Derivation of Escape Velocity Inconsistency

In summary: 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
  • #1
Michael King
10
0
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 [tex]\sqrt{2}[/tex] out:

First Derivation, from Newtonian Mechanics

I am ignoring vector notation for speed

[tex]F = \frac{GMm}{r^{2}} = ma[/tex]

[tex]a = \frac{GM}{r^{2}}[/tex]

However,

[tex]a = \frac{v^{2}}{r}[/tex]

Therefore,

[tex]\frac{v^{2}}{r} = \frac{GM}{r^{2}}[/tex]

Rearranging for v gives

[tex]v = \sqrt{\frac{GM}{r}}[/tex]
Second Derivation, from Energy

We can assume that by units,

[tex]E = \frac{1}{2}mv^{2} = \frac{GMm}{r}[/tex]

[tex]E = \frac{1}{2}v^{2} = \frac{GM}{r}[/tex]

[tex]E = v^{2} = \frac{2GM}{r}[/tex]

[tex]v = \sqrt{\frac{2GM}{r}}[/tex]

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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  • #2
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:


[tex]a = \frac{v^{2}}{r}[/tex]


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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  • #3
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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 
  • #7
There is another way to derive the escape velocity involving calculus and Newton's second law.
 
  • #8
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
 
  • #9
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 :biggrin:

[tex]F = m\frac{dv}{dt} = -\frac{GMm}{r^2} [/tex]
 
  • #10
Kurdt said:
Try it yourself :biggrin:

[tex]F = m\frac{dv}{dt} = -\frac{GMm}{r^2} [/tex]

hm...cancel out one of the m's and then say that [itex]\frac{dv}{dt}=v \frac{dv}{dr}[/itex],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?
 
  • #11
rock.freak667 said:
hm...cancel out one of the m's and then say that [itex]\frac{dv}{dt}=v \frac{dv}{dr}[/itex],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.
 
  • #12
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.
 
  • #13
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.
 
  • #14
thanks, I know
 

What is escape velocity inconsistency?

Escape velocity inconsistency refers to the discrepancy between the theoretical escape velocity of an object and its actual escape velocity in real life. Theoretical escape velocity is calculated using Newton's laws of motion, while actual escape velocity takes into account factors such as air resistance and the object's shape.

How is escape velocity calculated?

Escape velocity is calculated using the equation v = √(2GM/r), where v is the escape velocity, G is the gravitational constant, M is the mass of the planet or object, and r is the distance from the center of the object. This equation assumes a perfect spherical object with no air resistance.

Why does escape velocity inconsistency occur?

Escape velocity inconsistency occurs because the theoretical calculation does not take into account real-world factors such as air resistance and the non-spherical shape of objects. These factors can affect the actual escape velocity of an object.

How does escape velocity inconsistency affect space travel?

Escape velocity inconsistency can affect space travel by making it more difficult to accurately predict the amount of fuel and speed needed to escape the gravitational pull of a planet or object. This can lead to unexpected outcomes, such as spacecrafts not being able to reach their desired destination or using more fuel than anticipated.

How can escape velocity inconsistency be accounted for?

To account for escape velocity inconsistency, scientists and engineers can use more complex equations and models that take into account real-world factors. They can also conduct experiments and simulations to better understand the effects of air resistance and non-spherical shapes on escape velocity.

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