Firstly, I know that the equation for the escape velocity is $$v_{\text{escape}}=\sqrt{\frac{2\,GM}{r}}\tag{1}$$ and understand it's derivation.

The following is such a simple derivation; for a test body of mass $m$ in orbit with a massive body (assumed to be spherical) with mass $M$ and separation $r$ between the two body's centres. Equating the centripetal force to the gravitational force yields;

$$\frac{mv^2}{r}=\frac{GMm}{r^2}\tag{2}$$ which on simplificaton, gives $$v=\sqrt{\frac{GM}{r}}\tag{3}$$

What I would like to know is why eqn $(3)$ is not a valid escape velocity equation?

Or, put in another way, mathematically, the derivation in $(2)$ seems sound; yet it is out by a factor of $\sqrt{2}$. What is 'missing' from the derivation $(2)$?


As I mentioned in the comment below, just to be clear, I understand that equation $(3)$ will give the velocity required for a bound circular orbit. But to escape it should follow that the test mass has to move at any speed that is infinitesimally larger than $\sqrt{\frac{GM}{r}}$ such that $$v_{\text{escape from orbit}}\gt\sqrt{\frac{\,GM}{r}}$$

So in other words eqn $(3)$ gives the smallest possible speed for a bound circular orbit. I referred to this as the 'escape speed'; since speeds larger than this will lead to a non-circular orbit, and larger still will lead to an escape from the elliptical orbit.

So my final question is; do the formulas $(1)$ and $(3)$ actually give the highest possible speed not to escape orbit rather than the 'escape speed' itself?

Thank you to all those that contributed these answers.

  • $\begingroup$ you missed the energy required to take the test mass there $\endgroup$ – Rishi Kakkar Jan 28 '18 at 8:24
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    $\begingroup$ If you equate the forces, you will get orbital velocity. $\endgroup$ – Andrei Geanta Jan 28 '18 at 8:27
  • $\begingroup$ @RishiKakkar As I said already at the very beginning; I understand the derivation of $(1)$? What I am asking here is why is $(3)$ wrong. Does it seem that $\frac{mv^2}{r}\ne\frac{GMm}{r^2}$? $\endgroup$ – BLAZE Jan 28 '18 at 8:28
  • $\begingroup$ If escape velocity is what you want to find, then the forces are not equal because if they were, the test body will just orbit the massive body. The escape velocity is the minimum speed that the test body must have in order to escape the gravitational atraction of the massive body. $\endgroup$ – Andrei Geanta Jan 28 '18 at 8:33
  • $\begingroup$ @Andrei Okay then, perhaps I should have worded this a bit more precisely. If I change $v=\sqrt{\frac{GM}{r}}\longrightarrow v \gt \sqrt{\frac{GM}{r}}$ then equation $(3)$ will tell me which speed the body must move faster than to escape; is that better? $\endgroup$ – BLAZE Jan 28 '18 at 8:38

This answer addresses the edit made in the question.

Your question boils down to (correct me if wrong): if (1) $v_{escape} =\sqrt{2Gm/r}$ is escape velocity and (3) $v_{circular\;orbit} =\sqrt{Gm/r}$ is orbital velocity, then what would for example $v=\sqrt{1.5Gm/r}$ be? Or in other words, what happens with a speed higher than $v_{circular\;orbit}$ but lower than $v_{escape}$?

The answer is: an elliptical orbit.

(1) is derived for the orbital limit (it is assumed that the object will reach infinitely far away) and (3) is derived for a circular orbit (you used the circular centripetal force expression). A speed in between will distort the circular orbit as if escaping but then still coming back at some point completing the now non-circular, elliptical orbit.

The full range of possible speeds is:

  • $v=0$: No orbit (vertical fall).
  • $0<v<v_{circular\;orbit}$: "Vertical" ellipse
  • $v=v_{circular\;orbit}$: Circle
  • $v_{circular\;orbit}<v<v_{escape}$: "Horizontal" ellipse
  • $v=v_{escape}$: Orbital limit
  • $v_{escape}<v$: No orbit
  • $\begingroup$ Unless you are providing different answers to the same question, I think you should only provide one answer to the whole question, otherwise you are attracting the same votes for each part of the question. $\endgroup$ – sammy gerbil Jan 29 '18 at 11:26
  • $\begingroup$ @sammygerbil Agreed, but I see these two answer as different. $\endgroup$ – Steeven Jan 29 '18 at 11:47
  • $\begingroup$ They are different answers to different parts of the question, not different answers to the same parts of the question. In Minimum velocity required for Loop de Loop (Centripetal Force) there were 2 parts to the question (as there is here), but you only provided one answer. $\endgroup$ – sammy gerbil Jan 29 '18 at 11:57
  • $\begingroup$ @Steeven Okay, this answer is getting closer to what I would like to know. What do you mean by "Limit of orbits" in the second to last bullet? $\endgroup$ – BLAZE Jan 30 '18 at 5:22
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    $\begingroup$ @sammygerbil Since multiple answers is allowed on this site, nothing is wrong it two answers. How consistent I choose to be depends on how I woke up that morning $\endgroup$ – Steeven Jan 30 '18 at 14:15

First things first, in Newtonian mechanics, when an object travels around its host (e.g. a planet to a star), it follows an orbit that is a conic section: either an ellipse, parabola, or hyperbola. A circle is a special case of an ellipse. An ellipse is a bound orbit while the other two are unbound.

Your derivation assumes a circular orbit. However a perturbed circular orbit doesn't become hyperbolic - it becomes elliptical. In other words, if you take an object that's currently moving in a circle and get it to move a little faster, it doesn't shift to a hyperbolic orbit. It's still bound to the host.

The escape velocity is the minimum velocity needed for the object to become unbound. An object needs to move at $v \geq \sqrt{\frac{2GM}{r}}$ to be on a hyperbolic orbit.

  • $\begingroup$ Thanks for your answer; what is meant by "which is why the derivation of escape velocity calculates the work done"? 'work done' on/by what? $\endgroup$ – BLAZE Jan 28 '18 at 8:32
  • $\begingroup$ The work done on the test body by whatever launches it into space. $\endgroup$ – Andrei Geanta Jan 28 '18 at 8:36
  • $\begingroup$ I'll edit my answer to be more easily understood. $\endgroup$ – Allure Jan 28 '18 at 9:05
  • $\begingroup$ @BLAZE Does it make sense now? If not, which parts do you not understand? $\endgroup$ – Allure Jan 28 '18 at 9:13

The first equation you give, $v_{\text{escape, surface of Earth}}=\sqrt{\dfrac{2\,GM}{r_{Earth}}}$, is the escape velocity for an object which is given a certain amount of kinetic energy to escape from the gravitational attraction of a non-rotating Earth, or any other non-rotating massive body.

When a satellite is in orbit about the Earth it has some kinetic energy and also more gravitational potential energy that it had whilst on the surface of the Earth so you would expect that the escape velocity for a satellite in orbit around the Earth to be less than if the body starts from the surface.

Taking the zero of gravitational potential energy of the two body system as infinity the total energy of a satellite in circular orbit radius $r$ is

$$\dfrac 12 m v^2_{\rm orbit} - \dfrac{GMm}{r}$$

which on using your equation for the orbital velocity $v_{\rm orbit}=\sqrt{\dfrac{GM}{r}}$ gives the total energy of a satellite as $-\dfrac{GMm}{2r}$

From this orbit to make the satellite escape from the gravitational attraction of the Earth it must be given some extra kinetic energy $\dfrac12 m v^2_{\text{escape, in orbit}}$ such that

$$\dfrac12 m v^2_{{\text{escape, in orbit}}}-\dfrac{GMm}{2r}=0 \Rightarrow v_{{\text{escape, in orbit}}}= \sqrt{\dfrac{GM}{r}}$$

So your equation (3) is a valid equation for the escape velocity for a satellite in circular orbit around the Earth.

You may have noticed that satellite launching sites tend to be near the Equator?
This is because on a rotating Earth a satellite already has some kinetic energy which can contribute to the total energy which it requires to go into orbit or escape from the Earth as is explained in this article.

  • $\begingroup$ Thank you very much for this rigorous answer; there were many good answers, but in my opinion, this one is most useful to me since you made the connection with the formula $(3)$ I was questioning. Then you used it in the derivation to obtain $v_{{\text{escape, in orbit}}}= \sqrt{\dfrac{GM}{r}}$ which is what I needed to see. However, this equation gives the velocity required to be 'just' bound. To escape is it not better to write $v_{{\text{escape, in orbit}}}\gt \sqrt{\dfrac{GM}{r}}$? Kindest regards. $\endgroup$ – BLAZE Jan 28 '18 at 22:32
  • $\begingroup$ @BLAZE In your comment I am not sure what you meant in the last sentence? $\endgroup$ – Farcher Jan 28 '18 at 22:38
  • $\begingroup$ Hi, I have just added an edit to the end of my question; hopefully, this will explain it more clearly, thanks. $\endgroup$ – BLAZE Jan 28 '18 at 22:56
  • $\begingroup$ @BLAZE The quoted speed will make the satellite escape from the Earth's gravitational attraction possibly along a parabolic trajectory with the satellite "reaching" infinity with no kinetic energy and the satellite will not return to Earth. Any greater speed will result in a finite amount of kinetic energy when the satellite "reaches" infinity possibly along a hyperbolic trajectory. $\endgroup$ – Farcher Jan 29 '18 at 10:20

[...] for a test body of mass $m$ in orbit [...]

Your equation is not giving escape speed, because the object has not escaped. Escaped means that it is not "caught" in an orbit anymore. You are literally assuming that it is still in orbit.

What you are deriving is an expression for the object while it is in orbit - so you get the orbital velocity. If you want escape velocity you must do the derivation in a situation where the object has escaped.


When you equate the forces, as you did, you get a valid expression for the speed of an object in stable (circular) orbit. However, the escape velocity is the velocity needed to take an object from $r=0$ to $r = \infty$. That's why you need to use the energy equations.

The total work done for an object to escape orbit will be: $$ W = \int_{r_0}^{\infty}\frac{GMm}{r^2}dr = \frac{GMm}{r_0} $$ The kinetic energy obtained from the escape speed must thus be equal to the work done: $$ \frac 12 m v_e^2 = \frac{GMm}{r_0} $$ And so $$ v_e = \sqrt{\frac{2GM}{r_0}} $$


Note that an object at escape velocity always escapes, regardless of direction (assuming it doesn't hit the planet). Simply put, this is because it has too much energy to possibly stay in an orbit.

With that said, you calculated the speed that keeps you in a circular orbit. It should be self-evident that you can't be in a circular orbit and escaping the planet at the same time. So clearly that can't also be the escape velocity.


protected by Qmechanic Jan 28 '18 at 11:43

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