Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

An object is moving on circular orbit with constant speed v. Orbit has inclination α relatively to plane A. Viewer is on the A plane in the centre of circle (well, it's projection to A).

Will the projection of object to A (moving on ellipse) have the same angular velocity as the object itself?

I'm trying to track objects on circular orbits. When orbit has zero inclination, tracking is simple: angular speed = 360°/Period. But I get errors with objects on inclined orbits (I see them only on some points like π/2 and such).

share|improve this question
    
Imagine the case when $\alpha=\frac{\pi}{2}$, then the motion would be up-down sinusoidal motion on $\mathbf{A}$, but the angle would be constant (apart from flipping from $0$ to $\pi$ every half-orbit). –  Alyosha Jan 27 '13 at 21:15
add comment

1 Answer 1

up vote 2 down vote accepted

If $\alpha=0$, the motion of the particle can is

$$x=\cos(t), y=\sin(t)$$ Imagine $\alpha$ rotates the circle clockwise in the $y-z$, so that the $x$-axis is the 'pivot'. enter image description here $$x=\cos(t), y=\sin(t)\sin(\alpha), z=\sin(t)\cos(\alpha)$$

The angle on plane $\mathbf{A}$ is $\arctan(\frac{y}{x})$, which is $\arctan(\tan(t)sin(\alpha))$.

$$\frac{d}{dt}\arctan(\tan(t)\sin(\alpha))=\frac{1}{(\tan(t)\sin(\alpha))^2+1}\frac{\sin(\alpha)}{\cos^2(t)}=\frac{\sin(\alpha)}{\sin(\alpha)^2 \sin^2(t)+\cos^2(t)}$$ Which obviously depends on $\alpha$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.