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

Question

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 $11.2\,\mathrm{km/s}$.

Answer 1

Escape velocity is defined in an inertial frame. It's the velocity required to escape a gravitational source centered at a given point. For Earth, that velocity is 11.2km/s.

However, when you talk about throwing an object, you are not typically talking about velocities in an inertial frame. You are typically referring to velocities with respect to the surface of the Earth (that you're standing on). The Earth is rotating.

If you convert these velocities into an inertial frame centered on Earth (known as ECI, by the way), you'll see that it takes a weaker toss to the east to reach the 11.2km/s in ECI, and a stronger toss to the west to reach the same 11.2km/s in ECI. The escape velocity is the same, the only difference is whether the rotation of the earth helped or hurt you.

This is why rockets are generally launched to the east. By launching them in this direction, they benefit from the rotation of the Earth and require less fuel to reach orbit (or escape velocity).