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I'm in a situation where I need to determine an orbit's inclination. At my disposal I have all of the orbit's other Keplerian elements, and I have the vector of a single point on the ellipse. Is this enough information to derive the orbit's inclination?

Right now I'm enacting a "search," in which I gradually increase the orbit's inclination until the point I have is struck by the resulting orbital plane. Needless to say, I'd much rather be able to simply calculate the inclination directly.

Thank you!

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Yes, it is possible to determine the orbital inclination from the Longitude of the Ascending Node , and a position vector that is not the Ascending node or the Descending Node, because both the ascending node and your position are points in the orbital plane, and the vectors pointing to each from the body being orbited uniquely define that plane.

There are two possible inclination values, depending on whether the object's motion is prograde or retrograde around the orbit.

Given:

Parameter Symbol
Longitude of the Ascending Node $\Omega$
Position Vector $\vec{r}$

Define $\vec{n}$ as a vector pointing towards the Ascending Node.

$$\vec{n} = (\cos \Omega, sin \Omega, 0 )$$

$\vec{H}$ is a vector that would point in the same (or opposite) direction as your specific angular momentum vector, $\vec{h}$. Its magnitude isn't important, but the line it defines is.

$$\vec{H} = \pm \vec{n} \times \vec{r}$$

Choose the positive value if the angle from the node to the position vector, measured in the plane of the orbit, along the direction of travel, is less than 180°, the negative value if it is more than 180°

Your orbital inclination is:

$$ i=\arccos {H_{z} \over \left|\vec{H}\right|}$$

Where $H_{z}$ is the z-component of the vector $\vec{H}$.

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    $\begingroup$ Ah! Thank you.. so simple, just take the cross product to get the orbit normal. I'll test it out tomorrow and let you know how it goes. $\endgroup$
    – Keegan
    Oct 10 at 0:20

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