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Turn a sinusoid into a elliptic orbit 
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#1
Sep2013, 11:38 AM

P: 366

I am trying to get my head around these equations. I am not sure they are correct. My logic is an orbit exists at a starting point (x0,y0,z0) with a starting velocity at time zero (vx0,vy0,vz0) changing with time (dx/dt, dy/dt, dz/dt). The gravity is (d^2x/dt^2, d^2y/dt^2, d^2z/dt^2). How do you turn a sinusoid into a elliptic orbit?
x = (1 / 2 * G * M / r ^ 2 * Cos(h) * Cos(p)) * t ^ 2 + vx0 * t + x0 y = (1 / 2 * G * M / r ^ 2 * Sin(h) * Cos(p)) * t ^ 2 + vy0 * t + y0 z = (1 / 2 * G * M / r ^ 2 * Sin(p)) * t ^ 2 + vz0 * t + z0 where r=(x^2+y^2+z^2)^0.5 h=atan(y/x) p=acos(z/r) 


#2
Sep2013, 07:38 PM

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You know how this works if the plane of the orbit is the xy plane right? However, I think you have been too general in your setup. Gravity is usually a central force  always directed to some point  with a magnitude that depends on the distance to that center. You write that down and apply Newton's Laws. There are several possible shapes  the stable ellipse is usually quite tricky to hit on by trail and error. 


#3
Sep2213, 02:11 AM

P: 366

Which is the correct postulation in Newtonian 2 Body solution:
h=atan(y/x) p=acos(z/r) or h=atan(vy/vx) p=acos(vz/vr) where vr = (vx^2+vy^2+vz^2)^0.5 


#4
Sep2213, 03:28 AM

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Turn a sinusoid into a elliptic orbit
Depends what you want h and p to represent.
The former are the angles to the position and the second to the velocity. They do not appear to represent any kind of postulates, and are not specific to the twobody problem. 


#5
Sep2213, 06:37 AM

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The correct solution was found by Kepler. Why do you persist in avoiding that solution? 


#6
Sep2213, 06:46 AM

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#7
Sep2313, 04:56 AM

P: 366

Then how do you find the equations for x,y,z, xdot, ydot, zdot, rdot thetadot, phidot, xdoubledot, ydoubledot, zdoubledot, rdoubledot, thetadoubledot, and phidoubledot with nonuniform acceleration?



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