Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have several confusions regarding escape velocity. I am sure I am missing something(s) obvious or maybe I am taught wrong.

  1. Lets say we throw an object of any mass at exactly escape velocity of earth calculated from $v^2=\frac{2GM}{r}$ which is almost $11 \text{km s}^{-1} $ but I am talking about exact escape speed. That ball initially has $KE=\frac{1}{2}mv^2$ and $PE=mgh$. Wikipedia says and I quote

    In physics, escape velocity is the speed at which the kinetic energy plus the gravitational potential energy of an object is zero.

    How is that possible?

  2. As $F=\frac{GmM}{r^2}$ no matter how much the particle travels away from earth's surface it will always be accelerated towards earth. With increasing $r$ the $F$ will decrease but it will never reach $0$. That means that there will be no point where the particle will stop and will continue to move with slower and slower speed will never reaches zero. Am I right?

  3. A request: Please explain exactly what happens to particle's $KE$ and $PE$ at different points such as at $r=0$ and at $r=\infty$.
share|cite|improve this question
Comment on the first part: $U = mgh$ is a result coming from a first order approximation of the full Newtonian gravitational potential, with which the escape velocity is calculated. Dropping the approximation gets rid of (what I think is your) problem. – ACuriousMind Sep 1 '14 at 15:41
I think this perhaps should have been split into more than one question... but since it's been answered, I've just edited the title to improve it. – David Z Sep 2 '14 at 3:11
up vote 7 down vote accepted
  1. Gravitational potential energy is usually measured as a negative value. We do this because an object that is so far away from a gravity well that it practically is unaware of it shouldn't be considered as having any potential energy. So as $r\to\infty$, $PE\to0$. As an object falls into a gravity well, it loses potential energy, so gravitational PE is a negative value. Since energy is conserved, if you can sum the KE (a positive value) and the PE (a negative value) and the result is $0$, that means there is enough KE for the projectile to reach $r\to\infty$, where $PE=0$ just as $KE=0$. At that point, the object never returns. Thus that KE defines the escape speed. Also, $KE=\frac{1}{2}mv^2$ as you wrote, but $PE=-\frac{GMm}{r}$. The equation $PE=mgh$ is only valid near Earth's surface and it represents the change in potential energy from the surface, not the total potential energy (which is why it is positive, not negative).

  2. If $KE+PE=0$ then at $r=\infty$, the speed will be zero. However, in a more practical sense, you are correct. An object at escape speed in a universe with just one gravity source and the object will never reach zero speed, it will simply move slower and slower forever.

  3. At $r=\infty$, the PE goes to zero. The KE then becomes the only contributor to the total energy. so whatever the total energy of the object is, that's its KE. $r=0$ is a complicated case. For a point source of gravity, the PE would become infinite, but (other than a black hole, which isn't covered by Newtonian mechanics), there is no such thing as a point source of gravity. In the usual case, the Force of gravity disappears at $r=0$, but since moving away from that position would still constitute a gain in PE (as you'd experience an opposing force), it can get more complicated to talk about the PE at or near $r=0$. Generally, beneath the surface of an object, the PE becomes dependent on the distribution of the mass of the object. The KE is again dependent on the total energy. But there is nothing very special at $r=0$. In fact, there are no particularly interesting points anywhere. At every point, the total energy is $E=PE+KE$, PE is a negative value that approaches 0 as $r\to\infty$, and KE is whatever is left over such that total energy is conserved.

share|cite|improve this answer
About the special case where r is close to 0, I would add that usually M is also close to 0, else you can't reach it. (it may look satisfying in the formula for calculating v) – Ghislain Bugnicourt Sep 1 '14 at 16:30
@GhislainBugnicourt M isn't always close to zero, you could be inside the gravitating body (like a dyson sphere made entirely of neutrons or just boring a hole through a planet) – Jim Sep 1 '14 at 16:54
@JanHudec Serious clarification: are you rebutting my comment or saying I should mention M goes to 0 in point 3 (agreeing with Ghislain)? For the former, OK, I agree. For the latter, I think that would make the formula for v more misleading as it would indicate it takes less speed to escape from beneath the surface – Jim Sep 1 '14 at 19:37
@Jim: If you are below the surface, the material above you cancels some of the gravity and the forces perfectly balance in the centre, so effectively $M$ goes to $0$. – Jan Hudec Sep 1 '14 at 19:37
@Jim: Well, under the surface one would have to substitute proper formula for gravity inside a body. Trying to talk about what happens without that is somewhat misleading, yes. – Jan Hudec Sep 1 '14 at 19:42

Potential energy for a point mass (and also for a sphere) is not $PE = mgh $ (this is special case for a uniform field) but rather: $$ PE = - \frac{GMm}{r}$$

where G is the gravitational constant, M and m are both masses and r is the distance between the masses. (in the case of a sphere, the distance is to the centre of the sphere)

Can you see the answer to your question yourself now?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.