# What is the kinetic energy when an object reaches escape velocity?

Josielle Abdilla
What is the kinetic energy equal to during the escape velocity? Henceforth, what is exactly happening at the escape velocity in terms of gravity?

Staff Emeritus
2022 Award
What is the kinetic energy equal to during the escape velocity? Henceforth, what is exactly happening at the escape velocity in terms of gravity?

Escape velocity is simply the velocity required for an object to escape the gravitational pull of a body, ignoring complications like air resistance and other gravitational-influencing bodies (Sun, Moon, other planets, etc). There is no 'during'. Escape velocity is not an event.

Nothing is happening to gravity in the context of escape velocity. The object is simply traveling so fast that the gravitational acceleration of the larger body cannot decelerate the object at a fast enough rate to ever pull the object back to the surface.

davenn
Homework Helper
Escape velocity is simply the velocity required for an object to escape the gravitational pull of a body, ignoring complications like air resistance and other gravitational-influencing bodies (Sun, Moon, other planets, etc). There is no 'during'. Escape velocity is not an event.

Nothing is happening to gravity in the context of escape velocity. The object is simply traveling so fast that the gravitational acceleration of the larger body cannot decelerate the object at a fast enough rate to ever pull the object back to the surface.
I would add that escape velocity decreases as altitude increases.

If you have an object exactly at escape velocity at the Earth's surface, it will still be exactly at "escape velocity" when it it 100 miles up. It will be going slower. Escape velocity 100 miles up is a bit slower than at the Earth's surface

If a quick calculation serves, escape velocity is cut approximately in half once an object is at 12,000 miles altitude (16,000 miles from the center of the Earth). Escape energy, aka potential energy deficit, scales inversely with distance from the center of the gravitating body. Quadruple the distance and you've cut energy by 1/4. So you've cut escape velocity by 1/2. The object will be moving half as fast as it was at the surface. And it will still be at escape velocity.

If it were to reach infinity without interference, an object that started at escape velocity at the Earth's surface would be at a dead stop. [Which is still equal to escape velocity because it would have already escaped]

Last edited:
Staff Emeritus
Gold Member
The KE of an mass at escape velocity is such that when its added to the gravitational potential energy of the mass, the result is zero.
In other words:
$$\frac{mv^2}{2} - \frac{GMm}{r} = 0$$

Thus:
$$\frac{mv^2}{2} = \frac{GMm}{r}$$
$$v^2 = \frac{2GM}{r}$$

$$\frac{mv^2}{2} - \frac{GMm}{r} = 0$$
$$\frac{mv^2}{2} = \frac{GMm}{r}$$
$$v^2 = \frac{2GM}{r}$$