Tom MS said:
I'm taking high school level AP calculus right now, but I'd be happy to see the solution! In terms of differential equations, I only know how to solve simple linear differential equations and separable differential equations.
Also, I just realized. Why couldn't I just use conservation of energy?
Yes, you can use conservation of energy, but that's not all you need. I'll show you both methods. I've got slightly different notation:
##M## is the mass of the large body, ##R## is the radius of the large body, ##r_0## is the starting position of the falling mass, ##r_1## is the final position of the falling mass, where ##r_1 \ge R##, and ##r## is the variable position of the falling mass. And, ##r'(0) = 0## meaning the falling mass starts from rest.
First, you have the differential equation for ##r##:
##r'' = -\frac{GM}{r^2}##
The first trick is to multiply both sides by the integrating factor ##2r'## to give:
##2r'r'' = -2r'\frac{GM}{r^2}##
##\frac{d}{dt}(r')^2 = \frac{d}{dt}(\frac{2GM}{r})##
##(r')^2 = \frac{2GM}{r} + C = \frac{2GM}{r} - \frac{2GM}{r_0} = 2GM(\frac{r_0 - r}{rr_0})## (using the initial position to find ##C##)
This gives us the first equation:
##(r')^2 = 2GM(\frac{r_0 - r}{rr_0})## Equation (1)
The alternative, as you suggested, is to use conservation of energy:
##-\frac{GMm}{r_0} = -\frac{GMm}{r} + \frac{1}{2}m(r')^2##
Which gives you equation (1) more easily.
The main trick is a non-obvious substitution:
Let ##r = r_0 cos^2\theta## hence ##r' = -2r_0(cos\theta sin\theta)\theta'##
Substituting this into (1) gives:
##4r_0^2(sin^2 \theta cos^2 \theta)(\theta')^2 = 2GMr_0(\frac{1 - cos^2 \theta}{r_0^2 cos^2 \theta}) = \frac{2GMsin^2 \theta}{r_0 cos^2 \theta}##
##cos^4 \theta (\theta')^2 = \frac{GM}{2r_0^3}##
##cos^2 \theta (\theta') = \sqrt{\frac{GM}{2r_0^3}}## Equation (2)
Now we integrate this with respect to ##t##, noting that at ##t = 0, r = r_0, \theta = 0## and at ##t = t_1, r = r_1, \theta = \theta_1##
##\int_0^{\theta_1} cos^2 \theta d \theta = \sqrt{\frac{GM}{2r_0^3}}t_1 + C##
##\frac{1}{4}[2\theta_1 + sin2\theta_1] = \sqrt{\frac{GM}{2r_0^3}}t_1## (as ##C = 0##)
This gives us essentially an intermediate solution for ##t_1##:
##t_1 = \sqrt{\frac{r_0^3}{8GM}}[2\theta_1 + sin2\theta_1]## Equation (3)
Where ##r_1 = r_0 cos^2 \theta_1## hence ##\theta_1 = cos^{-1} \sqrt{\frac{r_1}{r_0}}##
Now, of course, you can use trig identities to give ##sin 2\theta_1## in terms of ##r_0, r_1##:
##sin2\theta_1 = 2cos\theta_1 sin\theta_1 = 2 cos\theta_1 \sqrt{1- cos^2\theta_1} = 2 \sqrt{\frac{r_1}{r_0}} \sqrt{1- \frac{r_1}{r_0}}##
This gives you the formula that Janus posted:
##t_1 = \sqrt{\frac{r_0^3}{2GM}}[cos^{-1} \sqrt{\frac{r_1}{r_0}} + \sqrt{\frac{r_1}{r_0}(1- \frac{r_1}{r_0})}]## Equation (4)
Although, this is actually a more general formula for time to fall from any distance ##r_0## to any other distance ##r_1##.
There is a special case worth mentioning. If the object falls from a long way away and you model the large mass as a point mass, then ##r_1 = 0## and you get a good estimate for the time it takes the two objects to collide. The exact equation would, of course, have ##r_1 = R##:
##t_1 = \sqrt{\frac{r_0^3}{2GM}}[\frac{\pi}{2} + 0] = \pi \sqrt{\frac{r_0^3}{8GM}}## Equation (5)
