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Value of g at the center of the earthby Newtonsstudent
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#1
Jul614, 10:28 PM

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Is the value of g at the center of the earth zero or infinite ?



#2
Jul614, 11:11 PM

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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
Jul714, 06:22 AM

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Didn't Isaac Newton already work this problem?



#4
Jul714, 06:36 AM

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Value of g at the center of the earth



#5
Jul714, 10:19 AM

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Intuitively I know it's zero, but when I applied that to the gravitational law, I got confused.



#6
Jul714, 10:27 AM

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http://en.wikipedia.org/wiki/Gravity_of_Earth#Depth 


#7
Jul714, 10:29 AM

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Well g wont be exactly zero, since earth doesnt 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 isnt 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. 


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Jul714, 12:48 PM

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#9
Jul714, 02:10 PM

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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. 


#10
Jul714, 02:18 PM

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#11
Jul714, 10:28 PM

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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
Jul714, 11:06 PM

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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
Jul714, 11:30 PM

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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: ... 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 


#14
Jul1514, 07:26 AM

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#15
Jul1514, 07:33 AM

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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
Jul1514, 07:37 AM

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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 earthmoon 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
Jul1514, 07:39 AM

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#18
Jul1514, 07:58 AM

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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? 


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