Why Does a Falling Particle Take 9/11 of Total Time to Cover Half the Distance?

[PiratePhysicist]

1. Problem: A particle falls to Earth starting from rest at a great height (may times Earth's radius ). Neglect air resistance and show that the particle requires approximately 9/11 of the total time of fall to traverse the first half of the distance.

Homework Equations

$$F=\frac{-G M_e m}{r^2}$$

The Attempt at a Solution

$$F=\frac{-G M_e m}{r^2}$$
$$a=\frac{-G M_e}{r^2}$$
$$\frac{\partial v}{\partial r} v = \frac{-G M_e}{r^2}$$
seperate and integrate
$$\frac{1}{2}v^2=\frac{G M_e}{r}+K$$
$$v^2 = \frac{2G M_e}{r}+K$$
$$v^2(r_0)=\frac{2G M_e}{r_0}+K=0$$
$$v^2=2G M_e \( \frac{1}{r}-\frac{1}{r_0}\)$$
Then take square root, replace v with the time derivative of r, separate and integrate and you get this mess:
$$\int{\sqrt{\frac{r_0 r}{r_0-r}}\partial r}=\sqrt{G M_e}t$$
Which comes out to having a tangent in it, and in general very messy. Also when you plug the two values $$r_0$$ and 0, you wind up with an infinity. So any ideas where I went wrong?

 

[dlgoff]

Are you sure a=(dv/dt)v?

 

[PiratePhysicist]

It's not a = (dv/dt)v, it's a=(dv/dr)v
$$a = \frac{\partial v}{\partial r}*\frac{\partial r}{\partial t} = \frac{dv}{dt}$$
Chain rule!

 

[dlgoff]

Okay, sorry. Couldn't make out the r on this old computer monitor.