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Center of the Earth

by Cyrus
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Cyrus
#55
Sep12-04, 07:02 PM
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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 cant detect them, or there so large you cant use them experimentally. Newton couldent see the effects of two plannets by moving them close or far appart :-P. I just dont 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
#56
Sep13-04, 04:48 PM
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Quote Quote by cyrusabdollahi
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 cant detect them, or there so large you cant use them experimentally. Newton couldent see the effects of two plannets by moving them close or far appart :-P. I just dont 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
#57
Sep13-04, 10:17 PM
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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
#58
Sep15-04, 02:26 PM
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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
#59
Sep16-04, 09:44 AM
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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 ( thats 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
#60
Sep16-04, 01:43 PM
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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
#61
Sep16-04, 01:55 PM
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Quote Quote by cyrusabdollahi
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 ( thats 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
#62
Sep16-04, 09:50 PM
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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
#63
Aug3-09, 05:05 PM
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Quote Quote by chroot View Post
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
#64
Aug6-09, 05:18 AM
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Assuming a liquid planet, the pressure at the center is equal to the gravitational potential at the surface of the planet.
D H
#65
Aug6-09, 08:37 AM
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Quote Quote by stevo72 View Post
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.
Quote Quote by lalbatros View Post
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
#66
Aug6-09, 03:39 PM
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Quote Quote by D H View Post
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/m = Pa.m/m=J/m.
I was simply suggesting to use the Bernouilli equation for this question.
Cyrus
#67
Aug6-09, 04:25 PM
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Quote Quote by lalbatros View Post
Pressure is a density of energy.
Indeed: Pa/m = Pa.m/m=J/m.
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
#68
Aug7-09, 11:00 AM
P: 1,235
I was answering to stevo72, if you permit.
bm0p700f
#69
Aug9-09, 08:03 AM
P: 128
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
#70
Nov9-11, 05:45 AM
P: 1
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.
Doc Al
#71
Nov9-11, 06:31 AM
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Quote Quote by renek View Post
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.
Sorry, but this is completely incorrect. Only under certain circumstance can you treat the mass of the earth as concentrated at a single point. To find the net gravitational field within the earth, you must add up the individual contributions from each element of the earth's mass.

And this thread is years old.
A.T.
#72
Nov9-11, 09:09 AM
P: 4,019
Quote Quote by renek View Post
We will not experience no gravity.
So in which direction will the gravitational force vector point, at the center of the Earth?


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