Settle an argument
Neil Black

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

Don't all of your friends think the earth is flat?

Arthur Kalliokoski

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?"

Goalie Ca

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.

Dustin Dettmer

"In what direction would this person experience weight?"

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

OnlineCop

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! ;)

Arthur Kalliokoski
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.

Dustin Dettmer
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

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.

Goalie Ca

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.

Dustin Dettmer

According to that equation it would approach infinity.

Goalie Ca

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.

Dustin Dettmer

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

Nope. The equation never says that. Treating the earth as a point source is a convenient approximation people make.

verthex

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

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.

SiegeLord

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.

verthex

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

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

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
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
SiegeLord

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.

GullRaDriel

Are you playing "I got a bigger than yours" ? You should stop because I have the biggest ;-)

Dustin Dettmer

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.

LennyLen

Assuming the Earth's mass is evenly distributed around the centre, it's pretty intuitive that there as a net effect of zero gravity at the centre, as there's an equal force being applied in every direction, each cancelling out the effect of the one in the opposite direction.

Dustin Dettmer

Can you think of an experiment that proves this? Does gravity in fact cancel itself out?

Magnets can lock something in place that looks stable until you nudge it. Then it goes wild. How do we know that gravity is so different than magnets?

verthex
dustin dettmer said:

How do we know that gravity is so different than magnets?

Basically in electrodynamics it turns out that the divergence of the electric field is zero, and because magnetic fields do not have monopoles like electric fields do shows that magnetism and gravity are not similar, by virtue of the fact that gravity is monopolar. Basically the field equations that describe this subject can be very difficult to explain in terms of basic math, it requires vector calculus, plus I don't really know for a certain if there is a definite answer to your question in terms of explaining things apart from the foundations that describe the fields you want to talk about. Basically the math describes the subject and even though there has been no experiment to prove that gravity doesn't exist at the center of the earth, the equations of the fields I described "imply" such a condition! Physics as with any science is always extended by experiments as much as people can do them and if the equation works according to theory then its a good nights rest.

Dustin Dettmer

So you're saying in reality we don't know but this is our best guess, given modern theory.

SiegeLord

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

Glad you admitted to the size of your ignorance.

Quote:

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

I have a far better idea. We should send you down there and have you measure it. We'll have you stay there just in case it changes over time.

Dustin Dettmer
SiegeLord said:

We should send you down there and have you measure it.

Ah but then who will you compare sizes with? Better to keep me around so you can feel less ignorant.

Jonatan Hedborg
verthex

Oh yeah, the Lagrangian points, those are hard to work with mathematically but after a few examples they do get interesting, so there, gravity does have a net zero affect somewhere measurable.

Jonatan Hedborg

They were also discovered quite early (1770?). If this was worded as "is there a net effect of gravity at the center of the earth?" I do not know ;)

But Jules Verne wrote fiction, not facts. He probably did not believe that it would be possible to walk around at the center of the earth...

Goalie Ca

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

Says who? r in the x-direction, r in the y-direction, and r in the z-direction yield forces in each of those directions.

Quote:

It's also a convenient approximation the equation makes.

Where?? The equation says Mass at distance "r" away from you has this amount of force. When you are in the center then mass is located where? The entire mass of the earth is not located in a single point in the center.

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.

I stated that newton had it all figured out when he came up with GMm/r2 in the 1600s. Makes sense doesn't it? He probably studied different shapes and such.. and he did invent co-discover calculus.

Arthur Kalliokoski

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

Google for "washer method" +calculus

Thomas Harte

Speaking as someone who has recently been beside the sea, I do not believe that the world's mass is evenly distributed around the centre you'd attribute it when modelling as a sphere. Having read about geology, I know that mass is not evenly distributed according to strata. Having read about space, I know that the Earth is not an inertial frame. Furthermore, since the stuff in the middle of the Earth is fluid, the Earth doesn't even have a constant centre of gravity from which you can disregard most of the above.

verthex
thomas harte said:

Furthermore, since the stuff in the middle of the Earth is fluid, the Earth doesn't even have a constant centre of gravity from which you can disregard most of the above.

you're right! its just that the mass of the earth is megagigatons and your version of earths cog would probably have an uncertainty on the scale of tiny lengths, I can't quote any source however but it would be small.

Thomas Harte
verthex said:

its just that the mass of the earth is megagigatons and your version of earths cog would probably have an uncertainty on the scale of tiny lengths, I can't quote any source however but it would be small.

The tiny lengths you speak of are of at least the same order of magnitude as me. Ignoring that the original question is stupid even if you're naive ("what if you were at the centre of the Earth, but, ummm, somehow not affected by the flow or the pressure of the fluid down there, but the fluid is still down there for the purposes of gravity?"), there's the fact that the Earth and moon actually orbit each other, around their common centre of mass. So the answer is: if you were momentarily at the centre of mass of the Earth but otherwise unaffected by a large part of what dictates the centre of mass of the Earth, you'd be pulled towards the moon.

EDIT: I'll break it down:

We know today that at the Earth's center of gravity the Earth's gravity would have no net effect on an object.

"I've decided what conclusion I want to reach"

Quote:

However, me and my roommate are disagreeing about how long this has been known.

"We've decided that everybody else must agree with us"

Quote:

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".

"We can't see how, were one large body having no gravitational effect on us, any other body could either"

verthex
thomas harte said:

So the answer is: if you were momentarily at the centre of mass of the Earth but otherwise unaffected by a large part of what dictates the centre of mass of the Earth, you'd be pulled towards the moon.

I guess you didn't understand that the net force it zero when R=0 in the original equation. And I don't know if you've worked with uncertainties before but those comments are relatively small of which you speak and that is exactly where the uncertainty comes to play, whether the fluid is there or somewhere else, the scale of the earth dictates that it has a cog, which can possibly change because of fluid dynamics and this uncertainty is NOT ignored, its a part of every calculation that is made today in experimental physics. There is uncertainty in anything measured and there is a method that explains that. Obviously you don't read into science journals, there everything has uncertainty to some degree. I guess you could say science is naive for not propagating uncertainties but thats not the case.

Thomas Harte
verthex said:

Obviously you don't read into science journals

You've decided that ad hominem attacks are a good way forwards?

Quote:

I guess you didn't understand that the net force it zero when R=0 in the original equation.

There is no R in "the original equation". Assuming you mean r, which would be a surprising mistake from someone that is highly scientific, net force is undefined in the original equation. At best you mean that the original equation applies only where gravity is modelled as emanating from a point mass?

Quote:

I don't know if you've worked with uncertainties before but those comments are relatively small

One of the comments I made was that the moon and the Earth together make the major local gravitational field, and that the two orbit each other. Having looked it up, the centre of mass of the total system is about 4,700km from the centre of the Earth (3/4 of the Earth's radius) and isn't static because the Earth and the moon are constantly rotating.

Are you conveniently ignoring that, or are you arguing that 3/4 of the Earth's radius is "relatively small" when we're discussing an object the size of the Earth?

gnolam

Gravity map of the Earth, courtesy of NASA:
{"name":"PIA12146.gif","src":"\/\/djungxnpq2nug.cloudfront.net\/image\/cache\/3\/5\/357acab53f3b700d22b4b1c993f38486.gif","w":256,"h":256,"tn":"\/\/djungxnpq2nug.cloudfront.net\/image\/cache\/3\/5\/357acab53f3b700d22b4b1c993f38486"}PIA12146.gif
(http://photojournal.jpl.nasa.gov/catalog/PIA12146 in case the gif doesn't animate)

SiegeLord

You are wrong, Thomas Harte. The only time you can make an approximation that "centre-of-mass = point source of gravity" is when you are dealing with a single spherical shell (or really really far away from the component bodies, at a distance of a light year, you could probably take the center of mass of the entire solar system and treat it as a point source).

Or do you really think that during the high tide the gravity is 9 times stronger on the side of the earth that is closest to the moon compared to the side of the earth farthest from the moon? Plug in your 'centre of mass=point source' idea into the equation and you'll see that people must be able to jump quite a bit higher when there's high tide.

As verthex said, the effects you are clamouring are negligible.

Thomas Harte

At best you mean that the original equation applies only where gravity is modelled as emanating from a point mass?

SiegeLord said:

You are wrong, Thomas Harte. The only time you can make an approximation that "centre-of-mass = point source of gravity" is when you are dealing with a single spherical shell

Quote:

As verthex said, the effects you are clamouring are negligible.

The effect of a body of mass 7.36 × 1022 kilograms that is just 384,403 km away is negligible? That it's sufficient to move the centre of gravity of the two bodies combined by about 4783.5 km, but, you know, negligible on a much, much smaller scale?

Quote:

Or do you really think that during the high tide the gravity is 9 times stronger on the side of the earth that is closest to the moon compared to the side of the earth farthest from the moon?

You obviously fail to understand what I'm saying. The thrust of my argument is twofold: (i) the centre of gravity of the Earth is quite variable compared to the size of a human being; (ii) even if the human being were positioned such that the gravitational effect of the Earth were zero, there's another very large body very nearby that would prevent apparent weightlessness.

You seem to be saying that the second isn't true because the pull of the Earth over objects situated on the Earth is greater than the pull of the moon, which simply isn't relevant.

EDIT:

NASA said:

Gravity is determined by mass. Earth’s mass is not distributed equally, and it also changes over time.

Mentioned on gnolam's link. NASA seem to think that the change in distribution of mass is sufficient to need to be mentioned in the brief text following a diagram about how mass is not distributed evenly across the globe.

SiegeLord

Plug it in:

Mass of moon: 7.475 × 10^22 kg
Distance to moon fro the center of earth: 384403000 m
Acceleration due to gravity at the center of the earth due to the moon: 3.37*10^-5 m/s^2

That's 34 micrometers per second squared. Negligible.

verthex

Ok everyone let me honestly say that I dont try to disect every tiny argument word for word because THIS is not a scientific discussion. I've sat around with people that do this in a room all day and they don't do anything but pull up facts and figures and make me go crazy!

Thomas Harte
SiegeLord said:

Plug it in: ...

This suggests you really don't understand the problem with your tide argument. The Earth has a radius of about 6,371km. Of that, 2,260km is fluid, which extends continuously from about 1,220km from the centre. (EDIT3: 15% of the mass of the Earth is fluid, it seems)

If you want to plug numbers in, work out the effect of the moon on that, then the effect of that on the person in the centre.

EDIT: to sum up again, since you seem to be cutting and chopping my argument according to whim; the centre of mass of Earth is not fixed because a large part of the planet is fluid and there's a really big body nearby that affects the fluids on this planet. You would therefore not experience weightlessness even if you were magically and momentarily at the centre but somehow affected only by the gravitational effects of the things that actually are at the centre.

EDIT2: actually, that's not my argument. My argument is that a convincing argument that the centre of mass of the Earth is steady enough will need to explain this stuff away, rather than skimming over it as this thread has.

Arthur Kalliokoski

if you were momentarily at the centre of mass of the Earth but otherwise unaffected by a large part of what dictates the centre of mass of the Earth, you'd be pulled towards the moon.

No, because you'd share in the orbital motion around the common center (which is inside the earth, BTW). If the earth suddenly disappeared, you'd continue to "orbit" the moon, but the moon would pretty much take off in a straight line, (except for solar gravitation) unconcerned with your paltry hundred kilograms. As the moon receded, you'd also take on a solar orbit.

there's a really big body nearby that affects the fluids on this planet

That's the point of tides, the high tide opposite the moon is there because the water is too far away to share equally in the moon's gravitational pull, but the mechanical connection of the earth in the moon/earth orbit tends to sling it off centrifugally. Vice versa the near side facing the moon. But in the center, those would cancel out, and if it didn't, then the entire orbit would be affected.

OnlineCop

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!

SiegeLord said:

Glad you admitted to the size of your ignorance.

SiegeLord said:

We should send you down there and have you measure it.

Ah but then who will you compare sizes with? Better to keep me around so you can feel less ignorant.

Thomas Harte, taking your comment completely out of context, you can see how this adds all sorts of immaturity to this topic:

The tiny lengths you speak of are of at least the same order of magnitude as me.

;D

SiegeLord

My argument is that a convincing argument that the centre of mass of the Earth is steady enough will need to explain this stuff away, rather than skimming over it as this thread has.

I just calculated the effect of the moon alone on the person at the center of the Earth, it was negligible. I can integrate over that plot gnolam posted (do you by the way know the scale of that plot? The gravity ranges by at most +/- 5 cm/s^2 over the surface of the earth (e.g. see this chart) ) and also arrive at some negligible answer. I won't of course, because it is apparent to me that it is negligible, any of those effects you call attention to would apply equally (or even in greater magnitudes) at the surface of the Earth where each one of us is sitting. Those effects are negligible, since I don't feel them. I won't feel them at the center of the Earth either, where they will be far far smaller.

Thomas Harte

No, because you'd share in the orbital motion around the common center

Good point! I'd made the error you probably think I did, i.e. switched inertial frames for forces but not carried velocity along with me.

Quote:

But in the center, those would cancel out, and if it didn't, then the entire orbit would be affected.

Right, but the entire orbit does change over time. I'm sure I read earlier, though I can't find it now, that the position of the Earth now and the position of the Earth exactly a year ago vary by as much as 10,500km (though I can't find the source again, so that figure may be wrong). I'm not saying the differences don't cancel themselves out in the overview, merely that nobody has made a sufficient argument to establish that they have close-to-zero effect on the scale of a person over a very short amount of time.

EDIT:

SiegeLord said:

I won't of course, because it is apparent to me that it is negligible, any of those effects you call attention to would apply equally (or even in greater magnitudes) at the surface of the Earth where each one of us is sitting.

Gosh, I wish you were listening.

(1) will an object that is non-negligibly affected by the gravitation pull of the Earth act in exactly the same way as one that is not affected by the pull of the Earth?

(2) therefore, is it valid to say that if you can observe no effect in objects that are affected by the gravitational pull of the Earth, that you would not be able to observe an effect were that pull removed?

By your token, I can argue that the tiny thrusters astronauts use to manoeuvre themselves when in orbit obviously have negligible effect on them because if I apply the same force to an object sitting next to me in the room then it doesn't move.

Whether or not my argument is false, making arguments that just ignore what we're talking about (ie, that something can't affect an object that is otherwise free of the pull of the Earth because it doesn't affect things that aren't free of the pull) isn't persuasive.

Arthur Kalliokoski

The OP said that "at the Earth's center of gravity the Earth's gravity would have no net effect on an object." If you're at the center of gravity, even momentarily, then you're equally surrounded by mass. And I believe the whole point was that your "weight" would diminish when approaching the center of the earth instead of increasing or remaining constant.

Thomas Harte

The OP said that "at the Earth's center of gravity the Earth's gravity would have no net effect on an object."

He also said "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", which brings in questions of whether sources other than the Earth could produce a [non-negligible] gravitational pull at the centre of the Earth.

People then made a whole bunch of statements assuming the centre of mass of the Earth to be a constant position, which is what I consider questionable.

EDIT: in fact, the Earth's centre of mass definitely does move relative to the Earth itself (source), so the question is whether this is non-negligible over a short space of time.

EDIT2: actually, that's a fantastic resource to make my point. I'll quote it in its entirety with emphasis added:

Quote:

It has been found that the residual (Chandler) motion of the Earth's rotation pole results from the forced translational motion of the Earth's rotation axis relative to the geographic axis. The Earth's rotation axis moves parallel to itself without changing the angle of inclination to the ecliptic plane. The translational motion of the Earth's axis of rotation is caused by the motion of the Earth's center of mass in the Earth's body in the range from 1 to 30 meters relative to the Earth's surface. The motion of the Earth's center of mass in space is due to the motion of the consistent inner core of the Earth in the liquid outer core under the action of the total (internal and external) gravitational field. Formulae for calculation of the trajectory of the Earth's centre of mass from astronomical observations are suggested. The comparison of our calculations and observational data on variations in the latitudes of places and acceleration of gravity has shown that they are in good agreement. Our model has been shown to adequately describe the physical process of motion of the Earth's centre of mass inside the Earth's body.

Though it doesn't give an idea of how much time the up-to-30m shift in the centre of gravity (ie, much bigger than a person) takes. If it's 1,000,000 years or whatever then clearly it isn't relevant.

SiegeLord

ie, that something can't affect an object that is otherwise free of the pull of the Earth because it doesn't affect things that aren't free of the pull

This has nothing to do with making measurements with the pull of the Earth present or not! 34 micrometers per second squared is negligible both with a background 9.8 m/s^2 and with a background 0 m/s^2! I never said there is no effect whatsoever, and I never claimed that this effect is negligible because of the background 'noise'. The effect is negligible because it is negligible, it is tiny! It'd be negligible even on the surface of the sun, or next to the black hole. For humans, such low acceleration cannot be perceived, no matter what other forces affect them.

EDIT: This, at first reading, suggests that the threshold for human perception of acceleration is on the order of several millimeters per second. Over 100x larger than the effects we are dealing with here.

Though it doesn't give an idea of how much time the up-to-30m shift in the centre of gravity (ie, much bigger than a person) takes.

Who cares, when you are 30 meters away from the center of the Earth, you still feel basically no gravity! If you carved out a 30 meter hole at the center, that whole hole would be in a microgravity environment. If you didn't carve out 30 meters, but left a sphere within that hole, that 30 meter radius sphere would not appreciably attract you (the acceleration due to gravity would be around 1*10^-4 m/s^2). If that bothers you, just carve out the final 30 meters. I'm sure the story referenced in the OP did not involve the protagonist trapped in a man sized coffin in the middle of the Earth.

The whole point of the derivation I linked to earlier was that once you have a shell of matter around you, it doesn't matter where in the shell you are, as long as you are inside it.

Thomas Harte
SiegeLord said:

This has nothing to do with making measurements with the pull of the Earth present or not!

SiegeLord said:

I can integrate over that plot gnolam posted ... and also arrive at some negligible answer. I won't of course, because it is apparent to me that ... any of those effects you call attention to would apply equally (or even in greater magnitudes) at the surface of the Earth

In your opinion, effects apply equally to something subject to the pull of the Earth (eg, something at the surface of the Earth) and to something not subject to the pull of the Earth (eg, something at the centre of mass) but your argument has nothing to do with "making measurements with the pull of the Earth present or not"?

SiegeLord said:

Who cares, when you are 30 meters away from the center of the Earth, you still feel basically no gravity! If you carved out a 30 meter hole at the center, that whole hole would be in a microgravity environment.

Possibly we're just talking at different levels of conceit? If you instantaneously released a body into a 30 metre hole at the centre of mass of the Earth then that body would move relative to its surroundings by the up-to-30 metres, because the inner core would move.

I can see your argument as working only if we pretend that the Earth is a rigid body.

weapon_S
Quote:

Settle an argument

SiegeLord

In your opinion, effects apply equally to something subject to the pull of the Earth (eg, something at the surface of the Earth) and to something not subject to the pull of the Earth (eg, something at the centre of mass) but your argument has nothing to do with "making measurements with the pull of the Earth present or not"?

Well, it's not an opinion. Linear superposition of forces is a pretty well accepted property of the universe (it's Newton's Second Law). The moon will pull at me the same amount no matter if the Earth is replaced by a black hole, or by Dustin in his subterranean probe. The law of universal gravitation only depends on the two masses involved, the person, and the body in question (the moon). So yes, it does not matter if the Earth is present or not... it's not in the equation, so the pull of the Earth has no effect.

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Possibly we're just talking at different levels of conceit? If you instantaneously released a body into a 30 metre hole at the centre of mass of the Earth then that body would move relative to its surroundings by the up-to-30 metres, because the inner core would move.

Sure. Movement != perception of the force. You'd observe the Earth moving around you, but if you are just floating, far enough from the wall and closed your eyes (thus relying only on your vestibular sense) you would not be able to tell that the Earth's core is dancing around.

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I can see your argument as working only if we pretend that the Earth is a rigid body.

It has to be rigid to some degree (otherwise the cavern in the centre would collapse on you). Assuming that wall exists, I don't care if the Earth was made entirely from water past that point, the law of gravitation will still apply identically.

Thomas Harte
SiegeLord said:

Well, it's not an opinion.

You're either being deliberately obtuse or genuinely misunderstanding what I'm saying. Given that you falsely accused me of having a "centre of mass=point source" outlook even though I'd already said "at best you mean that the original equation applies only where gravity is modelled as emanating from a point mass?", it's probably the latter.

You have argued that some forces cannot be significant on a body that is not affected by the gravitational pull of the Earth because they are not significant on bodies that are affected by the gravitational pull of the Earth. When challenged, your response is "This has nothing to do with making measurements with the pull of the Earth present or no".

Your logic doesn't work. It's like saying that a hairdryer can't exert a significant force on a bit of paper because when you used one on a piece of paper under a paperweight, it didn't blow away. Then, when challenged, you respond that the presence or absence of the paperweight has nothing to do with what you're saying.

The exchange, in full:

SiegeLord said:

it is apparent to me that it is negligible, any of those effects you call attention to would apply equally (or even in greater magnitudes) at the surface of the Earth where each one of us is sitting.

making arguments that just ignore what we're talking about (ie, that something can't affect an object that is otherwise free of the pull of the Earth because it doesn't affect things that aren't free of the pull) isn't persuasive.

SiegeLord said:

This has nothing to do with making measurements with the pull of the Earth present or not!

Samuel Henderson
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weapon_S said:

Settle an argument

Well, since it seems like people here can't agree that it is even possible for someone to be at the centre of the earth and have absolutely no gravitational forces acting on them I would say that poor Neil is going to have look elsewhere, or at least be willing to accept that the answer presented here is probably divided as well.

My two cents...

Newton came up with his theory and equations and stuff in 1687 (as mentioned) so I assume that at the time (1864) if anyone thought about it they would have used the equations and figured that gravity does not apply at the centre of the earth.

OnlineCop said:

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

Being a diver I laughed at this :)

SiegeLord

Your logic doesn't work. It's like saying that a hairdryer can't exert a significant force on a bit of paper because when you used one on a piece of paper under a paperweight, it didn't blow away. Then, when challenged, you respond that the presence or absence of the paperweight has nothing to do with what you're saying.

That because is of your manufacture. I brought up the example of being on the surface of the Earth because, as I said, we are sitting on it. It has nothing to do with the Earth pulling on you, I made the example so you yourself can easily verify the verity of my statements without digging down to the center of the Earth. The difference I am highlighting between the two locations is not the force of gravity exerted by the Earth, I am highlighting solely the difference of location, and the corresponding lack of difference coming from the effects of the Moon, the Sun and the imperfections of the structure of the Earth.

I am saying that if you don't feel those effects here, you definitely won't feel them 6000 km away from this location at the center of the Earth. 6000 km is nothing compared to the distance of the Moon from the Earth, and even smaller nothing compared to the distance between the Earth and the Sun. Those effects start negligible, and remain so.

The effects of the imperfections of the Earth are greater, because I feel like those imperfections would probably cancel themselves out in the center of the planet relative to their summed effect on its surface. Even if they don't, they are negligible to begin with, as gnolam's diagram shows.

Thomas Harte
SiegeLord said:

I am saying that if you don't feel those effects here, you definitely won't feel them 6000 km away from this location at the center of the Earth.

And you don't see any potential flaw in "if you're not conscious of something small while also under the influence of something big, then you definitely won't be conscious of it if I take the big thing away"?

If I take your question at face value then, yes, the effect of the Moon's gravity is observable here on the surface, especially to those that live near the coast. And, yes, I think that effect would be more pronounced if I were surrounded by a huge mass of fluid that wasn't as affected by the Earth's pull, internal to the heavy stuff.

I think that clearly states my opinion. Of course I'll continue to defend myself if anyone accuses me of being definitely wrong because I've applied an obviously false model, but otherwise I guess that's all I really need to say to complete my participation in this topic.

GullRaDriel

OnlineCop owned you :-)

I lawled like a mad cow, man !

_

SiegeLord

And you don't see any potential flaw in "if you're not conscious of something small while also under the influence of something big, then you definitely won't be conscious of it if I take the big thing away"?

The moon can pull me in the direction orthogonal to the direction the Earth pulls me, if the moon is near the horizon. In that perspective, I already can't feel the gravity of the Earth. Not that it matters, I already told you the magnitude of the acceleration due to the Moon, and I already told you the limits of human perception of acceleration. The former is far below the latter.

As for the tides... you are cheating by using a rather sensitive detector of gravity, far more sensitive than a human is. I don't contest that given a sufficiently accurate gravimeter, you'd be able to measure the pull of the Moon, Sun, the Andromeda Galaxy, and any insect that happens to fly around around your cavern. A human would be unable to detect any of those.

Matthew Leverton

Here's what I've learned so far: If you need an argument settled, don't ask on a.cc!

anonymous

Isn't there like two questions here: if you are at rest at the centre of the Earth, how fast will you accelerate towards the centre? How would it feel down there (same as in outer space?)?

alethiophile

A.cc: where at least the flamewars are about intellectual topics. :P

OnlineCop
anonymous said:

How would it feel down there (same as in outer space?)?

You'll start to rotate (gyrate) because of both the magnetic poles of the earth, and because of the rotation of the system. Since the earth, as mentioned earlier, is not uniformly distributed, you will have varying amounts of gravitation pull depending on the land (and water) mass above you.

anonymous said:

if you are at rest at the centre of the Earth, how fast will you accelerate towards the centre?

About as fast as someone on A.cc can get a date with an actual woman. It approaches infinity.

Neil Black

Here's what I've learned so far: If you need an argument settled, don't ask on a.cc!

Yes, that's the main thing I've learned, too. >:(

In response to Thomas Harte:

"I've decided what conclusion I want to reach"

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"We've decided that everybody else must agree with us"

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"We can't see how, were one large body having no gravitational effect on us, any other body could either"

I wasn't aware that the idea that an object would have no net gravitational effect upon another object at it's gravitational center was under contention. Apparently I was wrong, and you disagree.

As for the last quote, I should have said, "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 from the Earth at the time that Jules Verne wrote "Journey to the Centre of the Earth"." Obviously the moon, the sun, and basically every mass that isn't Earth would still be effecting us. But the effect would be negligible.

Wait, it is true that all objects with mass effect all other objects with mass gravitationally, right? Proportionally to their masses and inversely proportional to the distance between them, right? Or is that under contention as well?

Tobias Dammers

In an idealized situation, you are a point, and the Earth is a perfect sphere. The net gravity is zero. If your ideal self is not point-shaped, then each part of it that is not at the center would experience a very light force towards the center, but far too small for you to notice.
In the real world, you'd burn or melt long before you reach the center.

Neil Black

If your ideal self is not point-shaped, then each part of it that is not at the center would experience a very light force towards the center, but far too small for you to notice.

While my ideal self is not point-shaped, my real self seems to be getting more point-shaped, if you consider round to be point-shaped.

Anyway, the argument is settled for my purposes. Clearly we knew in 1864 that the Earth would exert effectively no net gravity on a person at its center of gravity. I win. 8-)

Tobias Dammers

my real self seems to be getting more point-shaped, if you consider round to be point-shaped.

No, point-shaped means having zero size in any spatial direction (and thus also zero volume and surface area). You are probably rather growing away from point-shapedness and approaching the Earth's shape (but let's hope you won't reach its actual size).

Neil Black

You are probably rather growing away from point-shapedness and approaching the Earth's shape (but let's hope you won't reach its actual size).

Actually now that I think about it I've been getting less Earth-shaped over the last month or so. Probably because I have to walk up The Hill at least twice a day to go to class. I do nearly a mile of walking daily, almost half of it uphill! And conversely, almost half of it is downhill, too. Then there's the level bits. I love the level bits.

Tobias Dammers

Then there's the level bits. I love the level bits.

I hate the level bits. Especially when it comes to making games. As soon as the game is ready to add levels and other content, I lose interest and drop the project.

Evert

Hmm... I'm late to the party and I didn't read every post. Presumably the answer has been given, but I'll just add in case no one mentioned this.

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?

There is no net gravitational acceleration at the centre of an isotropic mass distribution. The proof for this is quite simple and was given by Newton in the Principia in 1687.
Note that this does notsay that there is no gravity, just that the net acceleration due to gravity is zero.

This proof needs two assumptions: there is no other source of gravity (fairly good approximation on Earth; the Earth is very clearly the dominant source of gravity near here. Tides are irrelevant to this) and the mass distribution is spherically symmetric (ie, radial shells have a uniform density). This is also a fairly good approximation. True, the Earth is neither smooth nor spherical, but the deviations are small.

So no, there was no reason to say that there was gravity at the centre of the Earth in 1864. It's been a long time since I read Journey to the centre of the Earth, but as far as I recall, Jules Verne says as much at one point and makes no claim to the contrary. Note that the journey described in the book comes no where close to the centre of the Earth anyway.

Wait, it is true that all objects with mass effect all other objects with mass gravitationally, right?

Yes. Actually, all objects that have energy (which is all of them), since gravity acts on energy rather than mass.

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Proportionally to their masses and inversely proportional to the distance between them, right?

Mmmmmmmmmmsortofmostofthetimebutnotreally. See below.

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Or is that under contention as well?

Not at all. Gravity in general is described by Einstein's theory of general relativity, which is not F=GMm/r2 er but something more complicated. However, in the low energy/low velocity limit, it reduces to Newtonian gravity.

Arthur Kalliokoski

I always wondered about the Leaning Tower of Piza, the plumb bobs and spirit levels were thrown off when the earth's center of gravity, the moon, and planetary alignment all conspired together

Tobias Dammers

It's either the Leaning Tower of Pisa, or the Leaning Tower of Pizza. Guess which one came first.

OnlineCop

"Leaning" came first.

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