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Author Topic: is gravity at the centre of the earth the same as it is on the surface?  (Read 5628 times)

Offline steelrat1

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is gravity at the centre of the earth the same as it is on the surface?
in other words  if you was to drill a hole through to the centre of the earth and measure the Gravitational forces in steps along the way would it be the same as it
is on the surface of the planet or proportional to the mass passed along the way or is gravity infinite?


 

Offline imatfaal

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Gravity at the centre is zero, and varies with 1/r.  See Newtons Shell Theorem
 

Offline Geezer

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Gravity at the centre is zero, and varies with 1/r.  See Newtons Shell Theorem

... and this horse has been flogged to death, resuscitated, flogged to death again, resuscitated, then flogged to death yet again, resuscitated, then.......  ;D
 

Offline syhprum

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Was it not St Aquinas who answered as to what god was doing before he created heaven and Earth that he was creating a special hell for those who persisted in asking silly questions.
 

Offline maffsolo

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Was it not St Aquinas who answered as to what god was doing before he created heaven and Earth that he was creating a special hell for those who persisted in asking silly questions.

Is that why an empty square has nothing left, except 3 consecutive right turns?

The wheels on the bus go round and round
                                                         
 

Offline steelrat1

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OK syhprum ...take that as zero then, right if we could construct a round room at the dead centre of the earth and threw a ball  would it land or float?

« Last Edit: 15/11/2010 18:37:29 by steelrat1 »
 

Offline Geezer

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If you threw it radially towards the surface of the Earth, it would come back to you and head in the opposite direction and continue to oscillate back and forth through the center of the Earth. It would eventually come to rest at the center.

You'd get a similar result if you drilled a shaft right through the center of the Earth that came out at the opposite side. If you jumped into the hole, you would zip back and forth through the center then eventually come to rest at the center.
 

Offline Bill S

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Quote from: syphrum
Was it not St Aquinas

It was St Augustine
 

Offline steelrat1

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so if the ball was placed at the centre it would float then?  but if mass exerts gravity wouldn't the mass of the room walls exert a gravitational force and attract the ball to the wall as every bit of matter attracts every other bit of matter.?
 

Offline Bill S

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If every part of the surrounding wall exerts exactly the same attraction, which way would you want the ball to go?
 

Offline rosy

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so if the ball was placed at the centre it would float then?  but if mass exerts gravity wouldn't the mass of the room walls exert a gravitational force and attract the ball to the wall as every bit of matter attracts every other bit of matter.?
In principle, if instead of the earth we were talking about a perfect sphere of uniform density and thickness at an infinite distance from any other massive body, if the ball were placed at the centre, it would float. Although every part of the walls would be exerting a pull on every part of the ball, the way the maths works out is that the attractive forces balance out exactly and the ball will remain at rest or continue at a constant velocity until it reached the edge of the chamber and bounced off.
In practice, the earth is not isolated in space, and although intuitively I think there must be some infinitesimally small point within the earth at which the gravitational pulls of the different bits of the earth, of the moon, and of the sun (and all the other stuff) balance out exactly, even if you could hollow out a chamber in the centre of the earth such that there were no effective forces (and, as the earth isn't perfectly spherical, you couldn't), your ball would still be attracted to different extents by other massive bodies.

I think. But we have had endless circular arguments about this on this forum in the past. So I daresay someone will tell me I'm wrong.
 

Offline Geezer

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 or continue at a constant velocity until it reached the edge of the chamber and bounced off.


I don't believe it will maintain constant velocity. As soon as it departs from the center of mass of the Earth, it will experience a gravitational attraction towards the center of mass which will cause it to decelerate and return towards the centre of mass. It will then go through the center of mass and oscillate through it until it gives up its energy to friction. If there is no friction because there is a vacuum in our hypothetical hollow shpere, it will continue oscillating back and forth through the center of mass for quite a long time.
 

Offline imatfaal

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Geezer - nope; I think Rosy has it right.  Presuming symmetry (which we shouldn't, but we will); within a spherical shell the net attraction at all points within the shell is zero.  Even though objects exterior to the shell are attracted to the point of the centre of mass - those interior to the shell feel no gravitational attraction in any direction at any point.   It is counter-intuitive, but the maths works in two distinct ways as discovered by Newton and by Gauss.
 

Offline Geezer

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Geezer - nope; I think Rosy has it right.  Presuming symmetry (which we shouldn't, but we will); within a spherical shell the net attraction at all points within the shell is zero.  Even though objects exterior to the shell are attracted to the point of the centre of mass - those interior to the shell feel no gravitational attraction in any direction at any point.   It is counter-intuitive, but the maths works in two distinct ways as discovered by Newton and by Gauss.

Bummer! I stand corrected, again!  ;D
 

Offline peppercorn

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I would just like to point out that gravity increases with distance. So gravity is strongest at the poles (Earth has an oval shape), i.e you would actually weigh more if you were at the poles.

What? Increases?
Surely that's decreases with distance ???
http://en.wikipedia.org/wiki/Inverse-square_law#Gravitation
 

Offline Bored chemist

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I would just like to point out that gravity increases with distance. So gravity is strongest at the poles (Earth has an oval shape), i.e you would actually weigh more if you were at the poles.
I'd like to point out that it doesn't.
It decreases with distance- it follows the inverse square law.
 

Offline rosy

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Of course, actually having reread the OP I've realised that the question I answered wasn't the one asked.

Quote
is gravity at the centre of the earth the same as it is on the surface?
in other words  if you was to drill a hole through to the centre of the earth and measure the Gravitational forces in steps along the way would it be the same as it
is on the surface of the planet or proportional to the mass passed along the way or is gravity infinite?

If you drilled a hole (of a diameter which was negligible relative to the Earth's radius) all the way to the centre of the earth and measured the net pull of gravity in steps, what you would see is a progressive decrease in gravity to (more or less) zero at the centre. According to the shell theorem discussed above, if you went into the earth 10 miles, then you would be inside a 10 mile thick shell of the earth whose gravity would therefore cancel itself out and it would be as if you were on a planet of 10 miles smaller radius. As you carried on in the shell that cancelled out would get thicker and thicker, and the "effective" planet would get smaller, to zero at the centre.

This is going back to a first year undergrad physics course a while ago now, but unless I've messed it up:
Assuming that the earth's density is uniform throughout (which it isn't):

If an object is in the tunnel at a distance from the planet's centre r:

mass of the "effective planet" 
M reversible arrow r3

acceleration due to gravity
a reversible arrow M / r2  (toward the planet's centre)

therefore
a reversible arrow r

If the restoring force is proportional to the distance from the equilibrium position, you get simple harmonic motion (like small osciallations of a pendulum or a mass on a spring).
The result of this is that on a perfectly spherical planet of uniform density except for a tunnel through it, and given a straight, frictionless (and air-resistance-less) tunnel through it, if an object were dropped at the surface of the planet it would move faster and faster until it reached the centre and then gradually slow until it came to a stop at the opposite surface, when it would start to fall back down the hole. In a perfectly frictionless system (and I've just assumed a perfectly spherical, uniform planet, so why not assume a frictionless system too...) it would carry on shooting through the planet forever.

As a sidenote, I suspect this is the result Geezer had in mind further up-thread, but the fact that I completely misread the original question seems to have thrown him off...



 

Offline Geezer

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As a sidenote, I suspect this is the result Geezer had in mind further up-thread, but the fact that I completely misread the original question seems to have thrown him off...


I don't think I read it properly either, and I forgot how Shell Theorem works.  ;D
 

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