Escape velocity differential equation

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In a lot of calculus texts, there's an example of an application of differential equations involving escape velocity. They write acceleration is dv/dt but then write it as [tex]\frac{dv}{ds}\frac{ds}{dt}[/tex]

That just looks like an application of the chain rule but what is the meaning of dv/ds? inverted time? How would you use it in a related rates problem?
 

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  • #2
tiny-tim
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hi autodidude! :smile:

s is distance (and so v = ds/dt)

so dv/ds ds/dt = v dv/ds = d(1/2 v2)/ds

this is particularly useful when your data give you v as a function of s (rather than t)

and it's the reason why 1/2 mv2 keeps appearing! :wink:
 
  • #3
Bacle2
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hi autodidude! :smile:

s is distance (and so v = ds/dt)

so dv/ds ds/dt = v dv/ds = d(1/2 v2)/ds

this is particularly useful when your data give you v as a function of s (rather than t)

and it's the reason why 1/2 mv2 keeps appearing! :wink:
Is it just random that momentum=mv =d/dv(1/2mv2)?
 
  • #4
tiny-tim
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Hi Bacle2! :smile:

No …

i] if we assume conservation of energy and galilean relativity, then we can easily porve conservation of momentum

ii] if acceleration (or force) depends only on position, then md2s/dt2 = mv dv/ds = d/ds (1/2 mv2) = f(s),

so ∫ f(s) ds = ∆(1/2 mv2) which is why 1/2 mv2 keeps appearing :wink:
 
  • #5
Bacle2
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Ah, nice, Tiny Tim, thanks.
 
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Hmm...I'm still not sure on the physical meaning of dv/ds...velocity as a function of distance just doesn't make much sense to me physically but mathematically..
 
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tiny-tim
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are you any happier about d(v2)/ds ?
 
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No, unfortunately >.<
 
  • #9
tiny-tim
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multiply d(v2)/ds by 1/2 m and you get d(1/2 mv2)/ds

physically, that tells you how kinetic energy depends on position

if KE + PE is constant, it also tells you how PE depends on position (multiplied by minus-one)!

we very often know what either PE or KE is as a function of position, so differentiating wrt position makes quite a lot of sense :wink:
 
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^ Got it now, thanks very much tiny-tim!
 
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Sorry I just wanted to revisit this cause I'm up to a point in my math class where we're dealing with variable forces with weird functions like force as a function of time, as a function position, as a function of velocity and no physical motivation or meaning is given (I'm not even sure if all the units/dimensions add up). It's all formula manipulation and calculus 'tricks'

With [tex]a=\frac{dv}{dx}\frac{dx}{dt}[/tex], where did the physical motivation arise to write it like that? I saw in one textbook the author just multiplied dv/dx by dx/dx and moved it around.

What does it mean that force is the derivative of kinetic energy with respect to position? Could you please give me an example of when kinetic energy would depend on position without bringing in potential energy (or are they intimately tied in this respect/)
 
  • #12
tiny-tim
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What does it mean that force is the derivative of kinetic energy with respect to position? Could you please give me an example of when kinetic energy would depend on position without bringing in potential energy (or are they intimately tied in this respect/)
kinetic energy is always the integral of force "dot" displacement, ie mass times the integral of acceleration "dot" displacement

for a conservative force, the work done does not depend on the path taken, and so the integral depends only on the end-points: it is the final kinetic energy minus the initial kinetic energy …

ie, if the force is conservative, the kinetic energy depends only on position

(and so in that case only we can talk about potential energy

potential energy is defined as minus the work done by a conservative force)​
Sorry I just wanted to revisit this cause I'm up to a point in my math class where we're dealing with variable forces with weird functions like force as a function of time, as a function position, as a function of velocity and no physical motivation or meaning is given (I'm not even sure if all the units/dimensions add up). It's all formula manipulation and calculus 'tricks'

With [tex]a=\frac{dv}{dx}\frac{dx}{dt}[/tex], where did the physical motivation arise to write it like that? I saw in one textbook the author just multiplied dv/dx by dx/dx and moved it around.
written like that, there isn't really any physical motivation

written as a = vdv/dx, which is the same as 1/2 d(v2)/dx, it's the derivative of kinetic energy wrt position

can you give an actual example in which you can't see the motivation? :smile:
 
  • #13
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Thank you for the patience and the reply!


kinetic energy is always the integral of force "dot" displacement, ie mass times the integral of acceleration "dot" displacement

for a conservative force, the work done does not depend on the path taken, and so the integral depends only on the end-points: it is the final kinetic energy minus the initial kinetic energy …

ie, if the force is conservative, the kinetic energy depends only on position

(and so in that case only we can talk about potential energy

potential energy is defined as minus the work done by a conservative force)​
I think I need to learn more physics!

written like that, there isn't really any physical motivation

written as a = vdv/dx, which is the same as 1/2 d(v2)/dx, it's the derivative of kinetic energy wrt position

can you give an actual example in which you can't see the motivation? :smile:
So did a=dv/dx dx/dt originally come out of the chain rule manipulation? Or did it come out just as a way to show the relation to kinetic w.r.t to position? Again, I feel I need to hit the physics books to understand this better :p

--
e.g. where I don't understand the physics:

A mass of 5kg moves at a velocity v, initially it's moving at 10m/s and is 12m from the origin. We are given the force acting on a the particle as functions of time, velocity and position: F=-12t, F=-12v and F=-12x and are asked to find position as a function of time.

What exactly is the meaning of 12? Like if I had x(t)=30t, then I know that 30 is the velocity but with these weird force ones, they make no sense. Like F=-12t, the units of -12 is [tex]\frac{[kg][m]}{[s^3]}[/tex] (if I did it correctly).

It then gets very difficult to keep track of all the units when you integrate and hard to visualize what's going on, unlike motion with constant acceleration.

I suspect all this has to do with the energy stuff as well, right?
 

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