How can we improve the accuracy of acceleration calculations?

[salamander]

Hi! I'm quite new here, and I'm not sure wheter this is to go here or in the math section, so I'll just post it here since i guess nobody really care anyway.

I've been thinking of this for a while, but I can't seem to get it right. (I'm not that good at maths but I'm learning.)

Concider dropping a rock from a rather high altitude down to the ground. Now, using Newtonian theory, find an expression for the rocks' velocity as a function of time, that includes the fact that the gravitational attraction must become greater as the distance to Earth shrinks.

Can somebody give me a hint?

I know g=MG/r^2
What confuses me is how to integrate time in this expression,
since r=r0-gt^2/2

Finally, I don't know why but i just like these guys:

 

[Palpatine]

You need to set up a differential equation relating the distance from the Earth to the acceleration and solve it.

 

[HallsofIvy]

Letting r be the distance from the center of the earth,
m\frac{d^2r}{dt^2}= \frac{-GmM}{r^2}

That's a non-linear differential equation but we can use the fact that t does not appear explicitely in it: Let v= dr/dt. Then (chain rule):
\frac{d^2r}{dt^2}= \frac{dv}{dt}= \frac{dr}{dt}\frac{dv}{dr}= v\frac{dv}{dr}
so the differential equation becomes
v\frac{dv}{dr}= \frac{-GM}{r^2} and then separate:
v dv= -GM \frac{dr}{r^2}. Integrating:
\frac{1}{2}v^2= \frac{GM}{r}+ C.

(Notice that that first integral is the same as
\frac{1}{2}v^2- GM/r= C, conservation of energy, since the first term is kinetic energy and the second potential energy.)

That is the same as v= \frac{dr}{dt}= \sqrt{2\frac{GM}{r}+ C} which is also integrable. You can use the fact that GM/r₀²= g (r₀ is the radius of the earth) to simplify.