Why is $v=\sqrt{\frac{GM}{r}}$ not a valid equation for escape velocity?

Question

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.

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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 $(2)$?

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EDIT:

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.

Answer 1

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.

Answer 2

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

Answer 3

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.

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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.

Answer 4

[...] 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.

Answer 5

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 $$ v_e = \sqrt{\frac{2GM}{r_0}} $$

Answer 6

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.