# Derivation of Escape Velocity Inconsistency

by Michael King
Tags: derivation, escape, inconsistency, velocity
 P: 10 Hey, I'm a first year Astrophysics student, and in revising the Schwarzchild radius, I wanted to derive it and so I started by deriving the escape velocity from first principles, then rearrange to get the Schwarzchild 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?
 P: 145 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, thats 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...)
 P: 10 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.
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## Derivation of Escape Velocity Inconsistency

 Quote by Oberst Villa 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.
 Admin P: 21,399 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.
P: 145
 Quote by Janus 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.
 PF Patron Sci Advisor Emeritus P: 4,974 There is another way to derive the escape velocity involving calculus and Newton's second law.
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 Quote by Kurdt 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
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 Quote by rock.freak667 Really? How does one do it using that? Have only done it using kinetic energy and the gravitational potential
Try it yourself

$$F = m\frac{dv}{dt} = -\frac{GMm}{r^2}$$
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 Quote by Kurdt 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?
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 Quote by rock.freak667 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?