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So the earth's escape velocity since the earth is rotating is dependent on the latitude. As stated in Wikipedia here: For example, as the Earth's rotational velocity is 465 m/s at the equator, a rocket launched tangentially from the Earth's equator to the east requires an initial velocity of about 10.735 km/s relative to the moving surface at the point of launch to escape whereas a rocket launched tangentially from the Earth's equator to the west requires an initial velocity of about 11.665 km/s relative to that moving surface.

So what will be the function depending on the latitude and direction in which we are firing a rocket? Considering the equation for escape velocity from Earth's surface is : $$v_{\text{escape}}=\sqrt{\frac{2\,GM}{r}}\tag{1}$$

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Yes it will, you can see it this way:

Our planet revolves from west to east. As a result, when a rocket is launched from west to east on the equatorial plane, it benefits from Earth's spinning speed. A component of the Earth's rotational speed is added to the rocket's projection speed. When launched from east to west, the rocket must overcome the rotational motion of the Earth, hence greater speed is necessary.

Hope this helps! Do correct me if I'm wrong!

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  • $\begingroup$ Hi thank you for the answer, I am looking for a function where I can input the latitude angle. For instance, if Earth's rotational velocity at an angle is $$v_r = R\omega\cos\theta$$ where $$\theta$$ is the latitude angle. $\endgroup$ Nov 4, 2023 at 16:56

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