# Time for an object to fall to a larger one

1. Dec 11, 2013

### kaikalii

I know that for short distances from the earth's surface, x=-1/2gt^2+vt+x works fine for finding the time it takes for an object to fall a certain distance ignoring air resistance.
However, what if the distance is many times the earth's radius?
The only thing I can think of to start solving this problem is f''(t)=GM/(f(t))^2, but try as I might, I cannot solve that to evaluate the time and object takes to fall, say, a distance, r. How can I accomplish this?

2. Dec 11, 2013

### JPaquim

I'm assuming you're talking about a body whose mass is much smaller than the Earth's, dropped with zero velocity from an arbitrary distance from the Earth. If so, please check out this article or this one. Basically the body follows a degenerate ellipse trajectory, similar to the trajectory of the moon around the earth. The equations aren't as easy to solve as the mgh potential because they're non linear, but if you want the time in terms of initial position, you write down the energy per unit mass, which is a conserved quantity (and should be negative if you're effectively falling and not escaping the Earth): $E = E_0 = T + U = (dr/dt)^2/2 - \mu/r$, solve for dt and integrate: $$\int_{t_0}^t dt = t - t_0 = \int_{r_0}^r \frac{dr}{\sqrt{2(E_0+\mu /r)}} = \frac{1}{\sqrt{2\mu}}\int_{r_0}^r \sqrt{\frac{r}{1-\alpha r}}dr$$ with $\alpha = -E_0/\mu > 0$, if I didn't mess up the algebra. The integral may found in integral tables or WolframAlpha, or computed numerically.

For the more general problem where you can have non-zero velocity and a body with a large mass, read into the gravitational two-body problem, or the Kepler problem if you're not interested in large masses.

Last edited: Dec 11, 2013
3. Dec 11, 2013

### kaikalii

Thank you! This is exactly what I was looking for.

4. Dec 11, 2013

### JPaquim

I've added some stuff to my answer, maybe you'd like to check it out :)