Recent content by Trying

@ehild and gDavidov yes! I will never ignore that C again. Fixing that solved all my problems. Thanks ehild for such an in-dept explanation. And thanks everyone else for your input. I'll definitely frequent this forums from now on.

Thank you gDavidov, I will re-evaluate from there and see what troubles I run into next!

gDavidov, I agree. But I just skipped showing a few steps. v^2/2 = GM/(D-x) + C1 I just ignored C for the moment as it's not important I solve for it here: v = √(2GM/(D-x)) + C2 plugging in initial conditions, x=0, v=0, C2 = -√(2GM/D) And to D H I've been trying to make sense of what you said...

Ok I take back what I said before. I'm stuck. v^2/2 = GM/(D-x) + C so v = √(2GM/(D-x)) - √(2GM/D) v = dx/dt= √(2GM)(√D-√(D-x))/(√(D-x)√D) dx(√(D-x)√D)/)(√D-√(D-x)) = √(2GM)dt TI-89'd and walframalpha'd the left side to get: D[-√DLn(x) + 2√DLn(√(D-x)-√D) + 2√(D-x)] + D^(3/2)Ln(x)...

Thanks a lot ehild for your help. It's good to know I was atleast using the right approach. I can fix the rest now. Thanks again!

Ok. I'm not sure if this belongs in this section. This is a problem relying on physics 1 concepts, but I do believe it requires MultiV Calc (which I haven't taken) to solve. 1. The problem How long would it take a particle of distance D away from the center of the Earth to fall to the surface...