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

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

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arildno

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(R being the radius of the Earth)

The adjective is calculous by the way..

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appreciated.....

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

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

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HallsofIvy

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[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 d

[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.

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arildno

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