# Derivation of Escape Velocity Inconsistency

• Michael King
I had it explained to me beforehand, but I still don't see why it is not escape velocity, as I am pretty sure you need to maintain orbital velocity in order to achieve the minimum escape velocity.In summary, the conversation discusses different methods of deriving the Schwarzschild radius, a concept from General Relativity that describes the size of a black hole. The first method involves using Newtonian Mechanics and calculating the orbital velocity needed to maintain circular orbit, while the second method uses energy equations to determine the minimum velocity needed to escape a black hole without additional thrust. There is a discrepancy in the results of the two methods, with the second method yielding a factor of \sqrt{2} more than the first. However, this is due to

#### Michael King

Hey, I'm a first year Astrophysics student, and in revising the Schwarzschild radius, I wanted to derive it and so I started by deriving the escape velocity from first principles, then rearrange to get the Schwarzschild Radius. Child's play, really.

However, depending from where you come from, either from energy or rotational motion, you end up being a factor of $$\sqrt{2}$$ out:

First Derivation, from Newtonian Mechanics

I am ignoring vector notation for speed

$$F = \frac{GMm}{r^{2}} = ma$$

$$a = \frac{GM}{r^{2}}$$

However,

$$a = \frac{v^{2}}{r}$$

Therefore,

$$\frac{v^{2}}{r} = \frac{GM}{r^{2}}$$

Rearranging for v gives

$$v = \sqrt{\frac{GM}{r}}$$

Second Derivation, from Energy

We can assume that by units,

$$E = \frac{1}{2}mv^{2} = \frac{GMm}{r}$$

$$E = \frac{1}{2}v^{2} = \frac{GM}{r}$$

$$E = v^{2} = \frac{2GM}{r}$$

$$v = \sqrt{\frac{2GM}{r}}$$

Are my mathematics skills not up to scratch, or is it that for two equally valid derivations, we get two equally valid equations, that mean the same thing, but are not equal to one another?

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I suspect that in your first derivation, you are not calculating the escape velocity, but instead the velocity needed to keep the test particle in circular orbit of radius r. The 3rd equation, that's the bad one:

$$a = \frac{v^{2}}{r}$$

This is the centrifugal acceleration of a particle moving with speed v in a circular orbit of radius r - its not related to the escape velocity.

Further, I see a more fundamental problem with your approach. You are using Newtonian Mechanics, but what you aim at is the Schwarzschild radius, which is a concept based on the General Theory of Relativity. So you have to use the equations of General Relativity right from the start (Though I have no clue how to do this, sorry...)

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ah so the first derivation was for orbital velocity, whereas the second was for escape velocity... I hate it when people don't label things correctly, or when I cannot read.

Oberst Villa said:
Further, I see a more fundamental problem with your approach. You are using Newtonian Mechanics, but what you aim at is the Schwarzschild radius, which is a concept based on the General Theory of Relativity. So you have to use the equations of General Relativity right from the start (Though I have no clue how to do this, sorry...)

Actually, you can derive the Schwarzschild radius from Newtonian escape velocity by simply setting the escape velocity to c.

In the first equation, one had a balance of force (acceleration) between gravitational force/acceleration and centripetal force/acceleration.

In the energy equation, one simply equates the change in gravitational potential energy required to go from r to infinity with the change in kinetic energy at r and assuming zero kinetic energy at infinity, i.e. one coasts to a stop at infinity.

So escape velocity is the minimum velocity to escape without additional thrust after leaving from r.

Janus said:
Actually, you can derive the Schwarzschild radius from Newtonian escape velocity by simply setting the escape velocity to c.

Oooopsie... of course you are right and I was wrong, checked it at some other sources. Though I think its really strange - I would have thought that, if there is one place in the universe where Newtonian mechanics is not applicable anymore, then this would be near a black hole ! Strange ! Anyway, thanks a lot for the correction.

There is another way to derive the escape velocity involving calculus and Newton's second law.

Kurdt said:
There is another way to derive the escape velocity involving calculus and Newton's second law.

Really? How does one do it using that? Have only done it using kinetic energy and the gravitational potential

rock.freak667 said:
Really? How does one do it using that? Have only done it using kinetic energy and the gravitational potential

Try it yourself $$F = m\frac{dv}{dt} = -\frac{GMm}{r^2}$$

Kurdt said:
Try it yourself $$F = m\frac{dv}{dt} = -\frac{GMm}{r^2}$$

hm...cancel out one of the m's and then say that $\frac{dv}{dt}=v \frac{dv}{dr}$,then integrate both sides and I'll get the same beginning as if I start with ke=gravitational pe. That's the correct way to do it?

rock.freak667 said:
hm...cancel out one of the m's and then say that $\frac{dv}{dt}=v \frac{dv}{dr}$,then integrate both sides and I'll get the same beginning as if I start with ke=gravitational pe. That's the correct way to do it?

Yeah pretty much.

still doesn't change the fact that one is a root 2 factor out compared to the other though..

Deriving the Schwarzschild radius isn't the problem, regardless of method. I am just curious as to which escape velocity to use.

I am assuming from Newtonian Mechanics, that the velocity there is actually orbital velocity (the velocity it takes to stay in orbit)

By Conservation of Energy, it is the velocity needed to escape a body of mass M, with a radius of r.

So the Newtonian is actually irrelevant, as it appears that because it does not contain that vital '2', and also from the equations used, it does not match up the logic.

Using the circular motion equation you're deriving orbital velocity not escape velocity. That is why there is a factor of root 2 in there.

thanks, I know