# Why is the "escape velocity formula" the "escape velocity"?

1. Jan 11, 2016

### victorhugo

I'm doing my HSC physics course and I was playing around with the algebra in space unit and came upon something that confused me (note this isn't required to know by the syllabus except knowing that escape velocity is the velocity to escape from a planet's gravitational field)

Ok, so from the Law of Conservation of Energy we can agree that in any situation: Work done = Kinetic energy = Potential energy.
Ek=W=Ep
To get our escape velocity I imagined a situation where we ignore all air resistance and assumed that the acceleration due to gravity is the same from the surface of a planet down to its centre:
Ep at the surface of the planet = Work done to lift it from the centre = Amount of kinetic energy at its peak velocity.

(Where: mE = mass of Earth, mO= mass of object, rE = radius of Earth)

Since W=Ep, by expanding the formula for work: (mgr = ( mO [GmE/rE^2] rE )
We can simplify it to give:
G mE mO / rE = Ep

since Ek=Ep
( 1/2 mO v^2 ) = ( G mE mO / rE )

Knowing this we can solve for velocity:

v^2 = 2G mE / rE
this is the escape velocity formula.

According to the maths, this should be the highest velocity any object will acquire if dropped from the surface of Earth and accelerated towards the very centre, ignoring air resistance and any change in the value of g.
Therefore, this should also be the velocity required for an object at the centre of Earth to be initially launched at to reach the surface of Earth. Now is where I am "confused": How is this also the velocity required to escape from Earths gravitational field and fly into space without falling back down? How do we know this?

2. Jan 11, 2016

### Orodruin

Staff Emeritus
Escape velocity is generally referring to the velocity required to escape from the surface of an object, not the center. You can also define the escape velocity for different radii and you will then generally have different results depending on the radius.

The escape velocity is the same as the velocity obtained by an object falling from infinity (with initial velocity zero) because Newton's equations are time reversible, i.e., if you change the direction of time, the scenarios transform into each other.

3. Jan 11, 2016

### victorhugo

What I am confused about is why is this the escape velocity... What makes it so? As you can see, the thought experiment of an object falling towards the centre gives the same formula as the escape velocity, and so I believe there is a relationship there.

4. Jan 11, 2016

### PeroK

If you release an object from a radius $R$ it reaches the escape velocity (from radius $R$) when it has fallen to a radius of $R/2$.

5. Jan 11, 2016

### equi_librium

Another way to look at it is that although mE is constant, the mass that you're using to calulate Ek is only valid at the very surface of the earth. As an object falls from the surface of the earth to the center, the effective mass which contributes to the acceleration is decreasing as the object approaches the center (the mass of the earth above it no longer contributes to the force). In other words F due to gravity is zero at the very center of the earth. The mass used to calculate the force due to gravity cannot be assumed to be a point mass in your model. (see Orodruin above)

6. Jan 11, 2016

### Staff: Mentor

It's a definition. If you want to figure out the speed of an object dropped from one height when it reaches another, you pick certain radii. If you pick infinity as one radii, that's "escape", hence, "escape velocity".

7. Jan 11, 2016

### PeroK

You've also made the mistake of assuming that the gravitational potential at the centre of the Earth is 0. It should be a worthwhile exercise for you to calculate what it actually is. Or, at least, to explain why it is not 0.

An object falling to the centre of the (solid) Earth does not, in fact, reach the surface escape velocity.

Last edited: Jan 11, 2016