Gravitional force of a particle inside a sphere;

• Benzoate
In summary, it is being discussed that the gravitational force exerted on a particle inside a hollow symmetric sphere is zero. The proof for this is similar to the proof for a particle outside a symmetric sphere, with the exception of one detail. The variables used in the proof include theta, phi, and r, and the proof is done using polar coordinates. The proof involves integrating over phi and theta, and locating the point along the k-axis. It is mentioned that the G*m1*m2 factor can be dropped and added in at the end, and that it matters that the sphere is hollow, not solid. A resource is provided for further understanding of the proof.
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.]

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.

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.

Dick said:
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?

Dick said:
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?

Benzoate said:
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.

1. What is the formula for calculating the gravitational force of a particle inside a sphere?

The formula for calculating the gravitational force of a particle inside a sphere is F = (G * m * M) / r^2, where F is the force, G is the gravitational constant, m is the mass of the particle, M is the mass of the sphere, and r is the distance between the particle and the center of the sphere.

2. How does the gravitational force inside a sphere vary with distance from the center?

The gravitational force inside a sphere varies directly with the distance from the center. This means that the force decreases as the distance from the center increases.

3. What is the relationship between the mass of the particle and the gravitational force inside a sphere?

The gravitational force inside a sphere is directly proportional to the mass of the particle. This means that as the mass of the particle increases, the force also increases.

4. What happens to the gravitational force inside a sphere if the particle is at the center?

If the particle is at the center of the sphere, the gravitational force will be zero. This is because the distance between the particle and the center of the sphere is zero, and according to the formula, the force becomes zero.

5. Does the size of the sphere affect the gravitational force inside?

Yes, the size of the sphere does affect the gravitational force inside. The larger the sphere, the greater the gravitational force due to the increased mass. However, as the distance from the center increases, the force will decrease as explained in question 2.

Replies
1
Views
572
Replies
2
Views
792
Replies
7
Views
2K
Replies
7
Views
1K
Replies
10
Views
2K
Replies
26
Views
4K
Replies
3
Views
1K
Replies
1
Views
999