Calculating Particle Travel Time in Solid Sphere Under Gravity

In summary: 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.
  • #1
foxjwill
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0

Homework Statement


How would I go about finding the time it takes for a particle of mass [tex]m[/tex] 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 [tex]M[/tex].


Homework Equations


[tex]\mathbf{F}=-{3GM \over R^3} \matbhf{r}[/tex] (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?
 
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  • #2
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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  • #3
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
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
Oh yeah, you're right. Sorry about that.
 
  • #6
Dick said:
But where did that '3' come from in your force equation?

[tex]\rho = {M \over {4\over 3} \pi R^3}[/tex]
[tex]dm = \rho 4\pi r^2 dr[/tex]

[tex]\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}[/tex]

And that's not an F, it's an acceleration.
oops! >_<
 
  • #7
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
Dick said:
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.
 
  • #9
foxjwill said:
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'.
 

1. What is the concept of gravity in a solid sphere?

Gravity in a solid sphere refers to the force of attraction between the particles within the sphere. This force is dependent on the mass and distance between the particles.

2. How does the distribution of mass affect gravity in a solid sphere?

The distribution of mass within a solid sphere affects the strength of gravity. The closer the particles are to the center of the sphere, the stronger the gravitational force will be.

3. How does the radius of a solid sphere impact the strength of gravity?

The radius of a solid sphere also affects the strength of gravity. The greater the radius, the weaker the gravitational force will be, as the particles are further apart from each other.

4. Can the gravitational force within a solid sphere be different at different points?

No, the gravitational force within a solid sphere is the same at all points. This is because the particles are evenly distributed throughout the sphere, resulting in a uniform force of attraction.

5. How does gravity in a solid sphere compare to gravity on the surface of the sphere?

The gravitational force within a solid sphere is greater than the force on the surface of the sphere. This is because the particles on the surface have less mass and are further from the center, resulting in a weaker force of attraction.

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