# Gravitational force question

1. May 5, 2006

### squiffy

Hi, I posted this question a while back on these forums. Didn't get an answer though.

How long will it take for two identical masses to reach each other by their attractive gravitational force?
Their masses can be 1kg and the distance between them 1m to simplify it.

Do NOT assume that the force is constant. It is not, it's inversely proportional to r^2, the distance between them.
I'm stumped on this one.
Thanks.

2. May 5, 2006

### Curious3141

OK, quote the formula for gravitational force between two identical masses as a function of the distance between them.

The rest is not so easy. Do you know how to solve a second order ordinary differential equation ?

3. May 5, 2006

### Andrew Mason

It depends on how big they are. You haven't provided enough information. If the 1m is the distance between their centres of mass, you can figure out the force but you don't know how far they have to move (without knowing how far apart their outer surfaces are). If the 1m is the distance between their outer surfaces, you don't know how far apart their centres of mass are and can't calculate their gravitational force.

AM

4. May 5, 2006

### Curious3141

Good point. I was assuming point masses, which seems to be in keeping with the spirit of the question (which wants to simplify the question to the most generic terms).

In any case, for uniform spherical masses, the same method would apply, with different distances being taken as the bounds of integration.

Last edited: May 5, 2006
5. May 5, 2006

### Andrew Mason

But then you have infinite potential energy and the masses end up with infinite speed when they collide.

One way to approach this is to integrate $dt = dr/v$. v is determined by the change in potential energy as both masses approach the centre of mass (r=.5 m).

AM

6. May 5, 2006

### Curious3141

That's true, but that happens only at one instant. It may still be possible to solve for the finite time taken for contact to occur (though I haven't worked it out, so I'm not sure). Of course, the answer would only be valid in a classical framework (non-relativistic).

7. May 6, 2006

### SpaceTiger

Staff Emeritus
Hint: Take a look at Kepler's Third Law of Planetary Motion.

8. May 6, 2006

### Andrew Mason

The masses are not in orbit about each other. How does Kepler's Third Law apply?

Try using a momentum/energy approach. The change in potential energy must equal the kinetic energy of both masses as they approach the centre of mass:

$$\Delta U = Gm^2\left(\frac{1}{r'} - \frac{1}{r_0}\right) = mv^2$$

Using:

$$Fdt = mdv = mdr/dt$$ and $$F = Gm^2/r^2$$

$$dt = \sqrt{mdr/F}$$

$$t = \int_r^0 dt = \int_r^0 \sqrt{\frac{2}{Gm}r^2dr}$$

I'm not going to try to tackle that integral right now.

AM

Last edited: May 6, 2006
9. May 6, 2006

### Curious3141

The approach I was contemplating was a little more direct.

$$x^2\ddot{x} + Gm = 0$$

EDIT : this diff eqn is flawed, please see my later post.
That is solvable with a substitution $$\frac{dx}{dt} = v$$ but I haven't worked it through.

Last edited: May 6, 2006
10. May 6, 2006

### SpaceTiger

Staff Emeritus
You shouldn't be solving the problem for the student, but free fall between two point masses is an orbit with eccentricity of 1.

11. May 6, 2006

### Curious3141

I have to read the link more thoroughly, but this clinches it : http://en.wikipedia.org/wiki/Free-fall_time

12. May 6, 2006

### Curious3141

Bit of weirdness, when I use that expression, there's a pi in it. But that won't satisfy my d.e. (which must be right).

How to resolve the disagreement ?

13. May 6, 2006

### Curious3141

Well, I don't see a way out of this. Either my diff eqn (from first principles) is wrong, or the free fall time based on K3L is wrong.

14. May 6, 2006

### Andrew Mason

A very clever approach. I was thinking that because the masses are initially stationary, this approach would not work. The problem is that each mass orbits the centre of mass and the centre of mass is the midpoint (r=.5 m.).

I am a little rusty at solving differential equations. My integral solution works out by substituting $u = \sqrt{r}$

$$t = \int_r^0 dt = \int_r^0 \sqrt{\frac{2}{Gm}r^2dr} = \int_{u^{2}}^0 \sqrt{\frac{2}{Gm}}u^2du$$

$$\sqrt{\frac{2}{Gm}}\int_{u^{2}}^0 u^2du = \sqrt{\frac{2}{Gm}}\frac{1}{3}u^3$$

$$\sqrt{\frac{2}{Gm}} \frac{1}{3}u^3 = \sqrt{\frac{2}{Gm}} \frac{1}{3}r^{3/2}$$

If I have done the math properly, this means that it would take

$$\sqrt{\frac{2}{6.673e-11}} \frac{1}{3}.5^{3/2}seconds$$

which works out to about 6 hours. That doesn't seem right for some reason.

AM

Last edited: May 6, 2006
15. May 6, 2006

### Curious3141

Andrew,

Your solution satisfies my diff equation perfectly. So our methods and answers concur.

But both of them differ from the answer from Kepler's third law. I cannot figure out where that pi would come from, for example, in straight line motion.

I know 5.6 or so hours seems short considering how weak we're taught gravity is, but remember that we have no real life experience of small objects free falling into each other in deep space. Everything we're used to is completely overwhelmed by the Earth's gravity.

Last edited: May 6, 2006
16. May 6, 2006

### Andrew Mason

Good question! Perhaps the Kepler approach is over-simplified. After all, the second law (equal areas swept in equal times) does not apply since the area is 0. I don't see how $\pi$ comes into it on a straight line free-fall.

This would be a good question for someone to take on.

Right. This Cavendish type of experiment shows that with small masses a few cm away from each other, the period of oscillation of the pendulum is about 1000 seconds. So 20,000 seconds to go about 1/2 a metre is in the ball park.

AM

17. May 6, 2006

### nrqed

Why can you replace ''dv'' by ''dr/dt''?

18. May 6, 2006

### nrqed

The equation does not seem right to me. I might be missing something but x here is the separation between the two masses, right? But this equation says that the second derivative of the separation depends only on the acceleration of *one* of the two masses. In fact, the second derivative should take into account the acceleration of both masses.
Am I missing something?

19. May 6, 2006

### nrqed

I might be missing something (again!) but Kepler's third law applies only to the limit of one mass being much more massive than the other. In this example, the two masses are equal. Maybe it is just a matter of using the reduced mass instead but that is not so obvious to me..

20. May 6, 2006

### SpaceTiger

Staff Emeritus
Kepler's Third Law applies equally well to similar masses as to one much bigger than the other, you simply need to switch to the COM frame (we apply the third law to binary star systems all the time) . Also, there's nothing wrong with the masses being stationary -- an object on a radial orbit would be momentarily stationary (v=0) at aphelion.