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I have (x,y,z) initial positions and (vx,vy,vz) initial velocities in a coplanar orbit with respect to the reference frame. If I change the inclination of this orbit with the angle i, how I can calculate the new (X,Y,Z) and (VX,VY,VZ) without having argument of periapsis and Longitude of ascending node (the calculation for new (X,Y,Z) is here: https://en.wikipedia.org/wiki/Orbital_elements but it needs both argument of periapsis and Longitude of ascending node).

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  • $\begingroup$ If it's a "coplanar orbit with respect to the reference frame" does that mean your initial values for z and vz are zero? The reason I ask is if it is so, then if you make your inclination non-zero, your point can be the ascending node. You just redistribute some of the vx and vy into vx, vy, vz trigonometrically with cos(i) and sin(i). If Z and VZ are not zero in your starting condition, then I don't think your question has enough information to answer yet (but I'm not 100% sure about that). $\endgroup$ – uhoh Jan 29 '19 at 0:57
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    $\begingroup$ Yes, the initial values for Z and Vz are zero. Do you mean that in the conversion formula it's enough to put omega equal zero? (I mean both Ω, ϖ = 0) $\endgroup$ – dove Jan 29 '19 at 15:16
  • $\begingroup$ Your question asks about getting the position and velocity vectors with a new inclination angle $i$. I've posted an answer, give it a try. I'll check back in about 8 hours if you have more questions. $\endgroup$ – uhoh Jan 29 '19 at 16:43
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Consider an equatorial orbit in the $xy$ plane. Then $z=0$ and $v_z=0$.

How to obtain the initial positions and velocities of an inclined orbit?

How I can calculate the new (X,Y,Z) and (VX,VY,VZ) without having argument of periapsis and Longitude of ascending node?

You'd like to tilt the orbit out of the $xy$ plane by an angle $i$ but keep the distance $r=\sqrt{x^2+y^2+z^2}$ and the speed $v_0=\sqrt{v_x^2+v_y^2}$ and the position $(x, y, z)$ the same.

So just do it with trigonometry. Make the velocity in the $\hat{z}$ direction change from zero to

$${v'}_z=v_{xy} \sin(i)$$

$$v_{xy}=\sqrt{v_x^2 + v_y^2},$$

and reduce the other two velocities accordingly;

$${v'}_x=v_x \cos(i)$$ $${v'}_y=v_y \cos(i).$$

Double check that the new speed is the same as the old speed:

$${v'}_0=\sqrt{{v'}_x^2+{v'}_y^2+{v'}_z^2}$$

$${v'}_0=\sqrt{(v_x^2+ v_y^2) \cos(i)^2 + (v_x^2 + v_y^2) \sin(i)^2}$$

$${v'}_0=\sqrt{v_x^2+ v_y^2} = v_0.$$

Done. No Keplerian elements necessary.

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    $\begingroup$ I have to say, listening to you explain anything always helps... you manage to take it down to the base elements and strip all extraneous data in a way textbooks never will. $\endgroup$ – Magic Octopus Urn Jan 31 '19 at 20:28

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