# Fairly Simple Gravity Problem

1. Oct 14, 2007

### PiratePhysicist

I named my original post poorly, so this one is meant to just rename it, my work so far can be found in https://www.physicsforums.com/showthread.php?t=191219

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.

2. Relevant Equations:
$$F=\frac{-G M_e m}{r^2}$$

3. Attempt at the solution:

2. Oct 15, 2007

### learningphysics

I think what you need to do is, at this point:

$$v^2=2G M_e ( \frac{1}{r}-\frac{1}{r_0})$$

let r0 = infinity...

then continue from there.

although your derivation using separable diff. eq. was all correct... you can use conservation of energy to do that, unless you haven't covered energy yet....

3. Oct 15, 2007

### PiratePhysicist

Nope, tried something similar (redefined axises so $$r_0$$ was 0), here's what happens:
$$\frac{dr}{dt}=\sqrt{\frac{2 G M_E}{r}}$$
$$\sqrt{r}dr=\sqrt{2 G M_E}dt$$
$$\frac{2}{3}r^{\frac{3}{2}}=sqrt{2G M_E}t$$
$$t=\frac{2}{3}\sqrt{\frac{r^3}{2GM_E}}$$
So if you take the ratio of t(1/2r) and t(r) you get
$$\frac{\frac{2}{3}\sqrt{\frac{(\frac{r}{2})^3}{2GM_E}}}{\frac{2}{3}\sqrt{\frac{r^3}{2GM_E}}}$$
Which boils down to
$$(1/2)^\frac{3}{2}$$ which is equal to .353535... not .8181818181

4. Oct 15, 2007

### dynamicsolo

Won't there be a non-zero integration constant to be evaluated?

The original differential equation

$$\frac{dr}{dt}=\sqrt{\frac{2 G M_E}{r}}$$

describes an outbound trajectory, since the right-hand side is positive. Infall should require the insertion of a minus sign, no? In any case, dr/dt is only zero according to this equation for r -> infinity; then you have a separate can of worms in describing when you have fallen halfway from infinity. If you start from a finite distance out (or time-reverse the path, start from the Earth's surface, and look at the times when the particle reaches (r0/2) and r0, the integration is incomplete as it stands...

5. Oct 15, 2007

### PiratePhysicist

Erm, my long drawn out derivation, was meant to show that the take to infinity assumption doesn't work, but your explanation works better and doesn't require lots of LaTeX ^_^

So anyone else have any suggestions?

6. Oct 15, 2007

### dynamicsolo

Something seems suspect about that 9/11 value. You can treat the "radial infall" problem using Kepler's Third Law on what are called "degenerate orbits".

Compare the fall from r0 with a fall from (r0/2) by treating each as having a pericenter at r = 0 and the apocenters at r0 and (r0/2), respectively. Then the "semi-major axes" of these degenerate ellipses are a1 = (r0/2) and a2 = (r0/4). The respective times to reach r = 0 are the half-periods of these "orbits"

T1 = (1/2)·sqrt[ {4·(pi^2)/GM} · (r0/2)^3 ] and
T2 = (1/2)·sqrt[ {4·(pi^2)/GM} · (r0/4)^3 ] ;

in other words, basically what you found by integrating the energy result (and starting from the gravitational force equation takes you to the same place).

The ratio of the infall times is T2/T1 = (1/2)^(3/2) = 0.3536... , as you already found. (This is a standard astrophysical exercise.) The time to fall the last half is 0.354 of the time to fall the full distance, so the time to fall the first half is about 0.646 of the total infall time.

That former value is very close to 4/11 = 0.363636... , which makes me wonder if there was a handwriting issue and whether the writer confused the statement of the problem. (This would make the answer to the stated question about 7/11 of the total time.) I think your derivation is basically correct; the 9/11 didn't sound right to me from the start, but I thought maybe I was recalling something incorrectly...

Last edited: Oct 15, 2007
7. Oct 15, 2007

### PiratePhysicist

The 9/11 number comes from the book (Classical Dynamics of Particles and Systems 5th Ed Thornton and Marion)