# Gravity in solid sphere

• foxjwill
I don't think that's necessary. I don't think the particle has to be inside the sphere. It could be on the surface of the sphere or even inside the sphere but outside of the radius. But I don't think that's necessary.f

## Homework Statement

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 $$M$$.

## Homework Equations

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

## 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?

Yes, there would need to be a normal force. Assume it's traveling 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.

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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.

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.

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

But where did that '3' come from in your force equation?

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

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

And that's not an F, it's an acceleration.
oops! >_<

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.

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.

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.

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.

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'.