# Calculating height achieved under changing gravitional field

1. Feb 21, 2005

### relativitydude

For some reason, this is alluding me at the moment. We know the gravitional acceleration equation is g = GM/r^2, integrate that in respect to r to yield -GM/r

I thought I could use

?Y = VoT + .5AT^2

and subsitute GM(-1/Ro + 1/R) into A

For some reason this is not working, is my line of reasoning correct?

2. Feb 21, 2005

### Staff: Mentor

Not exactly:
$$g = dv/dt = - GM/r^2$$
$$v dv/dr = - GM/r^2$$
Now you can integrate with respect to r:
$$v^2/2 = GM/r + C$$

That's only good for uniformly accelerated motion.
No. What problem are you trying to solve? The height of a projectile as a function of time? That's not so simple.

3. Feb 22, 2005

### relativitydude

Thank you very much! :)