# Turn a sinusoid into a elliptic orbit

1. Sep 20, 2013

### Philosophaie

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)

Last edited: Sep 20, 2013
2. Sep 20, 2013

### Simon Bridge

The sinusoids parameterise the ellipse.
You know how this works if the plane of the orbit is the x-y 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. Sep 22, 2013

### Philosophaie

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. Sep 22, 2013

### Simon Bridge

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 two-body problem.

5. Sep 22, 2013

### Staff: Mentor

This is incorrect. $\vec x(t) = \frac 1 2 \vec a\,t^2 + \vec v_0\,t + \vec x_0$ is only valid for constant acceleration. The equation you used does not apply to an orbiting body because the acceleration of an orbiting body is not constant.

Neither one.

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

6. Sep 22, 2013

### Staff: Mentor

They are not correct.

7. Sep 23, 2013

### Philosophaie

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 non-uniform acceleration?

Last edited: Sep 23, 2013
8. Sep 23, 2013

### Simon Bridge

Start with a free-body diagram and apply Newton's Laws - applying boundary conditions.

From post #2: