Gravity at the center of the Earth

[Jozsef]

Imagine that you could dig a hole up to the very center of the Earth and sit there. What will you experience?
Complete weightlessness or the experience of being torn apart in every direction?

 

[Orodruin]

What are your own thoughts on the matter and what is your level of physics?

 

[A.T.]

Imagine that you could dig a hole up to the very center of the Earth and sit there. What will you experience? Complete weightlessness or the experience of being torn apart in every direction?

To feel "torn apart" you need a strong gravity gradient. In a spherical cavity at the center the gravity gradient is zero.

 

[WhatIsGravity]

Aside from the tidal forces of the moon, sun, the geography of the Earth, Jupiter and such... the tidal forces... there is zero gravity at the center of the earth. Trippy.

 

[phinds]

A.T. and WhatisGravity, seems to me that Orodruin's response was more in keeping with the spirit of this forum, of helping people learn to think for themselves rather than spoon feeding them answers.

 

[berkeman]

A.T. and WhatisGravity, seems to me that Orodruin's response was more in keeping with the spirit of this forum, of helping people learn to think for themselves rather than spoon feeding them answers.

+1.0

 

[WhatIsGravity]

Phinds, I disagree, but I do wish what you said was right.

 

[phinds]

Phinds, I disagree, but I do wish what you said was right.

Well, I know it's a battle for some of us (me at least) to keep in mind that we're here to help others, not to show off how much we know and I agree w/ you that plenty of us lose the battle sometimes but it's still worth trying to remember.

 

[berkeman]

Phinds, I disagree, but I do wish what you said was right.

It's important that you understand the philosophy of the PF. It's a great place to learn, and one of the reasons is the PF philosophy of helping folks learn how to learn. Please be sure to read through the PF rules (under Site Info at the top of the page). That certainly is the approach that is required by the rules in the Homework Help section of the PF, and it tends to spill over to the technical forums as well.

 

[PhysicistMike]

Imagine that you could dig a hole up to the very center of the Earth and sit there. What will you experience?

Let's model the Earth as being a geometric sphere but not having uniform mass density. However we will assume that the mass density is spherically symmetric.

Let's first determine the gravitational field at the center of the earth. This is easy to see by symmetry. Due to the symmetry of the sphere the direction of the force at the center cannot have any direction since it would break the symmetry. The only force able to satisfy that requirement is F = 0.

Now think of a force at r = R. This defines a sphere with radius R which is surrounded by another sphere. The field inside the sphere due to the outer sphere is zero since the field inside any spherically symmetric shell is zero. The field is now totally due to the matter inside the sphere r = R. Due to symmetry that force must be radial and since the gravitational field due to such a sphere is that of a point of the same mass then the force is radially inward. So all points on your body experience a force directed radially inward with varying strength of the force. That's called a tidal force. This kind of tidal force crushes you towards the center of the Earth and squeezes you along your sides.

I hope that helps.

 

[Orodruin]

So since the issue was mentioned, let me give my view on the matter:

There are a few ways of replying to questions like this. One would be to simply blurt out the answer, which I do not doubt that anyone who regularly attend this forum would be able to do. While this teaches the OP a bit of information, it gives no insight to why this happens.

A second way would be to give the OP the answer along with some argumentation as to why it is so. The problem with this is that the explanation I give may not be at an appropriate level for the OP to understand it. Is it reasonable to assume that someone who does not already know the answer to this question knows what a gradient is? I would not want to do so.

So, I chose the third option, asking the OP for his thoughts on the matter and his level of physics. A reply to these questions would have provided the OP with the opportunity to formalize any thoughts about it, given us an opportunity to start from his/her current knowledge and to correct any misconceptions about how (Newtonian) gravity works. Even if this is not the homework forum, I think it is a reasonable approach to this type of questions, there is really not much to discuss about it anyway.

"Give a man a fish and you feed him for a day. Teach a man to fish and you feed him for a lifetime."


Now, that being said ...

Let's model the Earth as being a geometric sphere but not having uniform mass density. However we will assume that the mass density is spherically symmetric.

Let's first determine the gravitational field at the center of the earth. This is easy to see by symmetry. Due to the symmetry of the sphere the direction of the force at the center cannot have any direction since it would break the symmetry. The only force able to satisfy that requirement is F = 0.

Now think of a force at r = R. This defines a sphere with radius R which is surrounded by another sphere. The field inside the sphere due to the outer sphere is zero since the field inside any spherically symmetric shell is zero. The field is now totally due to the matter inside the sphere r = R. Due to symmetry that force must be radial and since the gravitational field due to such a sphere is that of a point of the same mass then the force is radially inward. So all points on your body experience a force directed radially inward with varying strength of the force. That's called a tidal force. This kind of tidal force crushes you towards the center of the Earth and squeezes you along your sides.

I hope that helps.

The second part of this answer is a bit misleading as it suggests the person would be crushed (he will be when the cavity collapses, but from the pressure of the surrounding material and that is beside the point). If there would be a spherical cavity in the center of the earth, surrounded by spherical mass shells, then the gravitational field would be zero inside all of the cavity for the very same reasons you gave above. I would leave it at that in order not to confuse the OP.

 

[Jozsef]

Let's model the Earth as being a geometric sphere but not having uniform mass density. However we will assume that the mass density is spherically symmetric.

Let's first determine the gravitational field at the center of the earth. This is easy to see by symmetry. Due to the symmetry of the sphere the direction of the force at the center cannot have any direction since it would break the symmetry. The only force able to satisfy that requirement is F = 0.

Now think of a force at r = R. This defines a sphere with radius R which is surrounded by another sphere. The field inside the sphere due to the outer sphere is zero since the field inside any spherically symmetric shell is zero. The field is now totally due to the matter inside the sphere r = R. Due to symmetry that force must be radial and since the gravitational field due to such a sphere is that of a point of the same mass then the force is radially inward. So all points on your body experience a force directed radially inward with varying strength of the force. That's called a tidal force. This kind of tidal force crushes you towards the center of the Earth and squeezes you along your sides.

I hope that helps.

Many thanks to PhysicistMike. At least a clear and comprehensive answer. Your answer also helps me to better understand another issue : degeneracy pressure. Strange, how an insight in one problem opens yours eyes for another. But no doubt, this will be the mean objective of PF. For the record only, 'm nothing but a Belgian doctor - surgeon with an insatiable appetite for physics and cosmology and never "spoon fed". Respectfully, Jozsef

 

[f '(x)]

Here's your answer in a video with illustrations

 

[sophiecentaur]

Imagine that you could dig a hole up to the very center of the Earth and sit there. What will you experience?
Complete weightlessness or the experience of being torn apart in every direction?

I have only just found this thread and I just spotted this idea. I think it's worth commenting on, on its own.
That intuitive idea is quite a reasonable one until you realize that things are only 'torn apart' when the forces on their component parts are strong and, locally, in different directions. Whatever the fields inside something are (s long as it's not a black hole, where all bets are off) they will not vary very much at all from your head to your feet - so no strong forces in different directions. In any case, the effect of all those forces in different directions is to produce a net resultant of (near) zero. If you were inside a strong spherical container at the centre of an ideal symmetrical planet, the only relevant g forces would be those between the parts of your own body, they would tend to keep you together. If you were liquid, you would eventually form a spherical shape, in fact.

 

[Jozsef]

I have only just found this thread and I just spotted this idea. I think it's worth commenting on, on its own.
That intuitive idea is quite a reasonable one until you realize that things are only 'torn apart' when the forces on their component parts are strong and, locally, in different directions. Whatever the fields inside something are (s long as it's not a black hole, where all bets are off) they will not vary very much at all from your head to your feet - so no strong forces in different directions. In any case, the effect of all those forces in different directions is to produce a net resultant of (near) zero. If you were inside a strong spherical container at the centre of an ideal symmetrical planet, the only relevant g forces would be those between the parts of your own body, they would tend to keep you together. If you were liquid, you would eventually form a spherical shape, in fact.

Thanks to f'(x) and sophiecentaur. Jozsef

 

[deadcat]

Here's your answer in a video with illustrations

Why would he fall faster when he gets to the outer core? I have 2 reasons for thinking otherwise if anyone could explain how this works a little better.

1. The Earth above him at this point would pull on him, slowing him down I would think
2. At the surface, their is more mass below him. I would think this would equate to more intense gravity rather than less

I'm sure there is a good reason I would be interested to know more about it.

 

[Bandersnatch]

Why would he fall faster when he gets to the outer core? I have 2 reasons for thinking otherwise if anyone could explain how this works a little better.

1. The Earth above him at this point would pull on him, slowing him down I would think
2. At the surface, their is more mass below him. I would think this would equate to more intense gravity rather than less

I'm sure there is a good reason I would be interested to know more about it.

Recall the law of gravity. It's not only about how much mass is there, but also how far away from that mass you are.
Can you figure out how it alters the two statements you've presented above?

 

[A.T.]

Why would he fall faster when he gets to the outer core?

He would accelerate the most at boundary of the outer core, because the Earth is not of uniform density:

I have 2 reasons for thinking otherwise if anyone could explain how this works a little better.
1. The Earth above him at this point would pull on him, slowing him down I would think
2. At the surface, their is more mass below him. I would think this would equate to more intense gravity rather than less

This is one and the same reason, expressed in two different ways. But it doesn't take non-uniform density into account.

 

[Borek]

So far nobody mentioned http://en.wikipedia.org/wiki/Shell_theorem

 

[Jozsef]

He would accelerate the most at boundary of the outer core, because the Earth is not of uniform density:

This is one and the same reason, expressed in two different ways. But it doesn't take non-uniform density into account.

It is amazing to realize how a question, originally perceived as "stupid" ( see start of this thread) finally results in unchained replies, each of which with its own personal considerations. This diversity in approach is surely very instructive. This strengthens my experience that the mental approach to the same physical problem can differ for anyone only to result in the same conclusions. Again, many Thanks to everyone. Jozsef

 

[A.T.]

It is amazing to realize how a question, originally perceived as "stupid" ( see start of this thread)

Who said it was "stupid"? It's a very common one. See "Related Discussions" at the bottom of this page.

 

[sophiecentaur]

Why would he fall faster when he gets to the outer core? I have 2 reasons for thinking otherwise if anyone could explain how this works a little better.

1. The Earth above him at this point would pull on him, slowing him down I would think
2. At the surface, their is more mass below him. I would think this would equate to more intense gravity rather than less

I'm sure there is a good reason I would be interested to know more about it.

We are assuming that the Earth is arranged in spherical 'shells' (spherically symmetrical), which are not all of the same density. It's one step beyond the simple model of a uniform density sphere.
1.This is not what happens. The sum total of the effects of the shells of the Earth which are further out than the falling object is Zero. There is no net "pulling" in any direction. This 'Shell Theorem" is well established and works for gravity or electric charge, arranged in shells.
2. This is much more like what happens.

 

[Borek]

It is amazing to realize how a question, originally perceived as "stupid" ( see start of this thread)

Nobody ever ridiculed your question. What some of us did we pointed to those answering that the way they answered was inconsistent with the site policy (despite their answers being correct in the technical sense).

 

[Jozsef]

Nobody ever ridiculed your question. What some of us did we pointed to those answering that the way they answered was inconsistent with the site policy (despite their answers being correct in the technical sense).

Correct.
I 'am very grateful to you for being admitted to this wonderful forum. Jozsef

 

[PhysicistMike]

So far nobody mentioned http://en.wikipedia.org/wiki/Shell_theorem

I spoke of it in regards to the concept and results, just not with respect to the name. From the way the question was phrased I assumed that the person asking it didn't have an understanding of math at the level required to read a link such as that.

 

[PhysicistMike]

Many thanks to PhysicistMike.

You're most welcome, Jozsef. :)

For the record only, 'm nothing but a Belgian doctor - surgeon with an insatiable appetite for physics and cosmology and never "spoon fed".

Really? Nice! I just might have a lot of questions in the future for you, if you don't mind that is. :)

 

[Borek]

I spoke of it in regards to the concept and results, just not with respect to the name. From the way the question was phrased I assumed that the person asking it didn't have an understanding of math at the level required to read a link such as that.

You don't need math to understand the most important conclusions - and IMHO it is a result nice enough to know it has a name

 

[Jozsef]

You're most welcome, Jozsef. :)


Really? Nice! I just might have a lot of questions in the future for you, if you don't mind that is. :)

Always welcome with all your questions. To the extent of our current medical knowledge, you will be answered. Jozsef