# Gravity in solid sphere

1. Jun 13, 2008

### foxjwill

1. The problem statement, all variables and given/known data
How would I go about finding the time it takes for a particle of mass $$m$$ to travel in a straight line only under the force of gravity between any two points on the surface of a solid sphere of mass $$M$$.

2. Relevant equations
$$\mathbf{F}=-{3GM \over R^3} \matbhf{r}$$ (I derived this using integration and Newton's Universal Law of Gravitation)​

3. The attempt at a solution

Don't tell me how to do it, just give me a hint. I think what's really tripping me up here is how do you have an object traveling along a non-field-line curve only under the force of gravity? Wouldn't there have to be some normal force keeping the object on the curve?

2. Jun 13, 2008

### Dick

Yes, there would need to be a normal force. Assume it's travelling through a frictionless tunnel. Then just consider the tangential force. But where did that '3' come from in your force equation? And that's not an F, it's an acceleration.

Last edited: Jun 13, 2008
3. Jun 13, 2008

### gendou2

I think the trick is, it has to pass through the center.
In other words, it's a tunnel to China problem.

One hint that might not be obvious is that it takes the same time to fall from the surface to the center as it does to travel from the center out to the other side.

4. Jun 13, 2008

### Dick

Actually, it doesn't have to fall through the center. That would be really easy. The time is the same along any chord connecting two points on the surface.

5. Jun 13, 2008

### gendou2

Oh yeah, you're right. Sorry about that.

6. Jun 13, 2008

### foxjwill

$$\rho = {M \over {4\over 3} \pi R^3}$$
$$dm = \rho 4\pi r^2 dr$$

$$\vec{a} = -{\int_0^r \rho 4\pi G dr} \hat{r} = -\rho 4\pi G r \hat{r} = -{3GMr \over R^3} \hat{r} = -{3GM\over R^3}\vec{r}$$

oops! >_<

7. Jun 13, 2008

### Dick

You don't need to integrate anything. a=Gm/r^2. m=rho*volume=rho*(4/3)*pi*r^3 where rho is density. M=rho*(4/3)*pi*R^3. Just eliminate the rho. But I think in the way you worked it you forgot that integral of r^2=r^3/3. That's the 3 that should cancel the one you have.

8. Jun 14, 2008

### foxjwill

But it it's not Gm/r^2. That's only if the particle is outside the sphere. And the r^2 canceled out, so that's why I have it like that.

9. Jun 14, 2008

### Dick

You can do the problem by only considering the mass inside the sphere of radius r, since the mass outside contributes nothing. I have no problem with your answer except for that '3'.