Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page NEWTON Teachers Visit Our Archives Ask A Question How To Ask A Question Question of the Week Our Expert Scientists Volunteer at NEWTON! Frequently Asked Questions Referencing NEWTON About NEWTON About Ask A Scientist Education At Argonne Escape Velocity
Name: Carl W.
Status: N/A
Age: 40s
Location: N/A
Country: N/A
Date: 2001-2002

I am a parent and was discussing escape velocity with my sons. The discussion turned to the fact that you would in theory never get out of a planets gravitational field (assuming it diminishes with distance but does not vanish). This being the case, if an object was launched from the surface at a velocity, it would immediately begin to slow down and given enough time would reach zero velocity and at that point begin to fall back to the object and would not in fact escape. Is there something wrong with this argument? or is escape velocity more complicated than simply looking at one (actually both) objects gravity?

You are absolutely right in saying that the earth's gravitational field diminishes with distance, but does not vanish, and in saying that an object launched from the earth with a given velocity will slow down due to the earth's gravitational field.

However, given a high enough velocity, the object will continue to increase its distance from the earth indefinitely. The velocity which will just allow the object to go to infinity (in principle) is called the escape velocity. It can be calculated by finding the velocity which gives the object a kinetic energy just equal in magnitude to its (negative) potential energy at the earth's surface. This makes the total energy (kinetic plus potential) just equal to zero. Then when the object is infinitely far away when its potential energy and kinetic energy are both just about zero and so the total energy is still zero. Energy is conserved!

The velocity needed to escape is just about 11.2 km/sec or 7 miles/sec.

Best, Dick Plano...

You have the argument exactly right. Escape velocity is the speed that just barely gets you to infinity. Of course it would take an infinite time to get there -- not only because infinity is, uh, infinitely far away, but also because you're supposed to arrive there with zero residual speed.

Tim Mooney


Although, you do in fact always feel a little gravity from a planet, the planet will not necessarily be able to make you turn around. Since you are always moving further away, gravity gets weaker and weaker. Every second, the gravity force takes away less velocity. So long each decrease is smaller than the previous one, it is possible to never bring the rocket to a halt. One interesting example is 1 - 1/2 - 1/4 - 1/8 - 1/16 - 1/32 - .... With every term, more is removed from the original value of one. Still, every term takes away only half of what is left. The value will never actually equal zero. If the first term were made less than one, the value would reach zero. If the first term were two, the value would never reach one. This is similar to how escape velocity works. If the initial velocity is fast enough, gravity will keep causing the velocity to decrease without ever stopping the object.

Dr. Ken Mellendorf
Physics Instructor
Illinois Central College

Hi, Carl !!

In reality, there must be a velocity beyond it the body does not return to earth. Below this limit velocity, the body goes up, stops and returns to the earth. In this case, all the energy that the body has is kinetic energy, that is : E1 = (1/2)m.v^2 All this energy is converted in potential energy, or : E2 = m.g.h If you put E1 = E2, than you have : v = SQRT ( 2.g.h ). Through this expression, you determine the velocity of a body to reach a distance "h". In this case, it was employed the force due to gravity "g", or :

F = m.g.

If you employ the expression for gravitational force between two masses M and m you have :

F = G (M.m/R^2)

than it is possible to determine the potential energy of a body :

E2 = GMm/R.

At the infinitum, the velocity must be zero for this calculation. Let us put

E1 = (1/2) m. v^2
E2 = G Mm/ R

then you get :
v = SQRT ( 2GM/R ).

This is the scape velocity, which reaches 40.000 km/h. But - anyway - let us get back to your question, where you point out that : " it would immediately begin to slow down nd given enough time would reach zero velocity and at that point begin to fall back to the object and would not in fact escape". Well - in this case - this point is infinity. And that is why you can consider that the body will NOT return. Infinity is too far away from here...

Best regards

Alcir Grohmann

Click here to return to the Physics Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (, or at Argonne's Educational Programs

Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012
Weclome To Newton

Argonne National Laboratory