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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?

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

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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.

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  • $\begingroup$ 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? $\endgroup$ Feb 12, 2023 at 16:09

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