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Old Jul 16th 2016, 07:42 PM   #11
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Yes, I do need some diagrams. I might have to dust off my Inkscape drawing package. Anyway, I will attempt to describe it in words.

I am assuming we have two sets of xy axis. To try to avoid confusion I will call one set xy which is stationary and the other set x'y' which is rotating. Their origin coincides and one set is fixed and the other rotating. In the two body problem I am assuming that one mass is at the origin and the other is fixed onto the rotating x' axis some distance from the origin. (I should really say the axis is fixed to the moving mass, not the other way around. The moving mass can move along the x' axis otherwise I would be forcing a circular orbit. I wish to make no such restriction.) Thus the rotating axis is not rotating at constant angular velocity but coincides with the orbital position of the x'-axis. I will also choose two observers, one fixed to the stationary axis xy and I will refer to him as observer O and the other fixed to the rotating axis x'y' and I will call him observer O'

As you point out, when the orbit is exactly circular, then the gravitational force and the "centrifugal" force must be equal and opposite for an observer fixed to x'y'. For an elliptical orbit (seen by O) our observer O' sees an an oscillating force (presumably this "centrifugal" force) moving the second mass backwards and forwards a distance described by the difference between the apogee and the perigee.

In my simulation I tried to calculate this "centrifugal" force from angular momentum conservation (or a zero coriolis force) which I now realise was incorrect as these two forces are othogonal. Just thinking about it now can I compute it from the centripetal force F = mv^2/r using instantaneous tangential velocity and radius or is it more complicated than that ?

Last edited by kiwiheretic; Jul 16th 2016 at 07:44 PM.
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Old Jul 17th 2016, 10:50 AM   #12
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Conservation of momentum does not imply zero Coriolis force. What it does imply is that momentum is a constant, which means r^2 phi_dot is constant. Hence the derivative with respect to time of r^2 phi_dot =0, and thus:

d(r^2 phi_dot)/dt = 2r r_dot phi_dot + r^2 phi_double_dot =0, and finally:

2 r_dot Phi_dot + r phi_double_dot = 0.

This is the accelartion in the phi_hat direction, and it consist of two parts that cancel each other out: the Coriolis term and the angular acceleration term.

Regarding radial forces and acceleration, the equation is:

GMm/r^2 = m r_double_dot - m r phi_dot^2

That second term is centrifugal, and is equivalent to the mv^2/r that you asked about (assuming v is tangential velocity); but note that you can't ignore the radial acceleration term r_double_dot.
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Old Jul 17th 2016, 04:59 PM   #13
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Originally Posted by ChipB View Post
GMm/r^2 = m r_double_dot - m r phi_dot^2

That second term is centrifugal, and is equivalent to the mv^2/r that you asked about (assuming v is tangential velocity); but note that you can't ignore the radial acceleration term r_double_dot.
Tangential to the trajectory curve or perpendicular to the normal force? I'm trying remember if mv^2/r is strictly circular so that I would have to perform a vector decomposition into perpendicular and normal forces. (I may have to brush up on my vector calculus.)
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Old Jul 17th 2016, 07:48 PM   #14
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Originally Posted by kiwiheretic View Post
Tangential to the trajectory curve or perpendicular to the normal force?
I was picking up on your description of v - I assume what you meant s v is the component of angular velocity, meaning orthogonal to radial velocity.

Originally Posted by kiwiheretic View Post
I'm trying remember if mv^2/r is strictly circular so that I would have to perform a vector decomposition into perpendicular and normal forces. (I may have to brush up on my vector calculus.)
Yes, it's "circular," in the sense that the direction of v here is orthogonal to the radial vector. In polar coordinates there are two directional unit vectors: r_hat (which is radial) and phi_hat which is sometimes called "tangential." They are orthogonal to each other, much as in Cartesian coordinates x_hat and y_hat (sometimes written as i_hat and j_hat) are orthogonal. So yes, the object's'motion must be broken into these two directional components.

I suggest you go back and review post #2 in this thread - if there is something not clear in that please ask.
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Old Jul 23rd 2016, 11:23 PM   #15
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Ok, I think I sort of got it.

G M / r^2 = r_dot_dot - r phi_dot ^2

I worked thru the circular case as suggested with:

G M / r^2 = - r phi_dot ^2

I used L = r phi_dot and substituted:

G M / r^2 = - L ^2/r
=> GM = -r * L^2
=> G*M*t = -L^2 * Integral(r, dt)

Which looks reminiscent of Kepler's 2nd law so the result looks plausible.

Now for the full formula:

G M / r^2 = r_dot_dot - r phi_dot ^2
=> G M / r^2 = r_dot_dot - L ^2/r
=> r_dot_dot = G M / r^2 + L^2/r

This looks like a differential equation that I may be able to solve with separation of uvariables. I'll throw that into wolframalpha.com when I get a chance. Does that look like Im on the right track?

The subsequent plan is to then substitute r = L/phi_dot into the solution of the differential equation to get phi in terms of t and L. Then I have 2 parametric equations (in t) for the curve of the orbit.

This feels like a very brute force method though. Is there a simpler way of looking at this?

Last edited by kiwiheretic; Jul 24th 2016 at 11:24 AM.
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Old Jul 24th 2016, 03:46 PM   #16
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OK, I figured if I substituted the other way around it works a little better. I substituted L = r phi_dot into the equation for r instead of phi_dot so now I have . However I am still left with a quadratic in phi_dot. I am kind of at the limit of my knowledge of differential equations here.

Are there any other simplifying assumptions I can make?

Last edited by kiwiheretic; Jul 24th 2016 at 04:10 PM.
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