Originally posted by mtthw
Using the energy argument, you start from:
mv^2/2 - GMm/r = -GMm/R
Which gives...I think!...
int{r, R} sqrt[r/(1 - r/R)] dr = -t sqrt(2GM)
Not a nice integral to do by hand, though.
using trigonometric substitution I believe I have it here is how, and what I arrive at...if someone will check for correctness to this point I would appriciate it.
Int{r,R}sqrt[r/(1-r/R)]
basic triangle
Hyp: 1
OPP: sqrt( r/R )
ADJ: sqrt ( 1-r/R )
let the angle be represented by "A"
such that the new integral becomes:
sqrt (R) Int{ tanA dr }
for the differential dr:
sinA = sqrt(r/R)
r = R*(sinA)^2
dr = 2R*sinA*cosA dA
plug it in , and simplify.
2*R^(3/2)*Int{ (sinA)^2 dA } = R^(3/2)*Int{ (1 - cos(2A))dA}
= R^(3/2)*[ A - 1/2*sin(2A)]
then changing back to original variable r
= R^(3/2)*[ arcsin(sqrt(r/R)) - 1/2*sin(2 arcsin(sqrt(r/R)))]
just evaluate ( wasn't sure what limit was what) and simplify(which I have not investigated, but have a hunch will be moderately fruitful).
Eric
What the heck I did the evaluation ( assuming "r" was lower limit )
and come to
R^(3/2)*[ arcsin(sqrt(r/R)) - sqrt((r/R)(1- r/R)) - Pi/2 ] =-t sqrt ( 2GM )