Gravitional force of a particle inside a sphere;

[Benzoate]

Homework Statement

Show that the gravitational force exerted on a particle inside a hollow symmetric sphere is zero. [Hint. The proof is the same as for a particle outside a symmetric sphere, except in one detail.]

Homework Equations

F=m1*m2*G/R^2

The Attempt at a Solution

If a particle is going to interact with a sphere, My polar coordinates should be 3-dimensional.

variables will be theta, phi and r.

r hat will be in the direction of the r vector
theta ; rotation in the r-hat, k-hat plane , perpendicular to r-hat.
theta is the angle between the r vector and k vector.

phi-hat rotates about k-hat, in the i hat , j-hat plane
phi is the angle between i-hat and projection of the r vector in the i hat , j hat , plane r*sin(theta)*dphi.

dV= dr*(r*theta)*(r*sin(theta)*d-phi= lim(dr=>0) r^2*dr*(sin(theta)*dtheta)*dphi

sin(theta)*dtheta I think is the solid angle.

Since the horizontal components are i-hat and j-hat, by symmetry, the only component left is the k-hat component.

F=integral(from 0 to a) dr*integral(from 0 to pi) dtheta* integral(0 to pi)dphi*(-G*m(particle)*r^2*sin(theta)/R^2 * (b-r*cos(theta)/R)

R=sqrt((b-r*cos(Theta)^2+(r*sin(theta))^2)
Since symmetry is only around k axis, there isn't any dependence on phi

F=-2*pi*G*m(particle)k-hat((integral(from 0 to a)dr*integral(frome 0 to pi) dtheta integral(no upper or lower limits) r^2*sin(theta)/(R^3) *(b-cos(theta))

F=-2*pi*G*m(particle)k-hat((integral(from 0 to a)r^2*dr*integral(frome 0 to pi)*(b-r*cos(theta)*sin(theta)/(R^3)

R^2=(b-r*cos(theta))^2+(r*sin(theta)^2=b^2-2*b*r+r^2

2*R*dR=2*b*r*sin(theta)*dtheta

(b-r*cos(theta))=2b^2-2*b*r*cos(theta)=R^2+(b^2-r^2)/(2*b)

F=-(2*m*G*m(p)*rho/2b^2) integral(from 0 to a) r^2*dr
F=-(G*4*pi*a^3) *rho*m(particle)/(b^2) k-hat
=-(G*M*m(particle)/b^2 k-hat

Not correct.

I think my R is wrong.

e particle is inside the sphere not outside; Does it make a difference whether or not the sphere is hollow or solid?

Sorry for writing the integrals out in text form. I didn't know how to write out the integrals , whhere you would write out the integral and it would look like an integral you see in a math book.

 

[Dick]

Gack! No offense, but that's really hard to read. It's late, so I'll just give you a little advice. Drop the G*m1*m2 factor. It doesn't matter, it's just a constant. Put it in at the end. And yes, it matters that the sphere is hollow, not solid. Don't even integrate over r, just integrate over phi and theta. And, yes, just locate the point along the k-axis.

 

[Benzoate]

Gack! No offense, but that's really hard to read. It's late, so I'll just give you a little advice. Drop the G*m1*m2 factor. It doesn't matter, it's just a constant. Put it in at the end. And yes, it matters that the sphere is hollow, not solid. Don't even integrate over r, just integrate over phi and theta. And, yes, just locate the point along the k-axis.

Why does it matter if the sphere is hollow or a solid? Is it because a hollow sphere is not as dense as a solid sphere?

 

[Benzoate]

Gack! No offense, but that's really hard to read. It's late, so I'll just give you a little advice. Drop the G*m1*m2 factor. It doesn't matter, it's just a constant. Put it in at the end. And yes, it matters that the sphere is hollow, not solid. Don't even integrate over r, just integrate over phi and theta. And, yes, just locate the point along the k-axis.

So should I treat r^2 as a constant or should I eliminate r^2*dr altogther? Does r not matter because the particle is within the sphere?

 

[Dick]

Why does it matter if the sphere is hollow or a solid? Is it because a hollow sphere is not as dense as a solid sphere?

If the sphere is solid and the particle is inside, then the force is only zero at the center of the sphere. Sorry, I've tried several times to read your work and I'm not doing very well. Here's a fairly lucid exposition http://en.wikipedia.org/wiki/Shell_theorem See if you can follow that.