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TBoz
2004-Nov-19, 07:39 PM
As I move away from the earth, the effect of gravity is reduced. At the earth's surface, gravity is greatest on me. Will gravity decrease if I go below ground and closer to the center of the Earth?
If I could stand at the center of the earth, wouldn't I be weightless because there is no mass at the center but the earth's mass would be all around me? Would I be pulled in all directions and blow up like a balloon?

Anybody seen any web sites that discuss this topic?

Evan
2004-Nov-19, 07:49 PM
At the center of the Earth you would be weightless for the reason you surmise. In fact, if the Earth were made of a thin shell of neutronium and were otherwise completely hollow you would be weightless at all points within. You wouldn't blow up (supposing normal atmosphere).

pghnative
2004-Nov-19, 07:51 PM
For this discussion, let's assume that earth is perfectly spherical and has its mass perfectly distributed.

With these assumptions, if you could descend a magic elevator to the center of the earth, you'd weigh less and less as you descended. That's because it is only the amount of earth below you that has a net gravity force. When you got to the exact center, you would be "pulled in all directions equally", but this would result in weightlessness.

Edited to add: Drat you, Evan --- I need to type faster :D

fossilnut
2004-Nov-19, 09:55 PM
Physics isn't my strong point so forgive a naive question. Wouldn't gravity at the center of the Earth be equal in all directions only if the Earth itself was all of existence? What about the pulling of the Moon, Sun, and the rest of the Universe? The gravity acting on the Earth is not equal from all directions.

George
2004-Nov-19, 10:03 PM
Physics isn't my strong point so forgive a naive question. Wouldn't gravity at the center of the Earth be equal in all directions only if the Earth itself was all of existence? What about the pulling of the Moon, Sun, and the rest of the Universe? The gravity acting on the Earth is not equal from all directions.

Your thinking is correct. It is the sum of all the forces which produces the final result.

However, the force you feel from distance objects are quite weak. The gravitational force gets much weaker for every increase in distance. The inverse square law applies. If you double the distance of an object from you, you will feel (1/2)^2 (or 1/4) the force. Therefore, the distance of celestial bodies are not felt hardly at all.

TBoz
2004-Nov-19, 11:09 PM
I think that if you were in the middle of a hollow planet which consisted of material with black hole strength gravity, your body blow up like a balloon or would be pulled apart if the forces were close enough.

This is because (I think) that the gravity acting on one side of your body would be stronger than the far side (tidal effect). This would have minimul effect in the middle of the earth, but in my hypothetical black-hole-strength case, I believe your body would experience some expansion. Even in the middle of the earth, there would probably be some calculable expansion - theoretically anyway.

George
2004-Nov-19, 11:35 PM
I think that if you were in the middle of a hollow planet which consisted of material with black hole strength gravity, your body blow up like a balloon or would be pulled apart if the forces were close enough.

Wow. You just threw the cow into the "glass of milk" discussion. :)


This is because (I think) that the gravity acting on one side of your body would stronger than the far side (tidal effect). This would have minimul effect in the middle of the earth, but in my hypothetical black-hole-strenght case, I believe your body would experience some expansion. Even in the middle of the earth, there would probably be some calculable expansion - theoretically anyway.
You are correct in a big way. Getting close to a black hole has huge tidal consequences. Once you are in the center of the black hole, however, well...you won't really be able to say much at that "point" (pun intended - it's Friday) :wink:


BTW, welcome to the board. =D>

AT
2004-Nov-20, 12:00 AM
Yay! A physics questions I can answer! .... Wait.... Somone already did.

However, the last question:


I think that if you were in the middle of a hollow planet which consisted of material with black hole strength gravity, your body blow up like a balloon or would be pulled apart if the forces were close enough.

Well, yes or no. The part of me that has taken physics says "If it is a perfect sphere, and you are a point mass, then no." However, I don't know many people that are really point-masses.

Evan
2004-Nov-20, 12:01 AM
The gravity acting on the Earth is not equal from all directions.

Actually it is. The Earth is in free fall therefore the forces must be balanced.

TinFoilHat
2004-Nov-20, 12:32 AM
I think that if you were in the middle of a hollow planet which consisted of material with black hole strength gravity, your body blow up like a balloon or would be pulled apart if the forces were close enough.

You would be wrong. It doesn't matter how dense the material making up the shell is. The net gravitational effect sums to zero for every point inside the shell. You won't be pulled apart or blow up like a baloon, you'll be in free fall, feeling no gravitational effects at all.

Evan
2004-Nov-20, 12:53 AM
Think in terms of the rubber sheet analogy with a weight depressing it to make a funnel shape. For a hollow planet you could replace that weight with a ring. The field lines would slope down to the ring edge but inside the ring it would be perfectly flat.

mickal555
2004-Nov-20, 03:40 AM
Welcome to BABB TBoz,

I heard if you could dig a whole straight though the center on the earth and jumped in once you were at wou would have accerlerated to 11 km sec so therefore you would keep going and actully reach the other side only to fall back down again sounds like fun though.

Damburger
2004-Nov-20, 06:46 AM
From what I recall, gravity is greatest at the surface of Earth, dropping off according to the inverse square of the distance as you leave the surface. However, as you descend towards the centre of the Earth, gravity will drop off in a linear way instead.

This is from memory btw, I'd try and prove it but I've been up all night doing maths and I'm pretty much mathed out :-?

Brady Yoon
2004-Nov-20, 07:52 AM
Actually it is. The Earth is in free fall therefore the forces must be balanced.

I thought objects were stationary when forces were balanced.

Evan
2004-Nov-20, 08:08 AM
???

"An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force."

Newton's first law.

See here (http://www.physicsclassroom.com/Class/newtlaws/u2l1a.html)


The Earth is falling around the sun. It is following a straight line in the gravitational potential field of the space around the sun (including the myriad of other minor gravitational forces from planets etc.). I did neglect to mention that within a hollow sphere there would still be tidal forces since matter of any kind, even hypothetical neutronium, is not a "gravity shield". If you were drifting within a hollow Earth orbiting the sun the tidal forces (clarification: from external objects) would cause you to drift towards the center of mass, the center of the hollow Earth.

AT
2004-Nov-20, 04:11 PM
You would be wrong. It doesn't matter how dense the material making up the shell is. The net gravitational effect sums to zero for every point inside the shell. You won't be pulled apart or blow up like a baloon, you'll be in free fall, feeling no gravitational effects at all.

True, if you are a point mass; the net gravitational force on your body is zero; however, if you are on a scale where arn't a point mass, you'd have problems, as the net force on your left hand would be opposite than on your right. I'm told by Bio majors that connective tissue isn't infinitly strong, so.....

Glom
2004-Nov-20, 04:18 PM
Inside a spherical shell, the net gravitational force from the shell's mass is zero at all points on your body so tidal forces are not an issue. Tidal forces are the result of gravitational forces being different and different points on a body because the body takes up space and so parts are significantly further away than others, hence differential weight. However, in the spherical shell, it's all zero so no tidal forces.

Evan
2004-Nov-20, 04:50 PM
Glom is correct, the only tidal forces (as I mentioned above) are from other gravitational sources external to the shell.

TBoz
2004-Nov-20, 07:06 PM
Inside a spherical shell, the net gravitational force from the shell's mass is zero at all points on your body so tidal forces are not an issue.

So as I drifted around inside this hypothetical hollow sphere and moved real close to the "side," I wouldn't experience any more gravitational pull on the side of my body closest to that edge? It would seem that as I drifted close to a side, that now closer side would have an increased gravitational pull and the sides further away would have less pull, in essence "sticking" me to that side, or at least having SOME tidal effect on my body. This is all assuming no outside forces.

Where am I wrong?

Glom
2004-Nov-20, 07:29 PM
Consider a uniform spherical shell. A mass, m, is inside at a distance d from a small segment of the shell, dA, subtending a solid angle, dS, at the mass. dA = dS d. Let's say that the shell has a density, s. The gravitational attraction on m is Gm.s.dS.d/d = Gm.dS. In other words, gravitational attraction from a segment of the shell is independent of the position of the mass. You can think of it as the far side may have a weaker gravitational attraction per unit mass, but there's more mass.

TBoz
2004-Nov-20, 07:40 PM
You can think of it as the far side may have a weaker gravitational attraction per unit mass, but there's more mass.

That's the part of your explanation I understood . . . thanks for the clarification.

What a great forum I have stumbled onto.

Evan
2004-Nov-20, 07:49 PM
Here is a graphical illustration of why it is so. The closer you get to the shell the less mass there is "below" the line and the more above it. The line intersects the shell where the shell mass is pulling the object "left" and "right" in both cases.

http://vts.bc.ca/pics/grav1.jpg

Tobin Dax
2004-Nov-21, 07:44 AM
???

"An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force."

Newton's first law.

See here (http://www.physicsclassroom.com/Class/newtlaws/u2l1a.html)


The Earth is falling around the sun. It is following a straight line in the gravitational potential field of the space around the sun (including the myriad of other minor gravitational forces from planets etc.). I did neglect to mention that within a hollow sphere there would still be tidal forces since matter of any kind, even hypothetical neutronium, is not a "gravity shield". If you were drifting within a hollow Earth orbiting the sun the tidal forces (clarification: from external objects) would cause you to drift towards the center of mass, the center of the hollow Earth.

But still, Evan, your implication that all gravitational forces are balanced is incorrect. The earth's centrifugal force is is not gravitational (at least not in origin, coming from the spin of the molecular cloud), and this is one of the forces balanced by the others.

Evan
2004-Nov-22, 03:37 AM
What? Centrifugal "force" is not a force.

"spin of the molecular cloud"?????

Please clarify your objection.

Tobin Dax
2004-Nov-22, 04:19 AM
I meant to say centripetal force. :oops:

In any case, my objection is that not all forces acting on the earth are gravitational. Our motion around the sun is associated with another force, a non-gravitational force. Therefore, the gravity acting on Earth is *not* equal from all directions. This is the point that you made earlier which I am refuting. Think of a 2-body system. Gravity is definitely not the only force involved there. Add bodies and my point still remains valid.

Ut
2004-Nov-22, 04:22 AM
I meant to say centripetal force. :oops:

In any case, my objection is that not all forces acting on the earth are gravitational. Our motion around the sun is associated with another force, a non-gravitational force. Therefore, the gravity acting on Earth is *not* equal from all directions. This is the point that you made earlier which I am refuting. Think of a 2-body system. Gravity is definitely not the only force involved there. Add bodies and my point still remains valid.

Err, uhh...

Assuming both bodies in a 2 body system are electrically neutral, then gravity is the only force at play. Gravity is the centripital force you speak of.

I think the point you're trying to make, and that you're picking nits with, is that there are other bodies contributing to the net force.

Evan
2004-Nov-22, 04:33 AM
Centripetal, centrifugal. They are not forces. There is gravitation. There is the electroweak force. There is the strong force. That's it to our knowledge. The so called centripetal "force" is the consequence of an object being perturbed from the path it would take if not so perturbed. The Earth in orbit around the sun is not so perturbed.

Kaptain K
2004-Nov-22, 04:59 AM
Think of a 2-body system. Gravity is definitely not the only force involved there.
Wrong!!! [-X Momentum is not a force!

Tobin Dax
2004-Nov-22, 08:53 AM
I meant to say centripetal force. :oops:

In any case, my objection is that not all forces acting on the earth are gravitational. Our motion around the sun is associated with another force, a non-gravitational force. Therefore, the gravity acting on Earth is *not* equal from all directions. This is the point that you made earlier which I am refuting. Think of a 2-body system. Gravity is definitely not the only force involved there. Add bodies and my point still remains valid.

Err, uhh...

Assuming both bodies in a 2 body system are electrically neutral, then gravity is the only force at play. Gravity is the centripital force you speak of.

I think the point you're trying to make, and that you're picking nits with, is that there are other bodies contributing to the net force.

Well that's what caused my confusion, but I'm still wrong. :oops: (My brain must already be on Thanksgiving vacation.) A five-second force diagram in my head didn't balance with other planets involved, but I see my mistake now. That's what I get for not thinking it through.

omypelt
2004-Nov-22, 03:25 PM
Here is a plot of the earth's gravity as a function of distance from the earth's center, calculated using the PREM model of the earth's density distribution:

gravity.png (http://www.cse.unl.edu/~csiedell/other/gravity.png)

It's notable how constant the gravitational pull is through most of the mantle.

Evan
2004-Nov-22, 03:43 PM
This thread has me thinking. Is there any possible consequence from the fact that the core is essentially in a low gee environment? I can't think of any.

George
2004-Nov-22, 04:47 PM
This thread has me thinking. Is there any possible consequence from the fact that the core is essentially in a low gee environment?

Yes. It is a perfect analogy of my business role. I am under tremendous pressure but get no sense of any one direction. :-?

A Thousand Pardons
2004-Nov-24, 09:14 PM
This thread has me thinking. Is there any possible consequence from the fact that the core is essentially in a low gee environment? I can't think of any.
omypelt's info shows that the gravity at the surface of the core is about the same as at the surface of the earth itself. That is a consequence of the higher density of the core.

Evan
2004-Nov-24, 09:19 PM
Yes, it also produces a non-linear drop off through the mantle. What does that do to convection in the mantle?

A Thousand Pardons
2004-Nov-28, 06:49 PM
Yes, it also produces a non-linear drop off through the mantle. What does that do to convection in the mantle?
No, it stays fairly constant throughout the mantle, with small variations, down to the core-mantle boundary, where it is a little greater than at the surface of the earth.

Evan
2004-Nov-28, 06:53 PM
Constant is non-linear. Any ideas?

A Thousand Pardons
2004-Nov-28, 07:43 PM
Constant is non-linear. Any ideas?
Constant is linear. :)

But I was referring to the "drop off" anyway.

Matt McIrvin
2004-Nov-29, 12:15 AM
Physics isn't my strong point so forgive a naive question. Wouldn't gravity at the center of the Earth be equal in all directions only if the Earth itself was all of existence? What about the pulling of the Moon, Sun, and the rest of the Universe? The gravity acting on the Earth is not equal from all directions.

Your thinking is correct. It is the sum of all the forces which produces the final result.

However, the force you feel from distance objects are quite weak. [...]

Also, consider that the Earth is already freely falling under the influence of all those external gravitational forces. If you were at the center of mass of the Earth, you would be freely falling in exactly the same manner (assuming a perfectly uniform Earth). So you wouldn't feel them, except for the extremely weak tidal forces arising from the variation of the gravitational pull over the extent of your body.

Jerry
2004-Nov-29, 12:58 AM
If you drill a hole through center of the earth, and jumped in, how you would fall depends upon where the hole is. If it is near the equator, your motion in the direction the earth is spinning will pin you against the side in a hurry. You will slide down until your angular momentum is in equalibrium with the force of gravity, at which point you will kinda hang there in geosinc orbit inside the earth...

...if the hole is at one of the poles, you will fall, with your momentum carrying you up to the south pole...neglecting all the usually effects, like bouyancy...which are kinda important, so the dampening would kill the ride in a few thousand cycles.

JMV
2004-Nov-29, 01:08 AM
I've understood most of the explanations regarding gravity inside earth, but now you have to clear one thing up for me.

When an object is in orbit around the sun, it is NOT accelerating towards the sun making its path of travel curved but "It is following a straight line in the gravitational potential field of the space around the sun"

Because if it was accelerating, there would have to be some unbalanced force acting upon it, right? But there isn't, right? So how can object's trajectory curve without it accelerating? Does it have something to do with the choice of reference frames in gravitational fields or some other einsteinishly complex idea? I didn't get the quoted part in previous paragraph.

This is a whole new thing for me. Please explain.

martin
2004-Nov-29, 01:47 AM
I've understood most of the explanations regarding gravity inside earth, but now you have to clear one thing up for me.

When an object is in orbit around the sun, it is NOT accelerating towards the sun making its path of travel curved but "It is following a straight line in the gravitational potential field of the space around the sun"

Because if it was accelerating, there would have to be some unbalanced force acting upon it, right? But there isn't, right? So how can object's trajectory curve without it accelerating? Does it have something to do with the choice of reference frames in gravitational fields or some other einsteinishly complex idea? I didn't get the quoted part in previous paragraph.

This is a whole new thing for me. Please explain.

The answer depends on whether you use the physics of Newton or of Einstein. To understand the motion of a planet around the sun, both are sufficient (there is a small difference in the behaviour of mercury between the two systems of physics). In the physics of Newton, it is accelerating towards the sun because of the force of gravitation. In the physics of Einstein, there is no force of gravitation. The presence of mass like the sun makes the space in its neighbourhood no longer obey the ordinary geometry of Euclid. In this geometry, if you travel in a straight line, you can come back to where you started this journey. Then the orbitting object is not accelerating, but moving in a straight trajectory through a curved space. Choice of reference frames is for convenience, the behaviour of the object does not depend on this choice.

Martin

Zjm7891
2004-Nov-29, 01:51 AM
(In Newtonian Physics) The Earth is accelerating around the Sun, as acceleration is a vector quantiy and therefore has a direction and since the direction of the Earth's motion is changing it therefore has acceleration...

I think..

Edit: Crap someone beat me!

martin
2004-Nov-29, 03:36 AM
(In Newtonian Physics) The Earth is accelerating around the Sun, as acceleration is a vector quantiy and therefore has a direction and since the direction of the Earth's motion is changing it therefore has acceleration...

I think..

Yes, the acceleration is in direction of the sun. This is perpendicular to direction of motion if the orbit is circular. If the orbit is elliptical, then it can be not exactly perpendicular. This is by Newtonian physics, by Einsteinian physics, then there is no acceleration, the effect is accomplished by bending of space near the sun. In Newtonian physics, to take a simple case of circular orbit, then distance d is fixed. If the orbit is in x-y plane with the sun at x=y=0 (because the sun is much larger than earth, the effect of earth's gravity on sun is minimal), then:

x(t) = d*cos(2*pi*t)
y(t) = d*sin(2*pi*t)

gives position of earth, where t is time in years (sine and cosine functions take inputs in radians). Velocity vector is given by first derivatives:

x'(t) = -2*pi*d*sin(2*pi*t)
y'(t) = 2*pi*d*cos(2*pi*t)

Acceleration vector is from second derivatives:

x''(t) = -4*pi*pi*d*cos(2*pi*t)
y''(t) = -4*pi*pi*d*sin(2*pi*t)

Components of acceleration vector are just negative (multiplied with 4*pi*pi) of components of position vector. Since the sun is at origin the acceleration vector points straight to sun. The acceleration is perpendicular to velocity (x''*x'+y''*y' = 0). This perpendicularity fails when the orbit is elliptical. This circumstance is much more complicated; I do not know how to express position as function in time, but can express time and distance from origin taking angle as independent variable in polar coordinate system. Also, it is possible to use three dimensions, but the orbit still exists in a plane in three-dimensional space. So a third dimension makes the maths more complicated, but adds nothing in way of concept.


Edit: Crap someone beat me!

I am not beating anyone!

Martin

JMV
2004-Nov-29, 11:39 AM
I think I see it now. Thanks for your input.