Influence of gravitational force on earth

[ace123]

Homework Statement

Consider the thought experiment shown in the figure below. A tunnel is constructed from the North Pole to the South Pole of the Earth and a cylinder is placed within the tunnel at the South Pole.


Study the influence of the gravitational force of the Earth on the cylinder.
Describe the motion of the cylinder fully.

Homework Equations

Well since it says study the influence of gravitational force I'm assuming that Formula will be used.
Maybe their are other formula's that I should use but I'm currently unsure

The Attempt at a Solution

At first glance at this problem I thought that this would be a simple harmonic motion. With the cylinder oscillating from one end to other end and eventually comming to rest near the centre of the earth.

Not sure if this is the correct approach. This is not really a homework assignment because it's optional but I'm going to at least try to attempt this and I need some help. If anyone has some reading material related to this that would be really helpful. I don't care if it's a book or magizine or some internet article I'm a little pressed for time and I would appreciate any kind of help.

Thanks.

 

[ace123]

Here is the attachment of the picture.

 

[Andrew Mason]

Homework Statement

Consider the thought experiment shown in the figure below. A tunnel is constructed from the North Pole to the South Pole of the Earth and a cylinder is placed within the tunnel at the South Pole.


Study the influence of the gravitational force of the Earth on the cylinder.
Describe the motion of the cylinder fully.

1. Let R be the radius of the earth. Let the tunnel be the x-axis from going from x = -R to R. What is the acceleration due to gravity of the cylinder at a distance x < R? Hint: does the matter in the "shell" where x > r contribute net gravitational force on the cylinder?

2. Integrate that force from x = -R to x = R

AM

 

[ace123]

Thanks for responding. Not quite sure what you mean by "shell." If I'am understanding you correctly than the acceleration due to gravity becomes smaller until it's eventually zero.
Also I'm not sure he wants me to use integration since the majority of the class hasn't even taken calculus and it's suppose to be physics without calculus. But if their is no other way to approach this problem then integration will be fine.

Thanks again for the response.

 

[rohanprabhu]

Not quite sure what you mean by "shell."

When you consider a point x = r, such that r < R, the surface of the sphere having the same center as the Earth and a radius 'r' is the shell he is talking about.

 

[ace123]

Okay I understand "shell" now but what matter is he referring to. Wouldn't it be the same matter since it's a smaller sphere around the center. So it would merely be less matter.

what would this topic be called so I could research it and maybe try to understand it myself?

Thanks

 

[Doc Al]

At first glance at this problem I thought that this would be a simple harmonic motion.

It will turn out to be simple harmonic motion. The first thing you need to do is find the gravitational force as a function of distance from the center of the earth.

With the cylinder oscillating from one end to other end and eventually comming to rest near the centre of the earth.

No reason for it to come to rest as long as you ignore things like friction or air resistance.

Okay I understand "shell" now but what matter is he referring to. Wouldn't it be the same matter since it's a smaller sphere around the center. So it would merely be less matter.

What Andrew Mason is asking is: At some point a distance "r" from the center of the earth, does the portion of the Earth's mass that is at a distance > r from the center exert a gravitational force?

 

[ace123]

I don't think so. I mean the cylinder is surrounded by the 2 masses so they would each exert a force in the opposite direction on the cylinder. Is this what you meant?

I think the portion's of the Earth's mass at distance >r can be ignored and only the distance< r should be included. I'm just not entirely sure why.

I think I need to know the shell theorem for this problem. Is that correct?

 

[Doc Al]

I think the portion's of the Earth's mass at distance >r can be ignored and only the distance< r should be included.

Right!

I think I need to know the shell theorem for this problem. Is that correct?

That is correct.

 

[ace123]

Okay thanks I'm going to look it up.

 

[ace123]

So I read the article in wikipedia about shell theorem and still unsure of what to do. I mean I don't understand why portion's of the Earth's mass at distance >r can be ignored and only the distance< r should be included. Can someone explain this or is the only way to understand it is through calculus?

Also would the Earth be considered as a solid sphere? If so my professor said that I should assume that the Earth's density is constant.

 

[Andrew Mason]

So I read the article in wikipedia about shell theorem and still unsure of what to do. I mean I don't understand why portion's of the Earth's mass at distance >r can be ignored and only the distance< r should be included. Can someone explain this or is the only way to understand it is through calculus?

Also would the Earth be considered as a solid sphere? If so my professor said that I should assume that the Earth's density is constant.

The only way to really understand it is through calculus. You have to add the force of each element of mass in the shell on a point mass in the interior. And the only way to do that is by calculus.

You can qualitatively understand part of the shell theorem this way:

1. If the mass is at the centre, there is no force at all by symmetry, as the gravitational force from each element of the shell is balanced by an equal force in the diametrically opposite direction.

2. A mass located at a point other than the centre will lie on some diameter. So draw a diameter through the mass and then consider symmetrical rings of the shell about that diameter. By symmetry, you can see that there will be no lateral force acting on the mass (ie. lateral to the diameter) since each element of the shell is balanced by an equal force in the laterally opposite direction.

3. The question then, is whether there is a net radial force (ie. along the diameter through the point mass) when the point mass is located other than at the centre.

That is a little harder to show since the shell is not symmetrical about that point along the diameter. The only way to do that is to use calculus.

This is very similar to the law for electrical charge. A charge inside a hollow spherical uniformly charged shell experiences no electrical field. In fact, Gauss' law says that the field is determined only by the enclosed charge. It is quite analagous to gravity. The gravitational field is determined only by the enclosed mass.


Now, in answer to your question, since the force at a distrance r from the centre is determined only by the mass enclosed in a shell of radius r, ($F = GMm/r^2$) and since that mass is proportional to 1/r^3 (assuming density of the Earth is constant) work out the expression for force in terms of r.

If the force is proportional to displacement, what kind of motion do you have?

AM

 

[ace123]

That is a little harder to show since the shell is not symmetrical about that point along the diameter. The only way to do that is to use calculus.


AM

Could you possibly show this? Or tell me what it is so I can look it up on this wikipedia link http://en.wikipedia.org/wiki/Shell_theorem

Because I don't really understand why the force at a distance r from the centre is determined only by the mass enclosed in a shell of radius r since a shell exerts no net force on particles anywhere within its volume.

I also read on wikipedia that inside a shell:

$$F = GMm/r^2$$ for $$r> R$$

$$F= 0$$ for $$r<R$$

So how come for the Earth we ignore the mass at r>R and only use the mass r<R.

Also how did you get the mass to be 1/r^3

Thanks for all this btw.

Edit: I do know calculus but I just wanted to know if their was a way to not use calculus.

 

[Andrew Mason]

Could you possibly show this? Or tell me what it is so I can look it up on this wikipedia link http://en.wikipedia.org/wiki/Shell_theorem

I'll have a look. But it is just a matter of dividing a sphere into infinitessimally wide slices along a diameter passing through the point mass (inside the sphere) and adding up the gravitational effect of each.

Because I don't really understand why the force at a distance r from the centre is determined only by the mass enclosed in a shell of radius r since a shell exerts no net force on particles anywhere within its volume.

You know that there is gravitational attraction between two masses and the gravitational attraction is GMm/r^2 where r is the distance between the centres of the masses. So that part is just the law of universal gravitation.

There are gravitational forces inside the sphere. It is just that they sum to 0. There is no gravitational effect on a mass inside a sphere of uniform density becuase the gravitational contributions of the atoms in the sphere on a point mass inside the sphere offset each other in total. You can see by symmetry why that does not occur if the point mass is outside the sphere.

I also read on wikipedia that inside a shell:

$$F = GMm/r^2$$ for $$r> R$$

$$F= 0$$ for $$r<R$$

So how come for the Earth we ignore the mass at r>R and only use the mass r<R.

R is the radius of the earth. In your tunnel, r is never greater than R. The statement [itex]F= 0[/tex] for [tex] r<R[/itex] applies only to a hollow sphere.

Also how did you get the mass to be 1/r^3

For simplicity we assume the density to be the same throughout the earth. So the mass is proportional to volume. What is the volume of a sphere of radius r?

AM

 

[Andrew Mason]

Please note the following correction to my earlier post: I meant to say, of course, that mass is proportional to r^3 nor 1/r^3.

AM

 

[ace123]

Okay just wanted to make sure that it wasn't the same as for a hollow sphere. The volume is $$(4\/Phi/3)r^3$$. So the mass over the volume will give us the density correct?

 

[ace123]

I think I almost got it. What do i have to do after i find the gravitational force at a distance r? You said integrate -R to R and that will give me the sum of the force correct? Isn't there more i have to do to show it's a simple harmonic motion? Or will the final result be that the force is proportional to the displacement

 

[Andrew Mason]

I think I almost got it. What do i have to do after i find the gravitational force at a distance r? You said integrate -R to R and that will give me the sum of the force correct? Isn't there more i have to do to show it's a simple harmonic motion? Or will the final result be that the force is proportional to the displacement

You can ignore my comment about integration. Just look at the force as a function of r.

You can also look at it from an energy point of view, but it is not necessary. If you integrate the force x distance, from -R to 0 you get the kinetic energy of the cylinder at the centre of the earth. If you integrate from 0 to R you will get the potential energy in the cylinder at r=R.

AM

 

[ace123]

That's great I'll probably do both and see what i get. Thanks for everything

 

[ace123]

I have a final question I thought harmonic motion is when the force is directly proportional to the negative of the displacement but that is not what i got my force to. Meaning I don't have a negative sign. This would still be a simple harmonic motion though because only the restoring force has to be proportional to the negative displacement. Is that correct?

 

[Andrew Mason]

I have a final question I thought harmonic motion is when the force is directly proportional to the negative of the displacement but that is not what i got my force to. Meaning I don't have a negative sign. This would still be a simple harmonic motion though because only the restoring force has to be proportional to the negative displacement. Is that correct?

You just have to define what you mean by negative force. The force is toward the centre of the earth. If you take the centre of the Earth as the origin and consider any outward radial direction to be positive, the force is always negative.

AM

 

[ace123]

That's what I wrote hopefully it's enough. Thanks