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Settle an argument
Neil Black
Member #7,867
October 2006
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We know today that at the Earth's center of gravity the Earth's gravity would have no net effect on an object. However, me and my roommate are disagreeing about how long this has been known. The question in debate is whether or not it was scientifically accepted to say that there would be gravity at the center of the Earth at the time that Jules Verne wrote "Journey to the Centre of the Earth".

Was it a scientifically valid theory (or hypothesis) to say that there would be gravity, in the sense of being able to walk around normally, in 1864?

Matthew Leverton
Supreme Loser
January 1999
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Don't all of your friends think the earth is flat?

Arthur Kalliokoski
Second in Command
February 2005
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I'd think any reasonably intelligent individual would be able to deduce there is no gravity at the center of the earth. I got detention in 8th grade for contradicting the teacher when he applied the "diminishing gravity with altitude" formula in reverse for calculating the weight of a person at the center. I really got slammed when I asked "In what direction would this person experience weight?"

“Throughout history, poverty is the normal condition of man. Advances which permit this norm to be exceeded — here and there, now and then — are the work of an extremely small minority, frequently despised, often condemned, and almost always opposed by all right-thinking people. Whenever this tiny minority is kept from creating, or (as sometimes happens) is driven out of a society, the people then slip back into abject poverty. This is known as "bad luck.”

― Robert A. Heinlein

Goalie Ca
Member #2,579
July 2002
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Let's assume of instant that this has been known since Newton published this formula on gravitation in 1687 (after we knew earth was round).

<math>F = G \frac{m_1 m_2}{r^2}</math>

Now.. here's the funny part. What does F become as you get to the centre of the earth? If you say "undefined/infinite" then you fail math.

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Bah weep granah weep nini bong!

Dustin Dettmer
Member #3,935
October 2003
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"In what direction would this person experience weight?"

Ah a trick question. I was going to say down but clearly it is left.

OnlineCop
Member #7,919
October 2006
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F == "out".

So once you got to the center of the Earth, all of the mass that surrounds you would pull on all parts of your body (equally) until you got ripped apart.

In space, you get taller by about an inch because there is no compression on your spinal column. In the center of the earth, it would pull your spinal column apart (using the above logic).

So it stands to reason that you will get taller, the higher mountains you climb, and you'll shrink, the closer to sea level you got, but once you stuck your head under the ocean, you'll start growing again.

You don't get the "bends" from surfacing too quickly. You get them because your body is getting compressed back to its previous height! ;)

I made a VR game!

Arthur Kalliokoski
Second in Command
February 2005
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Goalie Ca said:

Let's assume of instant that this has been known since Newton published this formula on gravitation in 1687 (after we knew earth was round).

Now break up the earth (surrounding you) into a multitude of small spheres (atoms, if you like) and calculate that formula for each one, taking direction into account.

“Throughout history, poverty is the normal condition of man. Advances which permit this norm to be exceeded — here and there, now and then — are the work of an extremely small minority, frequently despised, often condemned, and almost always opposed by all right-thinking people. Whenever this tiny minority is kept from creating, or (as sometimes happens) is driven out of a society, the people then slip back into abject poverty. This is known as "bad luck.”

― Robert A. Heinlein

Dustin Dettmer
Member #3,935
October 2003
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Goalie Ca said:

Now.. here's the funny part. What does F become as you get to the centre of the earth? If you say "undefined/infinite" then you fail math.

Wouldn't r become 0, making it undefined?

Arthur Kalliokoski
Second in Command
February 2005
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That formula assumes all the mass is on the same side of you. The gravity would diminish following the mass of the sphere which is between you and the center of the earth. For example, if you were 10 meters from the center, you'd be attracted to a 20 meter diameter sphere of significantly compressed matter (probably gold and uranium) as if this sphere were free in space and you were standing on it.

“Throughout history, poverty is the normal condition of man. Advances which permit this norm to be exceeded — here and there, now and then — are the work of an extremely small minority, frequently despised, often condemned, and almost always opposed by all right-thinking people. Whenever this tiny minority is kept from creating, or (as sometimes happens) is driven out of a society, the people then slip back into abject poverty. This is known as "bad luck.”

― Robert A. Heinlein

Goalie Ca
Member #2,579
July 2002
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Wouldn't r become 0, making it undefined?

Arthur nailed it. You forgot to consider that M is not a point but is in fact scattered all around you.

edit: anyways.. being a scientist and using equations I will go by the date of the discovery of the equation as the date we knew the center of the earth has 0 gravity. Isomorphism FTW.

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Bah weep granah weep nini bong!

Dustin Dettmer
Member #3,935
October 2003
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According to that equation it would approach infinity.

Goalie Ca
Member #2,579
July 2002
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According to that equation it would approach infinity.

No it wouldn't. From inside the earth you can no longer approximate it as a point source but rather as a collection of particles. Since the earth is symmetric about the center each particle's gravitational pull cancels out.

What you need to do is integrate GMm/r^2 for the entire volume!

edit: without calculus speak.. you need to calculate F using that equation for each atom of dirt in the earth and them sum it all together.

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Bah weep granah weep nini bong!

Dustin Dettmer
Member #3,935
October 2003
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We are arguing semantics.

This equation assumes a single radius between two objects. To do what you're describing requires a different equation.

Maybe you mean to say "in reality it does not approach infinity" which is very different than "this equation does not approach infinity."

Goalie Ca
Member #2,579
July 2002
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Nope. The equation never says that. Treating the earth as a point source is a convenient approximation people make.

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Bah weep granah weep nini bong!

verthex
Member #11,340
September 2009
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Well, basically if you dropped a penny down a hole that connects the surface of the earth going through the center, the penny would eventually go into a harmonic oscillation described by an SHO equation, and would keep oscillating along the path of the hole due to inertia as long as there is no friction and of course Coriolis forces remain absent so as to not cause the penny from hitting the walls of the hole. Remember the Coriolis effect is due to the Earths spinning and thats what causes Hurricane spin. The SHO stands for simple harmonic oscillator. My guess is that gravity only exists for regions of space not at the origin, a Heaviside step function could explain the field outside the origin and a Dirac delta function would describe the origin. I have a BS in physics and I might be completely wrong, but thats what I learned in school in a class called theoretical mechanics. And I really did not like solving Greens functions but thats the realm of math that describes such problems.

Neil Walker
Member #210
April 2000
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I think you need to wait for Evert for the definitive answer. As well as being a brain surgeon he's also a rocket scientist and astrophysicist.

Neil.
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SiegeLord
Member #7,827
October 2006
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What for? GoalieCa already explained how this works:

Goalie Ca said:

What you need to do is integrate GMm/r^2 for the entire volume!

This is a problem from high school physics, don't see what you are arguing... oh, it's Dusting Dettmer. Is there a field of science you are familiar with on at least a high school level?

In any event, here are two pages showing the integration procedure: calculating the force of gravity inside a shell and outside a shell. They sort of do it the hard way, but the easier way would require introduction of divergence theorem and the like.

You add those two results together to obtain the force of gravity on say a coin that falls through the Earth. The inside the shell answer applies to the part of the Earth that is a greater distance from the center of the Earth (all of the Earth if the coin is in the center of the Earth) and the outside the shell answer applies to the rest of the Earth (all of the Earth if the coin is above the surface).

Since Newton knew both the law of gravitation, and calculus, I'd say that I agree with Goalie Ca. I can't find a source for it, but I remember him calculating the outside the shell answer to allow for the point source approximation of the Earth, I am sure he calculated the inside the shell answer elsewhere in his books.

"For in much wisdom is much grief: and he that increases knowledge increases sorrow."-Ecclesiastes 1:18
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verthex
Member #11,340
September 2009
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I agree with Siegelord,

SiegeLord said:

You add those two results together to obtain the force of gravity on say a coin that falls through the Earth. The inside the shell answer applies to the part of the Earth that is a greater distance from the center of the Earth (all of the Earth if the coin is in the center of the Earth) and the outside the shell answer applies to the rest of the Earth (all of the Earth if the coin is above the surface)."

I also wanna mention that the Hare won and the Tortoise lost because his shell was heavy!

Neil Black
Member #7,867
October 2006
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I think you all misunderstood. My roommate and I are in perfect agreement that there is no net gravitational pull from Earth at the center of the Earth, the argument is over when humans first figured that out, before or after 1864.

verthex
Member #11,340
September 2009
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Well there was the Cavendish experiment which explored gravity and in 1797 he discovered this,

Quote:

"I deduced that the forces which keep the planets in their orbs must be reciprocally as the squares of their distances from the centres about which they revolve, and thereby compared the force requisite to keep the moon in her orb with the force of gravity at the surface of the earth and found them to answer pretty nearly." -- Isaac Newton, 1666
So Newton's original formula was:

where the symbol means "is proportional to".

To make this into an equal-sided formula or equation, there needed to be a multiplying factor or constant that would give the correct force of gravity no matter the value of the masses or distance between them. This gravitational constant was discovered in 1797 by Henry Cavendish.

NOTE: There is an equation missing thats a gif,

from link.
Basically the formula that he found has to satsify that there is no net force at R = 0 otherwise the equation would go to infinite.

Dustin Dettmer
Member #3,935
October 2003
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Goalie Ca said:

Treating the earth as a point source is a convenient approximation people make.

It's also a convenient approximation the equation makes.

SiegeLord said:

This is a problem from high school physics, don't see what you are arguing... oh, it's Dusting Dettmer. Is there a field of science you are familiar with on at least a high school level?

Ah so you're basing this on high school teachings. No wonder you're so confused.

Goalie Ca said:

What you need to do is integrate GMm/r^2 for the entire volume!

edit: without calculus speak.. you need to calculate F using that equation for each atom of dirt in the earth and them sum it all together.

Why would you need to integrate it. Please elaborate on this, I can't see why that would even help.

The problem I see here is you're not accounting for the direction of each particle.

Arthur Kalliokoski
Second in Command
February 2005
avatar

“Throughout history, poverty is the normal condition of man. Advances which permit this norm to be exceeded — here and there, now and then — are the work of an extremely small minority, frequently despised, often condemned, and almost always opposed by all right-thinking people. Whenever this tiny minority is kept from creating, or (as sometimes happens) is driven out of a society, the people then slip back into abject poverty. This is known as "bad luck.”

― Robert A. Heinlein

SiegeLord
Member #7,827
October 2006
avatar

Your posts, Dustin, don't indicate even a lowly high school level of understanding of this, especially when you say stuff like this:

The problem I see here is you're not accounting for the direction of each particle.

I suggest you learn about this to at least to that level, so at least you can form sentences that make logical sense. It's embarrassing.

"For in much wisdom is much grief: and he that increases knowledge increases sorrow."-Ecclesiastes 1:18
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GullRaDriel
Member #3,861
September 2003
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Are you playing "I got a bigger than yours" ? You should stop because I have the biggest ;-)

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Dustin Dettmer
Member #3,935
October 2003
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Are you playing "I got a bigger than yours" ? You should stop because I have the biggest ;-)

That may be true but at least mine's bigger than his!

Here's a PDF that seems to explain it.

Interesting... so if you cut the sphere up into circular slices than you can realistically calculate it.

What would be awesome is to see this tested. Send a probe down there and measure it for real.

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