A gravity calculusy kind of quesiton

  • #1
How would i go about writing an equation for the velocity of an object released at a high altitude above the earth which takes into account the sqaure increase in acceleration with displacement (s^2).

I posted it here because i'm aware it's more of a calculus question than physics.
 

Answers and Replies

  • #2
arildno
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Hmm, if s is your displacement from the SURFACE, then R+s will be your distance from the centre of Earth, agreed?
(R being the radius of the Earth)
The adjective is calculous by the way..
 
  • #3
appreciated.....

p.s. "The adjective is calculous by the way.." - :tongue: i know
 
  • #4
HallsofIvy
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Gravitational force is
[tex]\frac{-GMm}{r^2}[/tex]
where r is the distance from the center of the earth. If your s is height above the surface of the earth, M is the mass of the earth and R is the radius of the earth, then
[tex]\frac{d^2s}{dt^2}= \frac{-GM}{(R+s)^2}[/tex]

That's for something falling straight down, not an orbit, of course.
The best way to solve that differential equation is to let v= ds/dt, then not that d2s/dt2= dv/dt= (ds/dt)(dv/ds) (chain rule) = vdv/ds
[tex]v\frac{dv}{ds}= -\frac{GM}{(R+s)^2}[/tex]
so
[tex]vdv= -\frac{GMds}{(R+s)^2}[/tex]
That should be easy to integrate.
 
  • #5
damn it, i had a feeling it was going to be something that simple. Thanks a lot, you've put my mind at rest.
 
  • #6
arildno
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Which could also be gained with somewhat fewer contortions by multiplying the 2.order diff.ex with ds/dt, and integrate wrt. time.
 

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