One of the important things to remember in dealing with separable equations is that in principle, if we have r(t) then that tells us t® as well. There are some exceptions (we’ll see one below)

Here’s the full solution. I’ve enclosed it in a couple spoiler boxes if you want to reveal only part of it at a time. The first spoiler box deals with setting up the integrals; the other three deal with applying the initial conditions to get t®.

[spoiler]

First, let’s redefine our constants: let A = 2 C[sub]1[/sub] and B = 2 G M. Our differential equation is then

dr/dt = √( 2 C[sub]1[/sub] + 2 G m/r ),

which means that

dr/√( 2 C[sub]1[/sub] + 2 G m/r ) = dt

and so

∫ dr’/√( 2 C[sub]1[/sub] + 2 G m/r’ ) = t®

where the limits of the integration on the left-hand side are from r[sub]0[/sub] (the radius at t = 0) to r. Evaluating this integral then gives us t®, and we cross our fingers that we can invert it to get r(t). However, this integral has different forms depending on the sign of C[sub]1[/sub], which we’ll treat separately below. Note, however, that C[sub]1[/sub] is completely determined by r’[sub]0[/sub] (the bullet’s speed at t = 0) and r[sub]0[/sub], as shown in Post #2.[/spoiler]C[sub]1[/sub] = 0:

[spoiler]If C[sub]1[/sub] = 0, things are simple: our integral becomes

∫ √(r’ / 2 G m) dr’ = t®

and so

√2(r[sup]3/2[/sup] - r[sub]0[/sub][sup]3/2[/sup])/ (3√(G m)) = t.

This equation can be inverted to get r(t):

r(t) = (3 √(G m/2) t + r[sub]0[/sub][sup]3/2[/sup])[sup]2/3[/sup].

This case corresponds to the bullet being fired with a speed exactly equal to the escape velocity of the body the astronaut is standing on.

[/spoiler]C[sub]1[/sub] > 0: [spoiler]

For C[sub]1[/sub] > 0, the integral is

∫ 1/√( 2 C[sub]1[/sub] + 2 G m/r’ ) dr’ = t®

This integral is ugly, but can be shown to be

t® = [√(2 C[sub]1[/sub] r (2 C[sub]1[/sub] r + 2 G m)) - √(2 C[sub]1[/sub] r[sub]0[/sub] (2 C[sub]1[/sub] r[sub]0[/sub] + 2 G m)) - 2 G m sinh[sup]-1[/sup](√( C[sub]1[/sub] r/ G m)) + 2 G m sinh[sup]-1[/sup](√( C[sub]1[/sub] r[sub]0[/sub]/ G m))] /(2 C[sub]1[/sub])[sup]3/2[/sup].

“sinh[sup]-1[/sup]” here is the inverse of the hyperbolic sine function; it can also be written out in terms of logarithms. If you can invert this equation to get r(t), you’re a better mathematician than I am. This case corresponds to the bullet being fired with an initial speed greater than escape velocity.[/spoiler]C[sub]1[/sub] < 0:[spoiler]Here, we have to be careful: in this case, the integrand

∫ 1/√( 2 C[sub]1[/sub] + 2 G m/r’ ) dr’

is divergent at a particular value of r’. This particular r’ value corresponds to the highest point of the bullet’s trajectory: we’re looking in this case at a bullet fired with less than escape velocity, meaning that it’ll eventually get turned around and fall back onto the astronaut if he doesn’t get out of the way. However, if we take r to be below the apex of the bullet’s trajectory, we find that

t® = [-√(-2 C[sub]1[/sub] r (2 C[sub]1[/sub] r + 2 G m)) + √(-2 C[sub]1[/sub] r[sub]0[/sub] (2 C[sub]1[/sub] r[sub]0[/sub] + 2 G m)) + 2 G m sin[sup]-1[/sup](√( -C[sub]1[/sub] r/ G m)) - 2 G m sin[sup]-1[/sup](√( - C[sub]1[/sub] r[sub]0[/sub]/ G m))] /(2 C[sub]1[/sub])[sup]3/2[/sup].

Again, if you can invert this one, you’re a better mathematician than I am.[/spoiler]