Orbital velocity of a circular planet is $\sqrt{aR}$, where a is the centripetal acceleration, and $R$ is radius of the planet. With ${v}_{1}$ as the tangential velocity of the rotating planet at the equator.

On the non rotating body, suppose that the orbital velocity is ${v}_{0}$, and, for an object launched on the rotating body's "equator", that the orbital velocity will be in the form of ${v}_{1}+{v}_{2}$ (the body and the object both going counterclockwise). Now, I half-hypothesized ${v}_{0}={v}_{1}+{v}_{2}=\sqrt{aR}$ and ${v}_{2}=\sqrt{{a}^{\prime}R}$ where ${a}^{\prime}$ is the "true" rotating body's acceleration, able to be calculated from the rotating frame of reference as

${a}^{\prime}=a-\frac{{v}_{1}^{2}}{R}$

The logic was that from rotating body's reference frame, the object would be traveling at ${v}_{2}$, less than ${v}_{0}$ because of the centrifugal force, so ${v}_{2}$ has to be the orbital velocity if the gravity was "weakened" by centrifugal force.

Tried to solve for ${a}^{\prime}$ and comparing it to the value, got from rotating frame of reference, ending up with

${a}^{\prime}=a-(\frac{{v}_{1}({v}_{0}+{v}_{2})}{R})$

Something's not right, and if I had to choose, I would guess the ${v}_{0}=v1+{v}_{2}$, that acceleration is not the same on those two planets, but I don't know how it would change, or why.