I've been trying for days, but I just can't understand why escape velocities exist. I've searched the web and even this site, and although I've read many explanations, I haven't been able to truly understand them. Most of the explanations I've seen involve calculus; I only know very little calculus. Could somebody provide a more intuitive explanation?

Here's what I know & understand:

  • Escape velocity is the speed at which the kinetic energy equals the gravitational potential energy of an object
  • Escape velocity isn't the same as orbital velocity, a satellite never reaches escape velocity, escape velocity is only an initial velocity. And all the other common misconceptions.

What I do not understand, is why an object that has reached escape velocity will never return back to the planet it was launched from. To help you understand where I'm stuck, here's a short "proof":

  1. Our whole universe consists of only two bodies. A teapot of mass m, and a planet of mass M. M is many million times larger than m, so the gravitational force acting on the planet is trivial.
  2. We launch the teapot from the planet, giving it an initial velocity U (relative to the planet). U > escape velocity for that particular planet.
  3. Now, the only force acting on the body (teapot) is the gravitational force from the planet, resulting in a negative acceleration (relative to the planet). The force, and therefore the resulting deceleration, will be decreasing quadratically as distance increases. The negative acceleration will get very close to, but never quite reach, zero.
  4. Therefore, the speed of the teapot will never stop decreasing. The teapot will keep decelerating forever.
  5. We conclude that at some point, the speed of the teapot will reach zero. The teapot will then fall back to the planet, even though U was greater than the escape velocity.

I suspect that my implication (4) => (5) is false. Somebody please explain why, using as little Calculus as possible. Could this be similar to the Achilles and the Tortoise paradox?

Thanks in advance!

  • 1
    $\begingroup$ (5) is wrong. It takes an infinite time to reach 0 because the gravitational field is not uniform. $\endgroup$ Jun 13, 2013 at 9:16

6 Answers 6


I don't know if it will help you, but what you are missing is the basic insight of calculus if you want. This lack of understanding generated paradoxes since the time of the Greeks. See "Achilles and the tortoise" on Wikipedia.

The basic point is that you can sum an infinite number of "intervals" (real numbers) and obtain a finite result. For example if you sum $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$ you get 1 (perfectly finite). The idea is the same with the deceleration. Deceleration reduces the velocity a little bit in a certain time interval, than you are farther away, deceleration becomes smaller and reduces the velocity again but a little less then before, etc. The point is the the sum of all the small reductions of velocity is finite and if this sum is smaller than the initial velocity the velocity will never reach zero and the teapot will always be flying farther away never coming back.

For example if the initial velocity is 2 and the deceleration reduces the velocity in little steps like this $2 -\frac{1}{2} -\frac{1}{4} -\frac{1}{8} -\frac{1}{16}\cdots$ the final velocity is $2-1=1$, still positive! If it started with velocity less than $1$, the velocity would become negative and the teapot will fall back on the planet. I hope it helps your intuition, but study calculus, it is useful ;)

  • $\begingroup$ Nice answer, but it always helps to put in some paragraph breaks ;-) $\endgroup$
    – David Z
    Sep 7, 2012 at 20:56
  • 1
    $\begingroup$ 2-1-$\frac{1}{2}\cdots$ approaches zero not one. $\endgroup$
    – luksen
    Sep 7, 2012 at 22:49
  • $\begingroup$ Always my monkey-manners, I shouldn't write answers while eating bananas. Thanks to David for the edit and @lusken you are right of course, based on the previous paragraph I meant to write 2 $-\frac{1}{2}$ $-\frac{1}{4}$ ... $\endgroup$ Sep 8, 2012 at 5:13
  • $\begingroup$ Yes, that actually makes sense! The example helped a lot, I get it now! Thanks! $\endgroup$
    – Chris
    Sep 8, 2012 at 18:06

5 does not follow from 4. I'll use only the time derivative to show this. For a simple counter-example to 5, consider an exponentially decaying velocity:

$v = V_0e^{-\alpha t}$

Clearly, the speed asymptotically approaches zero. What is the acceleration?

$a = -\alpha V_0e^{-\alpha t}$

So, even though there is always a non-zero deceleration, i.e., the object decelerates forever, the speed never reaches zero.

Now consider an object with the following velocity:

$v = V_\infty + V_0e^{-\alpha t}$

Again, the object decelerates forever and yet, the speed approaches a constant, non-zero value.


Here is a beautiful mathematical explanation The Problem of Escape Velocity [1]

In short, the height at which the speed of teapot will become zero, is very very large (~ infinity). Hence the bodies which leave the earth with escape velocity don't return, even though gravity might still act on them.

[1] Calculus: An Intuitive and Physical Approach (Second Edition) By Morris Kline

  • $\begingroup$ So what you're saying is, the speed of the teapot will become zero, but it will be gazillions of kilometers away by then? After reading George's answer, Ι think this might not be the case. The teapot will never stop or return to the planet. $\endgroup$
    – Chris
    Sep 8, 2012 at 18:12
  • $\begingroup$ No, I mean that the height of teapot at which its speed becomes zero tends to infinity. Please see equation (66) in the link I gave. $\endgroup$
    – user13107
    Sep 8, 2012 at 18:17

Velocity equation of a body thrown straight up from Earth surface:

$v^2(h)=(v_o^2 – v_e^2)+v_e^2\frac{R}{h}$

$v_o$ - initial velocity, $v_e$ - escape velocity, $R$ - Earth radius, $h$ - distance from Earth center $(h>R)$

1) If $v_o<v_e$ then there is $h$ at which $v=0$:

$\large {h_{max}=R\frac{v_e^2}{v_e^2-v_o^2}}$

2) If $v_o=v_e$ then $v>0$ at any $h$:


3) If $v_o=2v_e$ then $v$ can't be less then $\sqrt3\;v_e$:


P.S. Initial velocity equation can be easily derived from the equation of energy conservation

$\large {\frac{mv_o^2}{2}-\frac{GMm}{R}=\frac{mv^2}{2}-\frac{GMm}{h}}$

and formula of escape velocity:

$\large {v_e=\sqrt{\frac{2GM}{R}}}$


I will be assuming only that you know about energy conservation and, in particular, the formula for the teapot's energy $$E= \frac{1}{2} m v^2 -\frac{ G M m}{r}$$ where $v$ is the speed and $r$ is the distance from the teapot to the centre of the Earth.

We know this is constant. Consider an arbitrarily large distance $r$; if we can solve for $v$, i.e if $v^2>0$ given the energy and masses, then the teapot can reach this distance and, with nothing to stop it, will reach greater and greater distances. Since $\frac{ G M m}{r}$ tends to zero as $r$ increases without bound the condition is that $E>0$ in order for the teapot to escape. Fixing $r=R_{earth}$ and setting $E=0$---as this is the threshold for escape--- allows us to solve for the escape velocity at the surface of the Earth.

By this reasoning you can think of $E$ as the "kinetic energy at infinity": if the teapot still has some left then it can actually reach it.


Every field has a boundary condition in which it is valid . The region of field there is always energy stored . To overcome this energy we need some external energy (may be mechanical) . So this can happen only when there is external fore acting on the body which will change the KE of the body , helping the body to go out of the field and hence some velocity is present i.e. known as escape velocity .


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