# Escape Velocity from orbit

I am a bit confused about the escape velocity formula. I know that escape velocity is $$V_\text{escape} = \sqrt{\frac{2GM}{R_e}} .$$

My question is does this equation change when for example I want to send an object from orbiting the Earth to infinity? Because my teacher gave us a proof that escape velocity from the Earth orbit is $$V_\text{escape} = \sqrt{\frac{GM}{R_e + h}}.$$

But isn't this just the orbital velocity? If it's the same velocity as escape velocity how come the object even stays in orbit?

This is what my teacher’s proof looks like:

Sorry it's not in English.

So I want to know if his proof is right.

• Please avoid screenshots on this site. Type the text, and translate it to English. Thank you. Nov 8, 2022 at 0:15
• When $h=0$, shouldn’t the second formula become the first formula? Nov 8, 2022 at 0:23
• @Ghoster Re is the radius of the earth so the first equation is the escape velocity from the earth's surface. h is the height from the earth surface so the second equation is the distance between the satellite and earth's surface
– Luke
Nov 8, 2022 at 0:35
• When $h=0$ you are escaping from the surface. Nov 8, 2022 at 0:37
• the second equation is the distance between the satellite and earth's surface That statement makes no sense. The second equation is not a distance. It is an escape velocity from height $h$ above the Earth’s surface. Perhaps you meant “in the second equation…”. Nov 8, 2022 at 0:39

Sadly, the teacher made a mistake. In the teachers equation, the kinetic energy is $$mv^2/2$$ and the second term should be potential energy, that is, without the 2 in the denominator. The escape velocity at any distance from the center of the earth is $$\sqrt2$$ times the circular orbit velocity. The direction for escape does not matter (unless the object runs into the planet), but the so-called orbital velocity needs to be perpendicular to the radius from the center. Different angles and velocities give elliptical orbits.