# Escape velocity from a rotating body depending on latitude

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 : $$v_{\text{escape}}=\sqrt{\frac{2\,GM}{r}}\tag{1}$$

• 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. Nov 4, 2023 at 16:56