At Center of Earth: Weightless or Crushed?

[HallsofIvy]

I had read in a book that gravity decreases as you approach the center of the Earth because you feel only the Earth under you. What's all this about being pulled by gravity in all directions?

The gravity form the Earth below you as well as the Earth on the "other side" of the center pulls you down but the Earth above you pulls you up. It can be shown that the pull of the Earth above you exactly (assuming a uniform earth) cancels the pull of the Earth on the other side of the center so the only pull is that of the Earth still "below" you.

 

[Cyrus]

I have another good question, (at least I find it puzzling). I agree that the pull of the Earth on either side of you will be equal and opposite so you won't go anywhere. But to a rigid body, it WILL feel too forces trying to pull it appart internally. But at the same time, the body will also feel the force pushing inwards due to the pressure of all the material above it. Do you think these two forces would be exactly equal in magnitue and opposite in direction, and so you really would feel weightless, despite all the hydrostatic pressures? You wouldent be crushed due to the weight maybe.

 

[Doc Al]

I agree that the pull of the Earth on either side of you will be equal and opposite so you won't go anywhere.

The gravitational force on a body at the Earth's center will be zero.

But to a rigid body, it WILL feel too forces trying to pull it appart internally.

What forces are you talking about?

But at the same time, the body will also feel the force pushing inwards due to the pressure of all the material above it.

As this discussed throughout this thread.

Do you think these two forces would be exactly equal in magnitue and opposite in direction, and so you really would feel weightless, despite all the hydrostatic pressures? You wouldent be crushed due to the weight maybe.

I don't know what you mean by two forces. A body will feel the pressure of the Earth pushing inward, and the body's structure will provide the reaction force pushing back on the earth.

 

[Cyrus]

Well, there will be two forces due to gravity won't there. You will have the gravitational force of all the mass above you pulling you upwards, and you will also have a gravitational force of all the mass below you pulling you downwards. The sum of these two forces add to zero. But for a body at the center, it will not be accelerated in any direction since the net force is zero, but that does not mean that it won't feel a tug on each side where the force is present. I am referring to someones post where two people tug on your arms. The net force is zero, but both of your arms sure do feel a force in either direction. By analogy, the force tugging on your arms could be the force of gravity above and below you. You don't go anywhere, because the net force is zero, but won't you feel it on both sides.

 

[Doc Al]

Well, there will be two forces due to gravity won't there.

No, there will be NO force due to gravity.

You will have the gravitational force of all the mass above you pulling you upwards, and you will also have a gravitational force of all the mass below you pulling you downwards. The sum of these two forces add to zero. But for a body at the center, it will not be accelerated in any direction since the net force is zero, but that does not mean that it won't feel a tug on each side where the force is present.

It's not just that the net gravitational force on a body will be zero, it's that the net force on every tiny piece of that body is zero. There is no "tug", there is no gravitational force. Another way to say it: there is no gravitational field at the Earth's center, so no gravitational force.

I am referring to someones post where two people tug on your arms. The net force is zero, but both of your arms sure do feel a force in either direction. By analogy, the force tugging on your arms could be the force of gravity above and below you. You don't go anywhere, because the net force is zero, but won't you feel it on both sides.

A bad analogy. There is no gravitational pull tugging you at all.

What post are you referring to?

 

[Cyrus]

someone earlier said something about that in tab 1 of this post. You should see it there.

 

[Doc Al]

Perhaps you're thinking of Hurkyl's analogy in post #10? If so, he's saying something quite different.

 

[GOD__AM]

I imagine the specific gravity of liquid nickel is very high (8.902?). If you were to make it to the center of the earth, you would likely be forced towards the upper layers of the liquid mantle.

Edit: Solid inner core? ...my bad sorry.

 

[Cyrus]

hey doc al, maybe I am just interperting something incorrectly then. I thought that Newtons law states that $$F = \frac{GmM} {R^2}$$. Let's do a simple case. Let's say I am at the center of the earth, and there is just a column of Earth above my head and below my feet that all has the same uniform density. The weight that is trying to push down on my head will equal the weigh trying to squash me from me feet, I think we both agree on that point. And that force would just equal the sum of the mass at its distance, times the gravity at that distance, added together over the distance. That would be the force pushing me down from my head, likewise there would be that force equal in magniude present trying to push me from my feet as well.

Now for the gravity. I understand that the NET gravitational force I will feel will be any mass that is below me and is unacounted for above me. What i mean is that if i don't stand directly in the middle of that column of earth, there will be more Earth one one side of me ( either above or below). As a result, there will not be a proper balance of attraction between myself and all the particles of earth, so there will be a net force in the direction where there is the larger amount of earth, and I will feel like I want to fall in that direction.

If I were in the center, there would be equal mass and distance both above and below me, so I would not feel any net gravitational force. You said it does not have to do with net gravitational force, but I do not see how you can avoid it. I thought it is TOTALY dependent on having that net force cancel out. For every particle trying to pull me towards it above my head, there is the same amount of force trying to pull me towards the column of Earth below my feet, and for the reason, the NET force on me is zero, and I go nowhere. But that does not necesairly mean that i would not feel myself trying to be pulled both up and down does it?


Hmm, now after all this typing I kind of see what you mean .haahhaaha. I will post it anyways. (SIGHHHHHHHHHHH) . If every particle on my body is feeling as thought it is being pulled both up and down, then as a whole, each particle in my body relative to each other, don't want to go anywhere.

I think what I was thinking about would be a gradient across my body. Because I take up space, I could not really be a point masss at the center, therefore, Techincally speaking, I should feel some small component of force from my head down to my toes. The thing was that I was considering the entire space I occupied as the center. In reality the center is a point, so a rigid body like me can't be the center, by definition, some parts of my body will be off center, by 3 feet or so on each side, I am a tall guy :-). So the parts of my body that are off center should feel some force due to the unbalance of force above and below it, but were talking about maybe 3 feet difference at most, I would think this to be a very very very very small force difference, and in affect, so small I would still think that my entire body is weightless. Thanks Doc Al you were right. I did not think about the situation hard enough.


As a side note, Shouldent that single particle feel a force on each side trying to pull it appart like the anlogy I made earlier? Would this force be strong enough to rip electrons off of that point mass, and be a reason why the center of the Earth would have a liquid plasma? Or am I still looking at the situation incorrectly.

Edit: I just made the same exact mistake to this situation. Each electron relative to the atom will not experience a great force tugging at it, thus this cannot be the reason for the plasma. The pressure has to be the only result of the plasma. great pressure = great temperature = energy to rip electrons from their orbits and turn stuff into a plasma state.

 

[LURCH]

Perhaps the problem here is that all points within your body cannot be in the exact center of thwe Earth; some will be above it , and some below. Is that the heart of your question?

 

[Tide]

I think Cyrus is trying to describe tidal forces! :-)

They will be extremely weak however as the tidal force varies as the gradient of the gravitational force.

 

[Cyrus]

what is a tidal force? Its 12:55 am, I am studying general chemistry. I just went to 7-11 and bought a sobe energy drink and now I am wired and good to go for another two hours. Tommorow morning is going to suck! ;-P Anyhoo, I would think that if i were off center from the center of the Earth by, say one atom, then the only force due to gravity would be the equivelant of the force between me and that atom and our distance squared, since all the other mass' gravitaional affects can be ignored, since they have a counterpart on the other side of me. So it really would be a weak force even if i am off center by just a tiny amount no? I made this into a hot thread, 600 views yes!

 

[Tide]

If gravity is pulling more strongly on one part of an object than on another part of the object then the object is effectively being pulled apart due to the differential. That is the tidal force.

 

[Cyrus]

I see. Makes sense to me. :-)

 

[Doc Al]

Perhaps the problem here is that all points within your body cannot be in the exact center of thwe Earth; some will be above it , and some below. Is that the heart of your question?

Even so, imagine a small sphere centered at the Earth's center. Assuming the usual symmetry, within that sphere there will be NO gravitational field due to the Earth's mass outside the sphere. So the only gravity pulling on a body within that (imaginary) sphere will be due to the trivial mass within the sphere itself.

 

[Cyrus]

HI Doc AL, I have a question for you. You said that the only gravity pulling on a body would be within that sphere due to the mass within the shere itself. Could this also be equal to the mass outside the sphere that does not have a counter part, if we take the origion to be the place of displacement of the particle. I attached a pic to show you what i mean. We could do like you said, and account for each point mass inside the blue sphere and its distance from the center of mass. But is it not equally correct to say that we could consider the point that is off center of the sphere to be defined as the new center of the sphere. And this new center point could have a max radius that would be equal to radius of the big sphere, minus the displacement from the center of the big sphere. So that it would now be possible for me to draw a new sphere of smaller radius inside the bigger sphere, centered at the new point. So now all the mass inside my sphere of radius 22mi, has a counterpart on the opposite side to cancel out the force gravitiationaly. The only parts that don't have something to cancel them out would be the highlighted green area. If I did the point mass times the distance squared for each of these green point masses and added them up, shouldent I get the same value of gravitational force as if we did it your way, by using all the point mass inside the blue sphere?

 

[Doc Al]

I think I'm missing your point with your off-center sphere, so I can't comment. Is it a hollow sphere? An imaginary sphere? Beats me. Please note that all of these arguments about zero gravity inside a hollow cavity within the Earth depend on a spherically symmetric distribution of mass.

I'll repeat: Assume a spherically symmetric distribution of mass, of radius R. Now imagine an imaginary sphere of radius r (r < R) co-centered with the big sphere of radius R. The mass outside of the small sphere will exert no gravitational field within the small sphere. So any gravity experienced at any point within the small sphere is entirely due to the mass within that small sphere. (We are of course ignoring any other bodies in the universe.)

 

[Cyrus]

Hey Doc,

Did you click on the picture i attached that displays what I mean. For simplicitys sake let's just think of them as circles in a common plane. At that every point inside the big circle has a point mass that is equal.

According to what you said, my red dot, is a point mass that is not located in the center of the big circle. The black dot IS the center of the big circle. According to you, the gravitational force will equal to the sum of all the point masses inside the circle that I can draw around the red dot, with a center of the black dot. (this corresponds to the blue circle in the picture.) I hope this part is clear.

Now, what I am saying is that, what if i IGNORE the black dot, and consider my RED dot as the center of some arbitrary circle. Then the maxiumim circle I can draw with the center at the RED dot without going outside of my big circle will equal the radius of the big circle, minus the distance from the red dot to the black dot.

Now If I draw this circle, (based on my picture), it has a radius of 22miles. And what I am saying is that at the red dot inside this circle of 22 miles, there should be no gravitaional forces present because for every point inside the circle of 22 miles, there is the same point opposite it to cancel out any net force due to gravity.

So clearly, it can be seen that the only things that can't cancel out, will be the points that lie within the green area. And what I am wondering is that if I sum the distances squared times the GMm for each particle of the green area, and sum it for all of the points in the green space, should this not produce a gravitational force that is equal to the way we do it your way, by including the effects of all the points in the blue circle and adding them up.

It says 0 views on my picture so I think you did not see it in my previous post. Please click on it as it will make things a lot clearer than my confusing text ( sorry about that )

 

[Doc Al]

I get you now.

Yes, I'd say you are correct, assuming that the large sphere has uniform density. If you want to find the gravitational field at point x (red dot) then it is equal to the field from a sphere of mass of radius 3 miles (in your example). But you can certainly draw a sphere around the red dot and say that all mass inside the sphere will contribute no field at the red dot. So the sum of the field from the remaining mass in the big sphere must add up to the same. Very clever!

 

[Cyrus]

YAY, I was about to try and derive some form of a proof of this, but I fast realized that it would be a MAJOR pain in the butt to do LOL! Ill just take your expert word for it and save myself the trouble Another side note, I am just reading on coulombs law right now. And wow is it EXactly the same as Newtons law of gravity, well except for the sign convention. How is it that these two separate things are so AMAZINGLY simliar? I assume Newton came up with his formula first? Becuase coulomb had the added advantage of charging two spheres and seeing how the affect of the force is 1/r^2. but how on Earth could Newton determine its a 1/r^2 force? The forces of gravity are either so minute you can't detect them, or there so large you can't use them experimentally. Newton couldent see the effects of two plannets by moving them close or far appart . I just don't see where Newton got the idea for this formula through practical logic, I do see how coulmb could though. He had an experiment he could play around with.

 

[nrqed]

YAY, I was about to try and derive some form of a proof of this, but I fast realized that it would be a MAJOR pain in the butt to do LOL! Ill just take your expert word for it and save myself the trouble Another side note, I am just reading on coulombs law right now. And wow is it EXactly the same as Newtons law of gravity, well except for the sign convention. How is it that these two separate things are so AMAZINGLY simliar? I assume Newton came up with his formula first? Becuase coulomb had the added advantage of charging two spheres and seeing how the affect of the force is 1/r^2. but how on Earth could Newton determine its a 1/r^2 force? The forces of gravity are either so minute you can't detect them, or there so large you can't use them experimentally. Newton couldent see the effects of two plannets by moving them close or far appart . I just don't see where Newton got the idea for this formula through practical logic, I do see how coulmb could though. He had an experiment he could play around with.

It's actually quite simple (although quite hard to explain without a picture, so I'll just mention the basic idea...if you want more details I'll try to be more specific but I don't have a scanner to show you a hand drawing unfortunately). The basic idea is that Newton understood that an object in orbit around the Earth (or the Sun or whatever) is actually constantly falling. The Moon, for example, is in constant free fall. Knowing the distance to the Moon (The Greeks already knew that!) and that the sidereal month (I won't get into this for now) is about 27.3 days, it's easy to calculate that in one second, the Moon falls about 1.35 millimeter! (what I mean by it falls 1.35 mm is this: Imagine that the force of gravity would cease to exist. The Moon would then keep going along a straight line. But instead it follows a circle (approximately). Now follow the Moon for one second and compare the position it would have if it was flying off along a straight line and the position it actually has. It's easy to use Pythagora's theorem to find the 1.35 mm (it's crucial that it be a small time interval hence a small distance. With large values, say one day, the geometry would be more complicated).


What does this have to do with the 1/r^2 law? well, on the surface of the Earth, an object released from rest (an apple let's say!) falls 4.90 m (1/2 a t^2).

So the apple has an acceleration (acceleration is proportional to distance when released from rest) which is about 4.9/0.00135 = 3630. or about 3600 if I round off a bit. On the other hand, the Moon is 384 000/63 80 = 60 times farther from the center of the Earth than the apple.

But 3600 is 60^2! So being 60 times farther reduces the acceleration (so the force) by a factor of 60^2. Therefore, it's an inverse square law.


Actually, it's also easy to get the same result by using Kepler's third law and assuming circular orbits (a very good approximation for most planets). One simply has to use that the centripetal acceleration is v^2/r and one finds an inverse square law again.

Regards

Pat

 

[Cyrus]

Very Interesting! Thanks Pat. I find it more meaninful to learn ABOUT formulas than to learn formulas. Any trained monkey can use a formula. I find the beauity in how did someone manage to think up such a correct idea mathematically with such hard things to deal with. Those greeks were pretty smart. I think carl sagan was right in saying that if they had had more time, they would have made up process scientifically 1000 years. Its too bad that never happened, maybe warp 9 --> engage, would not be fantasty today :-).

 

[Gara]

I think what Doc Al is saying is the gravity below you would not "only" be pulling on your legs, while the gravity above you would not "only" be pulling your head.

 

[Cyrus]

Pressure is defined as force per unit area, but as we approach the infitesmal point mass with point area at the center of the Earth we would have pressure =force/zero,

but i guess the force would be infinite and the area would be zero, and we would have an indeterminate, of the form infinity over zero, so in oder to find the pressure we would have to use l'hospital ( that's EL hospital's :-) )rule, and that anwser would give us a definite anwser as we did the limit and what not, because obviously there is finite mass and so there will have to be finite pressure.

 

[chroot]

Cyrus:

You're making it way too complicated. At the center of the earth, there is no gravitational force. There is, however, enormous pressure.

- Warren

 

[marlon]

Pressure is defined as force per unit area, but as we approach the infitesmal point mass with point area at the center of the Earth we would have pressure =force/zero,

but i guess the force would be infinite and the area would be zero, and we would have an indeterminate, of the form infinity over zero, so in oder to find the pressure we would have to use l'hospital ( that's EL hospital's :-) )rule, and that anwser would give us a definite anwser as we did the limit and what not, because obviously there is finite mass and so there will have to be finite pressure.

OOOhhhh, wow,
this is real science...
cyrusabdollahi,
listen to the answer of chroot, though, he is right...

Remember that physics is constructed in order to make the world more easy, understand it...

regards
marlon

 

[Cyrus]

Sorry for asking , I will not post in this topic anymore what do you mean by real science? I think the question I posed in terms of an indeterminate is a valid, provided I were to have a function of the area and the force and knew the limits of my integral.

 

[stevo72]

Cyrus:

You're making it way too complicated. At the center of the earth, there is no gravitational force. There is, however, enormous pressure.

- Warren

Sorry but this is not entirely true.

Gravity is "perceived" to be generated by the attraction of one mass against another. A body at the centre of the Earth would be attarcted by, and to, every particle that makes up the Earth around it. Thus gravity is in effect in every direction. However the NET gravitational effect would be ZERO.

Now pressure is created by the acceleration of a body against another body. The NET acceleration created by gravity at the centre of the Earth is zero, for the reason given above. So if only gravity was the cause of pressure, the pressure would be zero. However, their is both heat and motion in the Earth's core caused by friction, chemical reaction and gravity in the layers of the Earth beyond the centre. These energy sources generate acceleration of Earth particles which would apply pressure unevenly to any object at the centre of the Earth. As a result the contents of the Earth core are in constant motion. If these other sources of energy where not in place then the natural shape a planet such as the Earth would take is a hollow shell since a gravity inflexion point would be reached at a certain distance below the surface where there would not be sufficient gravity to hold the particles in place.

 

[lalbatros]

Assuming a liquid planet, the pressure at the center is equal to the gravitational potential at the surface of the planet.

 

[D H]

If these other sources of energy where not in place then the natural shape a planet such as the Earth would take is a hollow shell since a gravity inflexion point would be reached at a certain distance below the surface where there would not be sufficient gravity to hold the particles in place.

Assuming a liquid planet, the pressure at the center is equal to the gravitational potential at the surface of the planet.

Two back-to-back nonsense posts to a five year old thread!

The first of the two is utter nonsense. You are ignoring the pressure from above, stevo72. The second doesn't make sense at all. Energy and pressure are different things with different units.

 

[lalbatros]

Two back-to-back nonsense posts to a five year old thread!

The first of the two is utter nonsense. You are ignoring the pressure from above, stevo72. The second doesn't make sense at all. Energy and pressure are different things with different units.

Pressure is a density of energy.
Indeed: Pa/m² = Pa.m/m³=J/m³.
I was simply suggesting to use the Bernouilli equation for this question.

 

[Cyrus]

Pressure is a density of energy.
Indeed: Pa/m² = Pa.m/m³=J/m³.
I was simply suggesting to use the Bernouilli equation for this question.

I already got an answer to this question, years ago. I don't know why you're using Bernoulli for my question, as I didn't ask anything with regards to Bernoulli...

 

[lalbatros]

I was answering to stevo72, if you permit.

 

[bm0p700f]

Inside a solid shell there is no gravitational field. This come out though analysing Newton law of gravitation. So at the centre of the Earth g = 0.

F = G M1M2/r^2

M1 = mass of planet = (4/3)*pho*r^3
pho is the density of the planet assuming its of uniform denisty. The fact that is not does alter the result handily.
So F = (4/3)*G*M2*r

as F = mg

g = (4/3)*G*r

so as r tends to zero so does g!

Are more complete description using calculus to derive the result from first principles can be at
http://hyperphysics.phy-astr.gsu.edu/hbase/HFrame.html

Also a neat consequence of this is if you had an evacuated tube running through the centre of the earth, wore a spacesuit and jumped the time it takes for you to fall through the Earth to the other side and back again happily equals the time it would take for another intrepid explorer to orbit said planet just above the surface (agian the atmosphere has to pumped away.

So any takers on the ultimate ride? Smooth a path around the moon clearing it of all mountains and bore hole through the centre. Now jump and be shot out of cannon. The real reason why NASA wants to return to the moon. Make it a theme park so the engineers can play. After all what's the point of engineering if you can't do silly things with it.

 

[renek]

Ok listen up all u dopey heads.
You are using the right formula but you all made the same mistake. You would not experience a force of zero lols

Let me demonstrate:

Using Newtons law of universal gravitation

F=GMm/r^2 (which i am sure we all know)

G=Universal Gravitational constant
M=Mass of earth
m=mass of you
r=Distance between you and the centre of the earth

The force of gravity experienced at the centre of the Earth equals to

= 6.67 × 10^-11 x (Mass of Earth) x (Mass of you)
__________________________________________________
(Distance between you and centre of earth)^2

Since distance between you and the centre of the Earth is zero, we end up dividing the huge number at the top by zero.
As we all know, anything divided by zero is INFINITY (as 0 goes into any number an infinite number of times)

Therefore magnitude of gravity at the centre of the Earth is infinite

When we are at the centre of the Earth we will experience INFINTE gravity (theoretically).

We will not experience no gravity.