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I know that for velocity conversions between ECI and ECEF there is an $\omega \times r_{ECI} $ term, such that the overall transformation is $v_{ECEF}=v_{ECI}-\omega \times r_{ECI}$. In my belief there shouldn't be an extra term in the acceleration and forces conversions. For example, for a rocket at firing with a certain thrust force in the ECI frame $F_{T,ECI}$, the equivalent ecef force will be $F_{T,ECEF}=\mathbb{C}_{ECI}^{ECEF} F_{T,ECI}$, where $\mathbb{C}$ indicates the DCM from ECEF to ECI. The same would be true for acceleration, hence, for thrust acceleration vector: $w_{ECI}=\mathbb{C}_{ECI}^{ECEF} w_{ECEF}$ and vice versa if you wanted to go from ECI to ECEF. This should also be true for position. Can someone verify this?

EDIT: By the way my question just assumes rotational motion about the z axis; so no polar motion, nutation etc.

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"Real" forces are the same in all Newtonian frames. Gravity is a real force in Newtonian mechanics. The best thing to do is to calculate gravitational acceleration from the perspective of an ECEF frame, and that can include nutation and even polar motion. Transform that acceleration to ECI, where you don't need to concern yourself with fictitious accelerations such as centrifugal and Coriolis accelerations, and integrate.

Earth centered inertial (ECI) is a frame that is not rotating with respect to some ideal inertial frame. It is not an inertial frame. The Earth is accelerating toward the Moon, the Sun, Venus, Jupiter, etc. due to gravity. You will need to concern yourself with what are called "third body effects" if you want to be even more accurate. Third body effects reflect the fact that ECI is not an inertial frame of reference. There is a boatload of literature on how to calculate these third body effects.

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    $\begingroup$ Just to expand on something which may be unclear to the OP from this answer: the OP wrote, "The same would be true for acceleration, hence, for thrust acceleration vector: $w_{ECI}=\mathbb{C}_{ECI}^{ECEF} w_{ECEF}$". As you say, this would be true for acceleration due to real forces, i.e., thrust and gravity. However, this is not true for the total acceleration, because the total acceleration in ECEF includes terms due to centrifugal and Coriolis forces, which do not exist in ECI. $\endgroup$
    – Litho
    Jun 1, 2022 at 9:20

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