linear momentum (relative to the center of m***) IS, but that's a poor excuse, since the question was clearly abour angular momentum eh!
Thank you for your corteous correction, Mr. Wormley.
keith stein
"Robert J. Kolker" robert_kol...@hotmail.com
It would fly off at a tangent.
Bob Kolker
mme ...@cars3.uchicago.edu
Mati Meron | "When you argue with a fool, me...@cars.uchicago.edu | chances are he is doing just the same"
mme ...@cars3.uchicago.edu
What?
Mati Meron | "When you argue with a fool, me...@cars.uchicago.edu | chances are he is doing just the same"
"Franz Heymann" notfranz.heym...@btopenworld.com
No. They will maintain ngular omentum by flying off at constant speed along a tangent.
l = r x p where l = angular momentum vector r = position vector of planet p = momentum vector of planet
"Franz Heymann" notfranz.heym...@btopenworld.com
...
Yes, according to the definition of angular momentum.
{:-)) Franz
"Franz Heymann" notfranz.heym...@btopenworld.com
And the same direction. {:-)) [snip] Franz
mme ...@cars3.uchicago.edu
Sure, why not. You may ask where you should care about it, but that's another story alltogether.
Mati Meron | "When you argue with a fool, me...@cars.uchicago.edu | chances are he is doing just the same"
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