Value of g at the center of the earth

In summary: Intuitively I know it's zero, but when I applied that to the gravitational law, I got confused.What does the strength of gravity depend on?How does that vary with depth?The strength of gravity depends on the mass and distance of the objects from the center of the Earth. It also depends on the shape of the objects.
  • #1
Newtonsstudent
13
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Is the value of g at the center of the Earth zero or infinite ?
 
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  • #2
Nice - how are you figuring it either way?

What would be the likely consequence on the surrounding masses if the gravity at the center is infinite do you think?

What does the strength of gravity depend on?
How does that vary with depth?
 
  • #3
Didn't Isaac Newton already work this problem? :rolleyes:
 
  • #4
Newtonsstudent said:
Is the value of g at the center of the Earth zero or infinite ?
Which direction would it point to, if it wasn't zero?
 
  • #5
Intuitively I know it's zero, but when I applied that to the gravitational law, I got confused.
 
  • #6
Newtonsstudent said:
Intuitively I know it's zero, but when I applied that to the gravitational law, I got confused.
It's only valid for point masses and outside of spherical masses. For the inside see this:
http://en.wikipedia.org/wiki/Gravity_of_Earth#Depth
 
  • #7
Well g won't be exactly zero, since Earth doesn't have a perfect spherical shape neither a constant mass density across all of its volume, however it will be close to zero. Assuming perfect symmetry that is perfect sphere and perfect equal mass density across the volume of sphere you can understand why it would be zero(because if it wasnt and it was pointing somewhere as A.T notices then it would violate the symmetry).

Now hm let me think why do you ask if it will be infinite. You thinking along the lines of "if a point particle (like say an electron) of mass m is sitting exactly at the center of the Earth then it would create an infinite mass density there and hence an infinite gravitational field" well this isn't the case though. Though the gravitational field can become as big as we want as we get closer and closer to the point particle (due to the law of inverse square), at exactly the point of the particle it would be zero. And the reason again is symmetry.
 
  • #8
Delta² said:
Though the gravitational field can become as big as we want as we get closer and closer to the point particle (due to the law of inverse square), at exactly the point of the particle it would be zero. And the reason again is symmetry.

Classically, the gravitational force is undefined at a point particle. That is another valid way to get out of a symmetry concern.
 
  • #9
At the very center of the Earth (and assuming a homogeneous earth), the gravitational pull in every direction is the same so the net gravitational force will be 0.

Although it is a little harder, one can show that at distance r from the center of the earth, r less than the radius of the Earth so you are "under ground", the gravitational force is exactly that given by the mass under you. The gravitational pull from all the Earth at distance greater then r from the center cancels.
 
  • #11
If one were to hollow out a perfectly spherical planet that also had equal a bit and then enter it, the pull of gravity would be equal from all directions if one was in the center of the cavity. However, the Earth is not a perfect sphere and has gravitational inconsistencies, so the net force would be slightly greater than zero.

I imagine you are confused because, if your center of mass and the Earth's center of mass are in the same position, the radius would be zero m, causing one to divide by zero (in Newton's equation).

However, I have a question. If one were to gradually did down towards the center of the earth, would the force of gravity gradually increase or decrease? Intuition tells me the force would decrease, but the science says otherwise. The mass between one and the Earth's center would decrease, decreasing the force, but the distance between one ant the Earth's center would also decrease, increasing the force of gravity in total because the radius is squared.
 
  • #12
RyanXXVI said:
If one were to gradually did down towards the center of the earth, would the force of gravity gradually increase or decrease? Intuition tells me the force would decrease, but the science says otherwise.

The gravitational force would decrease. Science does not say otherwise.

The mass between one and the Earth's center would decrease, decreasing the force, but the distance between one ant the Earth's center would also decrease, increasing the force of gravity in total because the radius is squared.

You're only mentioning the mass between you and the center, but not the mass that's increasing between you and the Earth's surface as you go deeper. It's a bit more than simply mass directly above and below since the Earth is a spheroid, but net gravity will decrease as you near the center.

As an interesting aside, did you know that (with a perfect sphere and uniform density), if you drilled a hole between any two locations, you could jump in and gravity would accelerate you the first half, decelerate the second half, and you arrive at the other end. Perhaps more interesting is that the hole needn't pass through the center of the Earth (sphere), and also that every trip, no matter what locations the hole connects, take the SAME TIME to complete.
 
  • #13
Newtonsstudent said:
Intuitively I know it's zero, but when I applied that to the gravitational law, I got confused.
You have to show how you applied the gravitational law.

For a perfect sphere of uniform density (we'll worry about how the Earth differs from that later), the force of gravity decreases by the square of the distance from the surface of the sphere.

Below the surface of the sphere, the gravity from all the mass radially above you cancels out - you really need calculus to show this - leaving only the effect of the mass below you. Since this mass decreases as you descend, the strength of gravity decreases too. At the center of the sphere there is no mass below you so the force of gravity there is zero.

This decrease is linear. g=GM/r^2 : r>R, g=GMr/R^3: r≤R where R=radius of the sphere, M=mass of the sphere, and r is the distance of a small mass from the center of the sphere.

The Earth, however, is not a uniform density sphere, and you are not a point mass.
The details will vary a bit (read: lots) because of that. For instance:

http://upload.wikimedia.org/wikipedia/commons/5/50/EarthGravityPREM.svg
... the dark blue "PREM" line can be thought of as the actual variation of g with r.
... the green line is how it would go if the density of the Earth were uniform.

But I suspect that the "uniform sphere" model is what you are trying to think about.

Further reading:
http://en.wikipedia.org/wiki/Gravity_of_Earth#Depth
 
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  • #14
TumblingDice said:
As an interesting aside, did you know that (with a perfect sphere and uniform density), if you drilled a hole between any two locations, you could jump in and gravity would accelerate you the first half, decelerate the second half, and you arrive at the other end. Perhaps more interesting is that the hole needn't pass through the center of the Earth (sphere), and also that every trip, no matter what locations the hole connects, take the SAME TIME to complete.

Now that's funny, because once you have arrived at the other end, the process should work the other way around, taking you back to the initial end just by the force of gravity, and then back to the other end and so endlessly. Does not this sound like a perpetual motion machine moving you constantly to and fro without any energy being spent?
 
  • #15
Speaking of which, what is it that makes inner core more dense than the outer layers, is it its own attraction towards the center, or is it pressure from the layers above, or maybe both?
 
  • #16
Gerinski said:
Now that's funny, because once you have arrived at the other end, the process should work the other way around, taking you back to the initial end just by the force of gravity, and then back to the other end and so endlessly. Does not this sound like a perpetual motion machine moving you constantly to and fro without any energy being spent?

The hole need not go through the Earth - the same goes for low Earth orbit.

It is perpetual in the sense of any perfect machine, capable of perpetual motion of the first kind; it is the action of additional forces such as air friction which requires adjustments, and hence the continual decay of the orbit.

For a more perfect example, consider the earth-moon system. How long has it been going? Why is it slowing down? Something always goes wrong over the long term; it is a fun physics project to figure out where the energy losses are in each such system.
 
  • #17
carrz said:
Speaking of which, what is it that makes inner core more dense than the outer layers, is it its own attraction towards the center, or is it pressure from the layers above, or maybe both?

You can start here: https://en.wikipedia.org/wiki/Inner_core
 
  • #18
UltrafastPED said:
The hole need not go through the Earth - the same goes for low Earth orbit.

It is perpetual in the sense of any perfect machine, capable of perpetual motion of the first kind; it is the action of additional forces such as air friction which requires adjustments, and hence the continual decay of the orbit.

For a more perfect example, consider the earth-moon system. How long has it been going? Why is it slowing down? Something always goes wrong over the long term; it is a fun physics project to figure out where the energy losses are in each such system.
Thanks, so if I understand correctly you would be a sort of "underground pendulum, ideally without friction", right.

Speaking of which, would a pendulum suspended in a frictionless environment (say inside a vacuum chamber if interstellar space is not void enough) keep oscillating forever?
 
  • #19
UltrafastPED said:

I'm afraid I don't see the answer there. I'll try to answer it then. It kind of looks like two faces of the same coin. I'm not sure if north upper layers are really pressuring north lower layers, but lower north layers are surely attracted to south upper layers. Therefore, outer layers are definitively required, that much I can say.
 
  • #20
carrz said:
I'm afraid I don't see the answer there. I'll try to answer it then. It kind of looks like two faces of the same coin. I'm not sure if north upper layers are really pressuring north lower layers, but lower north layers are surely attracted to south upper layers. Therefore, outer layers are definitively required, that much I can say.
You're making it overly complicated. Do you know why hot air goes up and cold goes down? It's the same reason.
 
  • #21
Bandersnatch said:
You're making it overly complicated. Do you know why hot air goes up and cold goes down? It's the same reason.

That's rather the opposite. In your example density is the cause, but with inner core density is the effect and acceleration is the cause. I'd say it does not compare.
 
  • #22
carrz said:
Speaking of which, what is it that makes inner core more dense than the outer layers, is it its own attraction towards the center, or is it pressure from the layers above, or maybe both?
Below is a plot of density as a function of distance from the center of the Earth, based on the Preliminary Reference Earth Model:

640px-RadialDensityPREM.jpg


Note that density generally decreases gradually with increasing distance from the center but is punctuated by some marked step changes. The gradual decreases in density result from decreasing pressure. The step changes represent physical differences in the constituent matter at those step change boundaries. The largest step change is at the core-mantle boundary. The Earth's core is primarily iron and nickel, which are rather dense even at the surface, and are significantly more dense at the immense pressures inside the Earth. The mantle and everything above is "rock".
 
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  • #23
D H said:
Note that density generally decreases gradually with increasing distance from the center but is punctuated by some marked step changes. The gradual decreases in density result from decreasing pressure. The step changes represent physical differences in the constituent matter at those step change boundaries. The largest step change is at the core-mantle boundary. The Earth's core is primarily iron and nickel, which are rather dense even at the surface, and are significantly more dense at the immense pressures inside the Earth. The mantle and everything above is "rock".

That's really interesting graph, but different materials make the question less clear. It's really about the pressure or net forces, rather than density itself.

Suppose a whole planet is made of water. We now look at three molecules of water half-way to the center, they are right on top of each other. How much the distance between the bottom and the middle molecule depends on acceleration of the middle molecule itself, and how much it depends on acceleration of the molecule above it?
 
  • #24
carrz said:
That's really interesting graph, but different materials make the question less clear.
That different materials differentiate by density is critical to understanding the makeup of a planet. All of the terrestrial planets have a differentiated core, and the gas giants most likely have one as well.

Suppose a whole planet is made of water. We now look at three molecules of water half-way to the center, they are right on top of each other. How much the distance between the bottom and the middle molecule depends on acceleration of the middle molecule itself, and how much it depends on acceleration of the molecule above it?
What acceleration? A planet or star is more or less in a state of hydrostatic equilibrium, where the gravitational force and the gradient in pressure are in balance. There is no acceleration. Understanding hydrostatic equilibrium is easiest for a star or the atmosphere of a planet, where there are no phase changes. With regard to your hypothetical planet that is pure water (which can't happen!), hydrostatic equilibrium explains why pressure and density increase with increasing depth.

What hydrostatic equilibrium by itself cannot explain is why even your pure water planet would have step changes in density. At some depth, that liquid water will become solid, and it won't be the kind of ice that floats. It will instead be ice VII, ice X, or ice XI. In fact, you will see multiple step changes in pressure as liquid water transitions to ice VII, then ice X, and then ice XI.

569px-Phase_diagram_of_water.svg.png
 
  • #25
D H said:
What acceleration?

Gravity acceleration. I should have said "force" I guess. So since every molecule is static, that means acceleration vectors cancel at every point inside a planet, not just the center?

On the other hand force (pressure) vectors vary at every point proportionally to depth, so the maxim force exerted will be on the atom in the very center?


A planet or star is more or less in a state of hydrostatic equilibrium, where the gravitational force and the gradient in pressure are in balance. There is no acceleration. Understanding hydrostatic equilibrium is easiest for a star or the atmosphere of a planet, where there are no phase changes. With regard to your hypothetical planet that is pure water (which can't happen!), hydrostatic equilibrium explains why pressure and density increase with increasing depth.

What hydrostatic equilibrium by itself cannot explain is why even your pure water planet would have step changes in density. At some depth, that liquid water will become solid, and it won't be the kind of ice that floats. It will instead be ice VII, ice X, or ice XI. In fact, you will see multiple step changes in pressure as liquid water transitions to ice VII, then ice X, and then ice XI.

Wow, that's more interesting than I could have imagined. I'll chew on it for a while now. Thanks. In the meantime, is what you said specific for water or would a planet made completely of gold, for example, have those sharp density steps as well? And, can hydrostatic equilibrium explain why would iron end up in the Earth's core and not some lighter or heavier elements, or random mix of them all? And also, why a planet made completely out of water can not happen, theoretically speaking of course?
 
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  • #26
carrz said:
I'm afraid I don't see the answer [in wikipedia]. I'll try to answer it then.
You realize that your question has been answered by post #5 right?
Or did you have a different question I missed?

carrz said:
Gravity acceleration. I should have said "force" I guess.
"acceleraton" is fine. "acceleration of gravity" is better. "free-fall acceleration of gravity" is best.
You could also have talked about gravitatonal potential (not to be confused with potential energy).

The force of gravity, by comparison, varies with mass as well as separation, so it is not as useful for your purposes: you be constantly have to qualify every statement.

So since every molecule is static, that means acceleration vectors cancel at every point inside a planet, not just the center?
Only for a uniform hollow spherical shell. Only inside the hollow part.

The calculation will have some cancellations in it inside, pretty much, any mass distribution though.

To appreciate it you really need calculus - you divide the mass up into lots of small sub-masses and work out the gravity, on a test mass, at the point of interest, due to each sub-mass by itself. Then add them all up (remembering that the force of gravity is a vector) and divide by the test mass. This gets you the acceleration due to gravity at that point.

Without calculus you can see the effect by doing the same calculation for different arrangements of discrete masses.
 
  • #27
Aside:
Gerinski said:
Now that's funny, because once you have arrived at the other end, the process should work the other way around, taking you back to the initial end just by the force of gravity, and then back to the other end and so endlessly. Does not this sound like a perpetual motion machine moving you constantly to and fro without any energy being spent?
No. In order to be a machine it has to do useful work. A perpetual motion machine, therefore, must do useful work in perpetuity with no energy input.

This is what is meant by "perpetual motion of the second kind". There is nothing intrinsically non-physical about perpetual motion of the first kind - motion in perpetuity without doing useful work. I'm not keen on these terms because they make me sound like a UFO nut. Presumably perpetual motion of the 3rd kind is where the machine becomes sentient and gives you advise about stopping nuclear proliferation, and the fourth kind is where you vanish into the Alaskan winter never to be seen again...

Anyway:

The initial energy input is that required to raise the falling mass to the surface of the "Earth" - we don't normally see that energy input because it was provided when the Earth was first formed. But more important is the "work in perpetuity" part.

To extract work from an object falling along such a tunnel as described, you could put a wheel (say) in the center of the path. As the mass passes, it strikes the wheel, turning it... you've seen water-wheels, I don't have to bore you with the details. There are other approaches, we could drop a magnet and put a coil at the center, it does not matter, anything that extracts some of the energy from the motion results in the falling mass not quite reaching the other end of the tunnel. Back and forth it goes with a lower amplitude each time until it ends up stationary at the center all energy exhausted.

Notice that this is under the impossible idealized conditions that there is no friction or other loses in the system to bother us.

Discussions of specific pmm proposals is banned - I hope this discussion is general enough.
It is legitimate, however, to ask how much energy can be generated this way?

Doing the maths: it is about 4x107J/kg (U/m=GM/R=g(R)R)

Energy used globally, from all sources, passed 500x1018J per year in 2012.

So, back of envelope, each dropped kg is enough to supply the entire Earth for about four and a half months... assuming total conversion. Of course the energy would not be available that quickly anyway but it's fun to compare. i.e. the original description implies dropping ("you") a human - who shall we drop? If we dropped the average US American (~90kg apparently) you get energy equivalent to 30-odd years at the 2012 consumption rate. The sales would probably just about offset the damages in the ensuing lawsuit. "Gee we'd like to rescue him but he's running my dishwasher..." See? Fun!
 
  • #28
That was very interesting, thanks!

Something else though, the original quote said:
"if you drilled a hole between any two locations, you could jump in and gravity would accelerate you the first half, decelerate the second half, and you arrive at the other end. Perhaps more interesting is that the hole needn't pass through the center of the Earth (sphere), and also that every trip, no matter what locations the hole connects, take the SAME TIME to complete."

I guess that this would only happen if the starting location is farther from the center than the end location. Intuitively I can not imagine that if you were dropped in a location close to the Earth's centre gravity would transport you to any location closer to the surface. Even if the connecting tunnel passed through the center, you would only reach up to the same distance from the center as you left from, never further.

And so, my consideration of you getting transported continuously to and fro would only happen when the two locations are at equal distance from the center, otherwise you will just stop at the location closest to the center, right?
 
  • #29
Gerinski said:
That was very interesting, thanks!

Something else though, the original quote said:
"if you drilled a hole between any two locations, you could jump in and gravity would accelerate you the first half, decelerate the second half, and you arrive at the other end. Perhaps more interesting is that the hole needn't pass through the center of the Earth (sphere), and also that every trip, no matter what locations the hole connects, take the SAME TIME to complete."

I guess that this would only happen if the starting location is farther from the center than the end location.
The end location has to be at equal or lower radial distance than the start location.
The quote is talking about any two locations on the surface of a sphere.
If the start was at the bottom of the ocean, and the end was at the top of a mountain, you'd still have a climb ahead of you.

This is where you have to keep your wits about you when reading stuff like this - a lot depends on context.

And so, my consideration of you getting transported continuously to and fro would only happen when the two locations are at equal distance from the center, otherwise you will just stop at the location closest to the center, right?
Conservation of energy would have you pop up no farther from the center than you started - assuming uniform mass distributions of course. This is correct.
 

What is the value of g at the center of the earth?

The value of g at the center of the earth is zero. This is because at the center of the earth, all the mass of the earth is pulling equally in all directions, resulting in a net gravitational force of zero.

How is the value of g affected at the center of the earth?

The value of g at the center of the earth is affected by the distribution of mass within the earth. If the earth had a uniform density, the value of g at the center would still be zero. However, if the density is not uniform, the value of g may be slightly different from zero.

Can we measure the value of g at the center of the earth?

No, we cannot directly measure the value of g at the center of the earth. This is because it is impossible for humans to reach the center of the earth, and even if we could, the extreme pressure and temperature would make it impossible to take accurate measurements.

Is the value of g at the center of the earth different from the surface of the earth?

Yes, the value of g at the center of the earth is different from the surface of the earth. On the surface, g is approximately 9.8 m/s², while at the center it is zero. This is because the distance from the center of the earth affects the strength of the gravitational force.

Why is the value of g at the center of the earth important?

The value of g at the center of the earth is important for understanding the structure and composition of the earth. It also helps us to better understand the effects of gravity on different parts of the earth, such as the surface and the atmosphere.

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