A gravity calculusy kind of quesiton

[RooftopDuvet]

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.

 

[arildno]

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..

 

[RooftopDuvet]

appreciated...

p.s. "The adjective is calculous by the way.." - i know

 

[HallsofIvy]

Gravitational force is
\frac{-GMm}{r^2}
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
\frac{d^2s}{dt^2}= \frac{-GM}{(R+s)^2}

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
v\frac{dv}{ds}= -\frac{GM}{(R+s)^2}
so
vdv= -\frac{GMds}{(R+s)^2}
That should be easy to integrate.

 

[RooftopDuvet]

damn it, i had a feeling it was going to be something that simple. Thanks a lot, you've put my mind at rest.

 

[arildno]

Which could also be gained with somewhat fewer contortions by multiplying the 2.order diff.ex with ds/dt, and integrate wrt. time.