Physics Help Forum Newtonian Orbital Calculations

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 Jul 13th 2016, 08:32 PM #1 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 550 Newtonian Orbital Calculations I haven't posted in this section before so I hope I have it correct. Apologies otherwise. I am attempting to make an orbital simulation for a two body newtonian gravitational problem. It's probably been done before but I wanted to use it as a starting point for extending to 3 body problems and beyond. I'm running into a problem with my usage of polar co-ordinates. I always figured that a central inverse square law force works basically the same in a rotating frame of reference. (I thought Noether's Theorem stated that the laws of physics are the same under translation and rotation.) However when I look at the wiki article for polar co-ordinates all these "fictitious" terms pop out. Specifically [; r'' = (\ddot{r} - r \dot {\phi})\hat{r} + (r \ddot{\phi} + 2 \dot{r} \dot{\phi}) \hat{\phi} ;]. Apparently this includes centrifugal [; - r \dot {\phi} ;] and coriolis [; (r \ddot{\phi} + 2 \dot{r} \dot{\phi}) ;] terms. Apparently I can ignore the coriolis terms because conservation of angular momentum sets this to zero but I am not so lucky with the centrifugal term. Can anyone explain this centrifugal term to me and why it exists? I am having trouble understanding it. I suspect I need to include it (somehow) to fix my simulation. P.S. I am using this Chrome extension for latex. Edit: Screenshot of my latex view for those who don't have the chrome extension installed: Last edited by kiwiheretic; Jul 13th 2016 at 11:52 PM.
 Jul 14th 2016, 04:54 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,336 A couple of minor corrections: the centripetal term is r phi_dot^2, not r phi_dot. Also, the coreolis factor is the term 2 r_dot phi_dot; the r phi_double_dot term is simple acceleration in the phi_hat direction. The terms you ask about are not "fictitious," but rather come about directly from the implications of using a rotating frame of reference. Unlike with Cartesian coordinates, in a rotating frame the unit vectors r_hat and phi_hat are rotating, and hence when you take the derivative with respect to time you must include the derivative of these unit vectors as well. For your simulation using simple F=ma you would have two basic equations for the classic two-body situation: 1. Radial acceleration, due to force of gravity: GM/r^2 = (r_dbl_dot - r phi_dot^2) 2. Tangential acceleration = 0: r phi_double_dot + 2 r_dot phi_dot = 0. Note that you can't ignore that second equation.
 Jul 14th 2016, 11:27 AM #3 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 550 Ok, but if I build conservation of angular momentum into the system is that sufficient to deal with the Coriolis term? Thanks for explaining the centrifugal force (or giving me a working formula) of which I was greatly confused. I was calling it "fictitious" because it's existence depends on ones choice of co-ordinate system. Last edited by kiwiheretic; Jul 14th 2016 at 11:35 AM.
 Jul 14th 2016, 12:28 PM #4 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,336 I believe it should work OK that way. I found the following site which uses Newton's Law of gravitation together with constant angular acceleration to derive the equation of motion for the orbit, see: https://en.wikipedia.org/wiki/Specif...gular_momentum The equation becomes as shown in the attached, where mu = G/M1 and h = angular momentum. Attached Images   Last edited by ChipB; Jul 14th 2016 at 01:28 PM.
 Jul 14th 2016, 02:01 PM #5 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 550 However when I take that result and put it into the differential equation r_dot_dot * r^2 in the hopes of getting a constant. Well, I tried that on wolfram alpha and see attached for my results. How can this be? Shouldn't the result you posted be a solution to that differential equation? Edit: Hmmm, I might see my mistake. I'm differentiating wrt phi and not wrt to time. I have to go and stand in the rain for a bit (catching bus) so haven't got time to plugin the adjusted equations into wolframalpha but will try later. Attached Thumbnails     Last edited by kiwiheretic; Jul 14th 2016 at 05:00 PM.
 Jul 15th 2016, 04:51 AM #6 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,336 Sorry, but I don't understand why you think that r_dbl_dot r^2 would be a constant - it's not. Think about an orbital ellipse - at apogee and perigee r is constant, so r_dbl_dot = 0. At other points in the orbit r is increasing or decreasing, so r_dbl_dot varies throughout the orbit, and hence r_dbl_dot r^2 is not constant.` **EDIT** Perhaps you meant to write that phi_dot r^2 should come out as a constant (i.e. angular momentum is constant)? Not sure how to derive that from the orbital equation I posted earlier. Last edited by ChipB; Jul 15th 2016 at 05:36 AM.
Jul 15th 2016, 11:54 AM   #7
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 Originally Posted by ChipB **EDIT** Perhaps you meant to write that phi_dot r^2 should come out as a constant (i.e. angular momentum is constant)? Not sure how to derive that from the orbital equation I posted earlier.
Yes, that's what I was driving at. Why is that not a consequence of equating Newton's law of gravitation with Newton's second law? Ie: GMm/r^2 = m*r_dot_dot. That would certainly explain why my calculations are at variance with what you would expect if you consider my starting point wrong!!

Jul 15th 2016, 12:17 PM   #8
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 Originally Posted by kiwiheretic Yes, that's what I was driving at. Why is that not a consequence of equating Newton's law of gravitation with Newton's second law? Ie: GMm/r^2 = m*r_dot_dot. That would certainly explain why my calculations are at variance with what you would expect if you consider my starting point wrong!!
GMm/r^2 =- m r_dbl_dot is for linear motion only, not an orbit. It's fine for calculating motion of a projectile fired straight up from a planet, where phi_dot and phi_dbl_dot are both zero. In the formula I provided earlier this would meam angular momentum (h) is zero.

One scenario you might want to consider is the simple case of a circular orbit, in which r_dot and r_dbl_dot are both zero. In this case the equation of motion is simply:

GM/r^2 = r phi_dot^2

For such an orbit the factor C/mu in the earlier equation I posted = 0, and you get Gm = r^3 phi_dot^2.

 Jul 15th 2016, 01:37 PM #9 Senior Member   Join Date: Jun 2010 Location: NC Posts: 414 I had hoped to get into this discussion "near its beginning." I've had typing physics discussions via www before about this. It has been near impossible to proceed without some manner of graghics. So I started to layout some references, but now see you two have rode over the horizon toward Hollywood. Alas! My understanding is there are three "forces at a distance." Gravity, ESF and EMF. D' Alembert coined "inertial force" (using vector algebra to move ma over beside Fg et al). Centrifugal, Centripetal and Coriolis are not forces. They are consequences of "observation from the an assumed non-accelerating frame" that is in fact accelerating. These created-correctional effects, of course, adjust for the human observation error. Physics has not cleaned this up very well. I know nothing about 2-Body problem. I wonder what reference that would be.. where's the observer. I'm interested in the small parts of this. I got a little "started." http://www.thermospokenhere.com/wp/0...do_forces.html Good Luck... I'll read along. Jp Last edited by THERMO Spoken Here; Jul 15th 2016 at 01:41 PM.
 Jul 15th 2016, 02:23 PM #10 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,336 The subject of just what Coriolis and centrifugal accelerations come from can be a bit confusing. I disagree with your statement that they "are consequences of 'observation from the an assumed non-accelerating frame' that is in fact accelerating. These created-correctional effects, of course, adjust for the human observation error." this may be true for a rotating frame of reference - for example the prevailing winds in the northern hemisphere on Earth blow from west to east because of our coordinate system that is tied to the rotating earth - i.e.our coordinate system of latitude and longitude is rotating. But I wouldn't say it's due to human error. And even if the coordinate system is not rotating you would still have Coriolis and centripetal terms to deal with - these come about directly from the decision to use polar coordinates and the consequences of differentiation using polar coordinates. Let me give an example: consider a hockey puck gliding without friction at constant velocity across an ice rink. In Cartesian coordinates we describe its motion as straight line, such as x(t) = x_0 + V_0t . But if we mark a polar coordinate system on the ice rink that same straight line must be described not simply with linear equations for r(t) and phi(t) but also with acceleration terms for both r(t) and phi(t). Hence you must consider the centrifugal and Coriolis terms even without a rotating frame of reference. **EDIT - I should clarify that in an inertial frame using polar coordinates the terms -r phi_dot^2 and 2 r_dot phi_dot look very much like the centrifugal and Coriolis accelerations for a rotating frame, and they are called the centrifugal and Coriolis terms even in an inertial frame, but they are not the same thing. I know, it's confusing! Last edited by ChipB; Jul 16th 2016 at 07:14 AM.

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