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# Orbital Challenge

09/17/2006 7:54 AM

Newtonian (or Keplerian, if you wish) elliptical orbits generally have angular as well as radial acceleration. Going beyond Kepler's "sweeping out equal areas in equal time", how is it that a purely radial force, like gravity, can produce an angular acceleration in a free falling object?

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#1

### Re: Orbital Challenge

09/18/2006 12:27 AM

Gravity is only a pure radial force for objects in circular orbits (Radial meaning pointing toward the center of curvature of the orbit). Objects in elliptical orbits don't have a single center of curvature, so the gravity is only radial at the apogee & perigee (or aphelion etc.)

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#2

### Re: Orbital Challenge

09/18/2006 2:24 AM

Ok, to prevent confusion, let's define the term 'radial' for this question. It is usual for a 2-body problem to choose the centre of mass (or barycenter) as the origin of the coordinate system. The radial distance of the orbiting object is the distance from the barycenter to the orbiting object. The radial gravitational force is thus also pointing towards the barycenter and not towards the centre of curvature of the orbit.

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#3

### Re: Orbital Challenge

09/18/2006 6:28 AM

Great Challenge, I guess if the only object under considderation was a singular planet in orbit round a singular star, we might expect a circular orbit. A Satelite like our sister moon for instance, has been over the eons, been subjected to gravitational influences from rather a complex set of multiple planetary interactions. Hardly a surprise that an 'elliptical' orbit, stabilises as a consequence of these interactions.

What every cosmologist, including Sir Isaac Nerwton, wants to discover, is why the eccentric orbit of our moon produces 'regular' lunar and solar eclypses at an almost exact interval of 18 years 11 days and eight hours. so that a solar eclypse returns essentially to the same place on our globe, every three saros's..... A mind-boggling coincidence. leading to speculations of the 'Outer Limits' kind. (did the little green men place it there as a calling card?)

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#4

### Re: Orbital Challenge

09/18/2006 6:57 AM

Just a short postscript. Quentin Wallop, Earl of Portsmouth who holds the 'Portsmouth Papers' that were regarded as far too 'esoteric' to be accepted for inclusion in the muniment collection of Sir Isaac Newton's tenure at Cambridge. In those appendix papers Sir Isaac speculates on his 6th Perturbation, and why it might be that compounding the astonishing combinations of coincidences that give rise to the 'Saros' The Total Eclypse of August 1999 should fall at an hour(not day!) that the 6th perturbation reached a near perfect integer fraction, and a rather revealing fraction at that. Sir Isaac was so disturbed by this extra coincidence, that he began to engage in 'metaphysical' investigations that Cambridge University were rather keen to suppress. I am sure Quentin would oblige any interested scholars who wished to inspect these 'rejected' documents. his Fax is +44 (0) 1256810048 (Estate Office)

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#6

### Re: Orbital Challenge

09/18/2006 7:23 AM

Quoting Alastair Carnegie: "I guess if the only object under consideration was a singular planet in orbit round a singular star, we might expect a circular orbit."

In fact circular orbits are rare and by nature not a stable condition. The slightest perturbation will convert a circular orbit into an elliptical one. The formation theories for planets also do not favour circular orbits.

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#17

### Re: Orbital Challenge

09/19/2006 5:28 AM

Perfectly correct Jorrie, that 'might' is a pretty big one! though the eliptical orbit again 'might' be fairly circular, but I stand to be corrected. It has fascinated me that itterative computer models have revealed that our own solar system, may in the past have 'destabilised' into a complex reorganisation on rare occasions. Instability builds up very slowly and then two planets come close enough to upset the proverbial "Apple Cart" to remind us of Sir Icaac's inspirational Apple.

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#5

### Re: Orbital Challenge

09/18/2006 7:17 AM

Jorrie, by angular acceleration, do you mean an orbit that has an eccentricity greater than zero, but less than one?

I am assuming that must be one of the attributes. You have defined radial acceleration, but angular must be a vector tangent to the orbit. Am I correct?

If so, then is the answer to your question simply that an orbital eccentricity greater than zero and less than one guarantees angular acceleration because the velocity of the orbiting body changes?

I think I am missing something, but I don't know what. ;-)

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#7

### Re: Orbital Challenge

09/18/2006 8:13 AM

Quoting 'Hero': "…by angular acceleration, do you mean an orbit that has an eccentricity greater than zero, but less than one?"

Yep, that's what elliptical orbits mean.

"You have defined radial acceleration, but angular must be a vector tangent to the orbit. Am I correct?"

Angular acceleration is simply the second differential of the angular position, which can have two components, say Phi and Theta.

"… an orbital eccentricity greater than zero and less than one guarantees angular acceleration because the velocity of the orbiting body changes?"

In a way, but velocity has three components: dr/dt, dPhi/dt and dTheta/dt. I am after the underlying principle for the "guarantees angular acceleration" you mentioned.

Sorry if all this sounds a tad pedagogical, but since you asked...

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#8

### Re: Orbital Challenge

09/18/2006 8:37 AM

Okay, if I may use a pun, it all "revolves" around the conservation of angular momentum, then.

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#10

### Re: Orbital Challenge

09/18/2006 8:39 AM

Deleted - I found the EDIT button on my comment above...

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#11

### Re: Orbital Challenge

09/18/2006 9:03 AM

Nice pun - and perfectly correct!

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#9

### Re: Orbital Challenge

09/18/2006 8:37 AM

Hi Jorrie, Hot Stuff! second derivatives, those can be a can of worms, to be sure.

Quore:- "In a way, but velocity has three components: dr/dt, dPhi/dt and dTheta/dt. I am after the underlying principle for the "guarantees angular acceleration" you mentioned"

I remember playing on an old tyre suspended from a rope. pull your legs in, and you got more dizzy. Is that the answer you are looking for? Thanks for the puzzle, I am off with another can of worms to go fishing, and think about it. Have an excellent day Sir.

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#12

### Re: Orbital Challenge a guess from a learner

09/18/2006 10:11 AM

ok I'm no mathematician, and this is all more than I should really attempt to solve, in the following I will give it my best shot with the help of the internet and my own nous.

from the wikapedia on angular momentum I find that..

proof that torque is equal to the time-derivative of angular momentum can be stated as follows:

The definition of angular momentum for a single particle is:

where "×" indicates the vector cross product. The time-derivative of this is:

This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv, we can see that:

But the cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma, we obtain:

And by definition, torque τ = r×F. Note that there is a hidden assumption that mass is constant — this is quite valid in non-relativistic mechanics. Also, total (summed) forces and torques have been used — it perhaps would have been more rigorous to write:

and by comparison we find at the same source the rate of change of angular velocity with respect to time.

Angular acceleration is the rate of change of angular velocity over time. In SI units, it is measured in radians per second squared (rad/s²), and usually denotes by the Greek letter alpha (). For constant values of angular acceleration, a rotating body conforms to the rotational equations of motion.

Additionally: = T / I. Where T is torque and I is moment of inertia.

NOTE:therefore the answer might be expressed as... and I expect more hilarity than a barrel full of of "Larus artricilla."

there fore angular acceleration /s²

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#13

### Re: Orbital Challenge a guess from a learner

09/18/2006 11:24 AM

Quoting mdbobbo: "...proof that torque is equal to the time-derivative of angular momentum can be stated as follows:"

It is nice engineering math, mdbobbo, but not quite applicable to orbits, though. Since angular momentum is a constant for an orbit (conservation of angular momentum), there is no time derivative of angular momentum applicable and thus no torque in the sense used in the Wikipedia article.

More to point of the topic, let us rather ponder this question: which other quantities are conserved in free fall orbits?

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#14

### Re: Orbital Challenge a guess from a learner

09/18/2006 1:04 PM

An object that falls through a vacuum is subjected to only one external force, the gravitational force.

The acceleration of the object equals the gravitational acceleration. The mass, size, and shape of the object are not a factor in describing the motion of the object. So all objects, regardless of size or shape or weight, free fall with the same acceleration.

the second law equation gives:

a = W / m = (m * g) / m = g

acceleration = weight over mass = (combined mass times gravitation) over mass = gravitational force

thanks NASA

all object fall at the same rate in a vacuum - galileo

http://exploration.grc.nasa.gov/education/rocket/ffall.html

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#15

### Re: Orbital Challenge a guess from a learner

09/18/2006 4:31 PM

I know the Lagrangian approach to orbital problems. The Lagrangian is the difference between the Kinetic and Potential energy (L=KE-PE). Orbit problems fall into a class of problems known as central potential problems. In this case we use gravitational force ~1/r2 to find the potential energy, which is equal to -V=F. The kinetic energy is simply 1/2 mv2 (XYZ coordinates). Converting to spherical coordinates, we get:

Notice that the equation has a ∂θ/∂t but no θ variable. This is called a cyclic variable. As a consiquence, one of the equations of motion for this Lagrangian is:

Which means that the angular momentum doesn't change over time. As well as angular momentum, energy is also conserved.

I think relativity says that spacetime is curved by a gravitational field and thus when a planet is orbiting, it's following the curvature of space time, but that might be too much of an over simplification.

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#16

### Re: Orbital Challenge a guess from a learner

09/19/2006 12:04 AM

Quoting Roger: "Which means that the angular momentum doesn't change over time. As well as angular momentum, energy is also conserved."

Spot-on.

"I think relativity says that spacetime is curved by a gravitational field and thus when a planet is orbiting, it's following the curvature of space time, but that might be too much of an over simplification."

Orbiting bodies follow spacetime geodesics rather than the curvature of spacetime.

BTW Roger, how do you get the equations in? Copy and Paste as pictures?

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#18

### Re: Orbital Challenge a guess from a learner

09/19/2006 10:28 AM

"BTW Roger, how do you get the equations in? Copy and Paste as pictures?"

Most of the time I'm copying and pasting as images. Sometimes I'll make them if I can't find the equation I'm looking for. Wikipedia has great stuff if you know what you're looking for, as does http://mathworld.wolfram.com/. I'm hoping there will be some sort of equation editor in the future. I think I've heard there will be.

Jorrie, what is a geodesic? I've heard the term before but I always forget what it is. I know it's some sort of curved line, but beyond that I'm lost.

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#19

### Re: Orbital Challenge a guess from a learner

09/19/2006 11:15 AM

Thanks Roger, I suppose one can make them with TEX and then copy and paste from a DVI-type file as pictures.

Geodesics are the path of least action. On earth, great circles are surface geodesics (shortest route). In 4-dimensional curved spacetime the path of least action is normally a freefall scenario, with space and time being considered. So a circular orbit's geodesic is not a circle, but a helix.

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#20

### Re: Orbital Challenge a guess from a learner

09/20/2006 11:28 PM

Lovely concise explanation Jorrie, thanks. It has reminded me of a book my Great Uncle David bequeathed me,(He was a cartographer) "Non Spherical Geodesics" as most of us know our planet is a 'grapefruit' P.S. the advantage of being a cartographer, is you can plaster the map with all the family's names.(OZ?) My daughter is named after a fresh water spring.

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