So I learn earth science in school, and today, I was taught about the variation of earth's gravity according to latitude, and I've been told it's because of the combination of the fact earth isn't completely round, and the centrifugal force of the earth's rotation.
Now, I've been trying to prove the effect of centrifugal force from inertial frame of reference, specifically from the north pole, like looking at a spinning top. Additionally, I thought about 2 circular planet, one rotating like ours, and one not rotating.
Knowing that 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, I made the following calculations.
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 $v_2=\sqrt{a'R}$ where $a'$ is the "true" rotating body's acceleration, able to be calculated from the rotating frame of reference as $$a'=a-\frac{v_1^2}{R}$$ 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.
So I tried to solve for $a'$ and comparing it to the value I got from rotating frame of reference, ending up with $$a'=a - (\frac{v_1(v_0+v_2)}{R})$$ 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.
