# Shouldn't the escape velocity of earth (with respect to earth) be less than $\sqrt{\frac{2GM}{R}}=11.2\,\mathrm{km/s}$

We know that the escape velocity of earth is, $$\sqrt{\frac{2GM}{R}}=11.2\,\mathrm{km/s}$$

Where $G=6.67×10^-11$ $M=\text{mass of earth}$ $R=\text{radius of earth}$

So if throw a object with velocity $11.2\,\mathrm{km/s}$ it should never come back on earth. But earth itself is rotating hence the object will also have this velocity. So it's velocity is greater than $11.2\,\mathrm{km/s}$ w.r.t to space.

So if I throw an object with velocity less than $11.2\,\mathrm{km/s}$ it should still not reach earth as earth's rotational velocity will add up into it.

Therefore isn't escape velocity of earth w.r.t to earth less than $11.2\,\mathrm{km/s}$.

• Yes in a sense, and that is why rockets are generally launched to the east. But to be picky, when one speaks of escape velocity, one takes the rotation of the planet to be zero. Otherwise, it would be different at different latitudes, and inclinations of the launch with respect to the equator. – garyp Jun 2 '17 at 18:40