Brenton Dixon

2022-07-22

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

edgarovhg

You had some problems of frames of reference when making your hypothesis. Say that the non rotating planet is $A$ and the other is $B$. The orbit velocity is $v$. So an object orbiting $A$ have a velocity with respect to the center of $A$ and to the surface is the same, and is ${v}_{0}=v$. For $B$, say the object's velocity with respect to $B$'s surface is ${v}_{1}$, and $B$'s surface with respect to $B$'s center(the velocity of $B$'s rotation at its equator), ${v}_{2}$. Then the object's velocity with respect to $B$'s center is ${v}_{1}+{v}_{2}$.
And since $A$ and $B$ are otherwise identical, ${v}_{0}={v}_{1}+{v}_{2}=v$.

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