Eccentricity changes of an orbit during out-of-plane burns

I am learning about Space exploration and I have the following question:

Does the eccentricity of the orbit remain unchanged during a manoeuvre that is perpendicular to the orbital plane?

• There's no need for the "Thanks!" at the end. Dec 21, 2022 at 14:39
• oops i did that a couple of times now, will keep that in mind for the future! Feb 12, 2023 at 23:10

An out of plane maneuver definitely changes where the eccentricity vector points. It might or might not change the magnitude of that vector.

Denoting $$\hat r$$ as the unit vector directed along the radial vector, $$\hat z$$ as the unit vector directed along the angular momentum vector, and $$\hat\theta$$ as $$\hat z \times \hat r$$, an impulsive out of plane maneuver results in a change in velocity of $$\Delta v \hat z$$. This changes the $$\hat r$$ and $$\hat z$$ components of the eccentricity vector but leaves the $$\hat \theta$$ component unchanged:

$$\Delta \vec e = \frac{-r v_r \Delta v\,\hat r + r\Delta v^2 \hat z}{\mu}$$

The change in eccentricity is thus

$$\Delta e = \sqrt{e^2 + 2\vec e \cdot \Delta \vec e + ||\Delta \vec e||^2} - e$$

This can be positive, negative, or zero.

• Thank you very much! I may have not completely understood it. So can the change in eccentricity be positive, negative or zero for a maneuvre that is perpendicular to the orbital plane? Feb 12, 2023 at 16:09