No gravity at center of earth= no pressure?

[Femme_physics]

Pfffff, you guys clearly haven't seen Journey to the Center of the Earth...

 

[sophiecentaur]

I have to admit that I had completely forgotten about the hollow middle.

 

[pete20r2]

IF THE MIDDLE WAS HOLLOW, and vented to the surface;
1. There would be increased pressure, going up in the atmosphere decreases pressure and going down increases, until you get to the center, passing through means your going up again and it decreases.
2. Due the the EVEN DISTRIBUTION OF MASS around you, every single mass (atom, particle, whatever you want to call it), is pulled in every direction evenly. You do not feel the force because it is balanced at every point in your body.

In a vacuum (good start), if there was a hole in the earth, through to the other side and I jumped down it, this is what would happen:
1. I would accelerate down until I crossed half way, this acceleration would be DECREASING as I approached the center and would be 0 as I crossed.
2. I would continue past the center, since I have a lot of speed and there is little acceleration.
3. The acceleration would now build up in the opposite direction to my velocity and I would slow to a stop with my head level to the ground on the other side of the earth, with my feet in the air.
4. I would fall and do it all again.

In air:
Same as before but with losses, I would certainly not make it to the other side, but I would converge on the center. As I get slower and slower every time I pass the center though, my rate of convergence decreases. This is because the acceleration at the center is 0 so there is nothing to keep me there and I would theoretically forever be oscillating about the center.

 

[sophiecentaur]

Perhaps you could explain why pressure necessarily disappears when the gravity does. Take a pressurised spacecraft into deep space. Will it lose it's pressure?
We're talking two entirely different effects.

 

[I like Serena]

IF THE MIDDLE WAS HOLLOW, and vented to the surface;

[...]

In a vacuum (good start), if there was a hole in the earth, through to the other side and I jumped down it, this is what would happen:
1. I would accelerate down until I crossed half way, this acceleration would be DECREASING as I approached the center and would be 0 as I crossed.

I hope your tunnel is in a curve, since the Coriolis force would otherwise be rather nasty! :)

 

[pete20r2]

I was thinking about coriolis as I was writing that, but thought it irrelevant to this discussion, thanks for making people aware of the facts, though.

 

[I like Serena]

I was thinking about coriolis as I was writing that, but thought it irrelevant to this discussion, thanks for making people aware of the facts, though.

Note also that you would not fall through the center of the earth, but more in an elliptic orbit with the geometric center coinciding with the center of the earth.

 

[sophiecentaur]

Take the North South route and avoid coriolis. But still get crushed and roasted to death before you've gone any significant distance down there.

Bear in mind that air pressure doubles every 5.5km downwards, so the air you went through would soon no longer be a gas on the way down.

 

[Drakkith]

Ok folks! Threads over! Nothing to see here! *fireworks are going off from a burning fireworks stand behind me* Nothing to see here! Move along! *Rabid elephant ninjas on fire are running rampant through the town behind me* Move along! Nothing to see here!

 

[pete20r2]

It's a bit like that.. HEY EVERYONE LOOK OVER THERE IT'S A HIGGS BOSON HERE TO RUIN EVERYTHING

 

[sophiecentaur]

Note also that you would not fall through the center of the earth, but more in an elliptic orbit with the geometric center coinciding with the center of the earth.

Elliptic? Are you sure? The attraction law is not inverse square and elliptical orbits are generated by inverse square. It may not 'not' be but I'm not sure. in the detail. If you were a whizz at analysis, you could, no doubt derive the actual curve and show that it is (or not) an ellipse.

 

[Studiot]

This whole discussion so much hypothetical dreaming anyway because it assumes the Earth is an isolated body.

Of course, this is not so. Indeed our tides depend upon the fact that the Earth - Moon system revolves around a common centre of gravity that is not at the centre of the Earth.

Any object (taking the Earth as sphere for simplicity) placed at the centre of mass this sphere will not experience zero gravity for this reason.

 

[sophiecentaur]

However, the time constants involved in any other interaction are much greater - by factors of 28, at least. So we are allowed to go along with it, I think.

ALso, you could do the same experiment with a pebble and an asteroid and still get the same result (assuming the density were similar).

 

[I like Serena]

Elliptic? Are you sure? The attraction law is not inverse square and elliptical orbits are generated by inverse square. It may not 'not' be but I'm not sure. in the detail. If you were a whizz at analysis, you could, no doubt derive the actual curve and show that it is (or not) an ellipse.

Yes, I am!
The attraction law is linear instead of inverse square.
Still an ellipse, but the center of attraction is not in one of the focus points, but in the geometric center. Cool huh!

(Of course that would be if the Earth were an isolated body, which it isn't, as Studiot pointed out. )As for the proof, isn't it enough that a spring or a pendulum makes a harmonic motion?

 

[cjl]

Yes, I am!
The attraction law is linear instead of inverse square.
Still an ellipse, but the center of attraction is not in one of the focus points, but in the geometric center. Cool huh!

Really? An inverse attraction law (1/r) creates an elliptical orbit with the body at the geometric center of the ellipse? I would be surprised if it were the case, but I haven't worked out the math...

As for the proof, isn't it enough that a spring or a pendulum makes a harmonic motion?

I don't see how that is sufficient. It does show that the object will oscillate harmonically given a straight path through the center, but it doesn't really say anything about 2-D orbits...

 

[I like Serena]

Really? An inverse attraction law (1/r) creates an elliptical orbit with the body at the geometric center of the ellipse? I would be surprised if it were the case, but I haven't worked out the math...

Not inverse - linear (~ r).

I don't see how that is sufficient. It does show that the object will oscillate harmonically given a straight path through the center, but it doesn't really say anything about 2-D orbits...

Parametric equation of an ellipse is:
x = a cos phi
y = b sin phi

(Need I say more?)

 

[cjl]

Not inverse - linear (~ r)

Ahh, good point. I wasn't thinking there.

Parametric equation of an ellipse is:
x = a cos phi
y = b sin phi

Yes, but I'm still not sure that is sufficient...

(I might try to work this out on paper in a bit, just to convince myself)

 

[I like Serena]

Yes, but I'm still not sure that is sufficient...

(I might try to work this out on paper in a bit, just to convince myself)

I you go from north to south as sophiecentaur suggested, you'll get a nice sin wave (if the Earth would be an isolated body ).

Now let's add a deviation in a perpendicular direction.
Hey, that will make a sin wave a well, which is independent and possibly shifted in phase and with a different amplitude...

 

[cjl]

I you go from north to south as sophie_centaur suggested, you'll get a nice sin wave (if the Earth would be an isolated body ).

Now let's add a deviation in a perpendicular direction.
Hey, that will make a sin wave a well, which is independent and possibly shifted in phase and with a different amplitude...

That does seem to make sense. That's an interesting result...

 

[I like Serena]

That does seem to make sense. That's an interesting result...

Cool huh?

 

[sophiecentaur]

OMG, it's a lissajou figure! Durrrrrrr, how dumb of me!
As I like Serena says, it's got to be an ellipse around the centre - not a focus. The timing round the ellipse is different from the timing round a planetary orbit, though- it's symmetrical for a start. No Keppler Law at work there.

 

[Malibuguy]

If there were a hollow center in the earth, then an object placed in the center would experience the gravity of the surrounding mass of the earth. Gravity would be pulling from all around. Sort of like someone pulling on your head and your feet and pulling on your right arm and your left arm.

 

[cragar]

If an object was placed in the middle of the Earth in a hollow cavity, it would not experience any forces they would cancel. Gauss's law.

 

[sophiecentaur]

If there were a hollow center in the earth, then an object placed in the center would experience the gravity of the surrounding mass of the earth. Gravity would be pulling from all around. Sort of like someone pulling on your head and your feet and pulling on your right arm and your left arm.

You are just making that up out of your imagination, I'm afraid. In a spherical hole in the centre of a sphere with a spherically symmetrical mass distribution there would be no forces at all on your body except those of mutual attraction between the various parts of your own body. You can't contradict Mr Gauss' law.

 

[Malibuguy]

Let's say we have several large masses very close together. An object placed in the center would experience the gravitational forces of all of the masses. The object might not have a net movement if the the gravitational forces were balanced but it would feel the pull from each mass

 

[Malibuguy]

What if we divided the Earth into 4 parts of a sphere. Now we placed them very close together. In an instant the four parts would come together again due to gravity. But before the instant they did come
Together-an object placed in the center would feel the gravitational pull of the four large masses. Gravity doesn't cancel itself out.

 

[Malibuguy]

The gravity effect would be pulling inward on all directions. large planets are spherical In shape in part due togravity pulling together at the center! Otherwise spinning planets would fall apart due to centrifugal forces

 

[Malibuguy]

So there is a lot of pressure at the center of the earth

 

[Doc Al]

Let's say we have several large masses very close together. An object placed in the center would experience the gravitational forces of all of the masses. The object might not have a net movement if the the gravitational forces were balanced but it would feel the pull from each mass

Assuming the masses were symmetrically placed about the center, the gravitational field at the center would be zero. So the object wouldn't 'feel' anything. For a spherically symmetric shell of mass one can show that the field is zero everywhere inside the shell.

What if we divided the Earth into 4 parts of a sphere. Now we placed them very close together. In an instant the four parts would come together again due to gravity. But before the instant they did come
Together-an object placed in the center would feel the gravitational pull of the four large masses. Gravity doesn't cancel itself out.

Sorry, that's wrong. Sure, the pieces would attract each other, but they would exert no net gravitational pull on a object placed at the center.

The gravity effect would be pulling inward on all directions. large planets are spherical In shape in part due togravity pulling together at the center! Otherwise spinning planets would fall apart due to centrifugal forces

Again, the pull of various parts of a planet on each other is not in question.

So there is a lot of pressure at the center of the earth

That's true.

 

[Studiot]

Originally Posted by Malibuguy
So there is a lot of pressure at the center of the Earth

That's true.

Are you quite sure?