Finding latitude of landing of projectile in Rotating Earth Assuming that the Earth is a uniform sphere of radius $R$, rotating about its axis with a uniform angular velocity $\omega$. A rocket is launched from the Equator in a direction due North. If it keeps on flying at a uniform speed $v$ (neglecting air resistance), the highest latitude that can be achieved will be what? This question came in my ming while studying coriolis force from Marion Thornton book. Every problem takls about deflection but none talks about final latitude. Someone told me formula to find final latitude is \begin{equation}
\frac{\pi}{2}-(\pi-2) \frac{\omega R}{v}
\end{equation}
But I am not sure and I don't even know how to derive this if angular velocity is put 0, the projectile will reach north pole according to this formula so it does make some sense.
 A: It's not too difficult to work out the problem in the general case. The highest latitude reached will just be the inclination of the orbit (or its supplementary angle if that's greater than $90^\circ$). If we consider an inertial frame (not rotating with the Earth), spherical trigonometry gives the relationship $\cos i = \cos \phi \sin \beta$. Here $i$ is the inclination of the orbit, $\phi$ is the latitude of the launch, and $\beta$ is the launch azimuth. The geometry is summarised in the diagram below.

We can relate $\beta$ to $\beta_{rot}$, the launch azimuth in a frame rotating with the Earth through regular trigonometry.
$$\tan \beta_{rot} = \frac{v_{orbit} \sin \beta - \omega R \cos \phi}{v_{orbit} \cos \beta}$$
In your case launch is due North from the equator, so $\beta_{rot} = 0$ and $\phi = 0$. Therefore,
\begin{equation}
\sin \beta = \frac{\omega R}{v} .
\end{equation}
This leads to
$$ i = \arccos \frac{\omega R}{v} .$$
A: Sometimes when reading how a problem is stated, you have to reconstruct one or more implicit features of the case.
With 'implicit features' I mean features that the problem author has in mind, but that are not mentioned explicitly.
The problem statement states the case as the rocket keeps on flying at uniform speed.
The only way to keep flying at uniform speed is by being in orbit.
I infer that an implicit feature of this case is that the rocket is not only launched, but that subsequently the rocket is inserted in circular orbit. The rocket becomes a satellite.
Then it's a matter of what the orientation is of the orbital plane of that satellite, and what the period of revolution is. That orientation may be perpendicular to the Earth's equatorial plane, or it may be at some angle to the Earth's equatorial plane.
So: when viewed as a problem about a rocket that is inserted in Earth orbit: the rocket doesn't come down again; it's in orbit.


You could choose to switch to a version of the problem slightly different from the above.
You could choose to try and find expressions for the case of a rocket that is not inserted into orbit. In that case the velocity is non-uniform. As the rocket climbs to its point of highest altitude the velocity component that is parallel to the direction of gravity is slowed down by gravity. Past the point of highest altitude gravity will increase the velocity again.
Handling that case requires knowing the angle with respect to the local horizontal of the initial launch angle, and the initial velocity.
So if you want to set up expressions for that version of the problem your expression must include launch angle, launch velocity, and the Earth's gravitational acceleration. (And if the rocket climbs to an altitude that is a significant percentage of the Earth's radius then you have to take into account that gravity is diminishes rapidly with distance; gravity is inversely proportional to the square of the distance to the center of gravitatinal attraction.)
