Transformation ECI to ECEF acceleration and forces

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.

• 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. Commented Jun 1, 2022 at 9:20