Wouldn't we weigh less on the equator than on the north and south pole due to centrifugal force? I know that it would be like 00.0000000000000343^4545454545 lbs lighter or something, but how would i figure this out?
We're talking about centrifugal force vs centripetal force... i think.
centrifugal force on earth.
Yes you would, but I think it has more to do with the fact that you are farther away from the earths center of mass at the equator than at the poles. Since weight is a function of mass and gravity and gravity is a function of the distance between mass centers, assuming your (or the earth's) mass doesn't appreciably change, the farther away from the earth's center of mass you are, the less you weigh.
Or somthing like that.
Or somthing like that.
Pretend that the Earth wasn't spinning on its axis and was motionless. The force of gravity holding you down firmly on earth is the same at the equator as it is at the North pole. (The minute variations in the magnetic field and the fact that the Earth isn't round is negligable) so you would basically weigh the same.
Now, if your standing at the equator when we restart the spin, Newtons first law of motion tries to make you go in a straight line... or trys to offset gravity and throw you off. At the North pole however, this wouldn't happen and its this effect which makes you lighter at the equator.
This means that your motion at the equator is about .07 mps squared but the acceleration due to gravity is about 22 mps squared so the difference is very tiny.... you would weigh about 0.2 to 0.5% less at the equator than at the poles.
Bettina
Now, if your standing at the equator when we restart the spin, Newtons first law of motion tries to make you go in a straight line... or trys to offset gravity and throw you off. At the North pole however, this wouldn't happen and its this effect which makes you lighter at the equator.
This means that your motion at the equator is about .07 mps squared but the acceleration due to gravity is about 22 mps squared so the difference is very tiny.... you would weigh about 0.2 to 0.5% less at the equator than at the poles.
Bettina
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Mobius
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I can wade in here...
Gravity varies from place to place about the earth, and in reasonably non-trivial amounts too. If you think about the Earth having \"one Gravity\" - that is only \"on average\". The presence of a lot of matter makes the gravity change quite a bit.
It's well known that seamounts (undersea mountains that don't break the surface) cause local increases in gravity, and these can clearly be \"seen\" from space with radar: above seamounts the ocean's surface may be 6-10 feet \"higher\" than just a coupde of klicks away, due to all the extra mass of the mountain.
Here's something I bet you didn't know too: the gravity created by the Earth's oceans is accelerating the moon, and causes it to shift further away from the Earth. From memory, it is receding at about 1-2 cm per year.
Here's how it works: The oceans, due to lag and friction, cause tides to precede the Moon in it's orbit. The extra mass of the tide runnign ahead of the moon exerts enough force to accelerate the moon minutely. This acceleration, of course results in a bigger orbit.
This is the same force which has \"tidaly locked\" the moon, such that the one face always points towards Earth. Some people think the mnoon doesn't rotate - but that is not true, it simply has a periodicity of 1:1, each time it completes a rotation around Earth, it also rotates once. Periodicity of 1:1 is quite common in the solar system, and several of Jupiters moons are similarly locked.
In terms of centrifugal forces acting on you at the equator - in fact there are none. The fact is that in 99.9% of cases the force actually experienced is \"centripetal\" and not centrifugal.
From Wikipedia: Centripetal force should not be confused with centrifugal force. The centrifugal force is a fictitious force that arises from being in a rotating reference frame.
Catch22 - yep, 9.8ms^2 is correct for most calculations - but that only applies to the surface of the Earth. As you descend into the Earth, or rise above it, acceleration due to gravity changes based on the inverse square of the distance from the center of the object. Descending into Earth is complicated, because the mass of earth above you also has gravity, which fights against the mass below you.
Bet; the magnetic field of Earth has no affect on you: as you aren't made of a ferrous metal. You can confirm this by getting put through an MRI: Magnetic Resonance Imaging machine - where the magnetic fields reach tens of thousands of Gauss - and it has no affect on your whatsoever.
Gravity varies from place to place about the earth, and in reasonably non-trivial amounts too. If you think about the Earth having \"one Gravity\" - that is only \"on average\". The presence of a lot of matter makes the gravity change quite a bit.
It's well known that seamounts (undersea mountains that don't break the surface) cause local increases in gravity, and these can clearly be \"seen\" from space with radar: above seamounts the ocean's surface may be 6-10 feet \"higher\" than just a coupde of klicks away, due to all the extra mass of the mountain.
Here's something I bet you didn't know too: the gravity created by the Earth's oceans is accelerating the moon, and causes it to shift further away from the Earth. From memory, it is receding at about 1-2 cm per year.
Here's how it works: The oceans, due to lag and friction, cause tides to precede the Moon in it's orbit. The extra mass of the tide runnign ahead of the moon exerts enough force to accelerate the moon minutely. This acceleration, of course results in a bigger orbit.
This is the same force which has \"tidaly locked\" the moon, such that the one face always points towards Earth. Some people think the mnoon doesn't rotate - but that is not true, it simply has a periodicity of 1:1, each time it completes a rotation around Earth, it also rotates once. Periodicity of 1:1 is quite common in the solar system, and several of Jupiters moons are similarly locked.
In terms of centrifugal forces acting on you at the equator - in fact there are none. The fact is that in 99.9% of cases the force actually experienced is \"centripetal\" and not centrifugal.
From Wikipedia: Centripetal force should not be confused with centrifugal force. The centrifugal force is a fictitious force that arises from being in a rotating reference frame.
Catch22 - yep, 9.8ms^2 is correct for most calculations - but that only applies to the surface of the Earth. As you descend into the Earth, or rise above it, acceleration due to gravity changes based on the inverse square of the distance from the center of the object. Descending into Earth is complicated, because the mass of earth above you also has gravity, which fights against the mass below you.
Bet; the magnetic field of Earth has no affect on you: as you aren't made of a ferrous metal. You can confirm this by getting put through an MRI: Magnetic Resonance Imaging machine - where the magnetic fields reach tens of thousands of Gauss - and it has no affect on your whatsoever.
It is.. I based my calculations on miles instead of meters so 22 (mile/hour) second = 9.83488 meters. I should have expressed it in meters.catch22 wrote:I was always under the impression that the acceleration due to gravity was 9.8 m/s^2 (~32 ft/s^2)
I know, but I've been reading a lot about the Quantum world, where fields influence gravity. I should have left that out.Mobius wrote:Bet; the magnetic field of Earth has no affect on you: as you aren't made of a ferrous metal. You can confirm this by getting put through an MRI: Magnetic Resonance Imaging machine - where the magnetic fields reach tens of thousands of Gauss - and it has no affect on your whatsoever.
Bee
Well put, Mobius. 
This Wiki article provides a good sum-up of the Earth's gravitational force and why/where it tends to vary. As it turns out, Bettina, you would be slightly lighter at the equator than the poles, and the ficticious \"centrifugal force\" is involved (this has to do with something called \"apparent weight,\" which is the normal force that the Earth exerts upward on you and makes you feel \"heavy\"), which is something I'd never specifically considered before. A factor on the side of \"actual weight\" is that, due to the \"bulge\" produced by the Earth's rotation, locations on the equator are actually somewhat farther from the Earth's center than the poles are; since the gravitational force decreases with the square of the distance, and since the Earth can generally be considered as a point mass when performing gravitational calculations, you'll experience less gravity at the Equator than the poles.
This Wiki article provides a good sum-up of the Earth's gravitational force and why/where it tends to vary. As it turns out, Bettina, you would be slightly lighter at the equator than the poles, and the ficticious \"centrifugal force\" is involved (this has to do with something called \"apparent weight,\" which is the normal force that the Earth exerts upward on you and makes you feel \"heavy\"), which is something I'd never specifically considered before. A factor on the side of \"actual weight\" is that, due to the \"bulge\" produced by the Earth's rotation, locations on the equator are actually somewhat farther from the Earth's center than the poles are; since the gravitational force decreases with the square of the distance, and since the Earth can generally be considered as a point mass when performing gravitational calculations, you'll experience less gravity at the Equator than the poles.
Yes Err. The moon is the center not the earth.
I know, but I wanted to make my explanation simple....and I don't need to be any lighter so I'll stay North.Top Gun wrote:As it turns out, Bettina, you would be slightly lighter at the equator than the poles
If you pretend to survive the heat, gravity would be balanced so you would be totally weightless.snoopy wrote:Mob. hinted at this- but think about the center of the earth- if you where at the exact mass center of the earth, what would you feel?
Bettina
Actually, the inner core is solid iron and nickel. You'd just have the little task of surviving enormous temperatures and pressures to worry about. 
And Bettina's right; if you did manage to make it down there and hollow out a little cocoon, you'd be just about completely weightless. The gravitational attraction of any portion of the Earth's mass would be balanced out by the attraction of the mass opposite it (minus a few irregularities). If anyone ever played the old Sierra adventure game Torin's Passage (made by Al Lowe), the concept might make more sense. (Bagpipes, anyone? 
)
And Bettina's right; if you did manage to make it down there and hollow out a little cocoon, you'd be just about completely weightless. The gravitational attraction of any portion of the Earth's mass would be balanced out by the attraction of the mass opposite it (minus a few irregularities). If anyone ever played the old Sierra adventure game Torin's Passage (made by Al Lowe), the concept might make more sense. (Bagpipes, anyone? )
no that'd be like a ball in a bowl. it's easy to keep a ball in a bowl since it self corrects - the ball always rolls towards the center.
but trying to balance a ball on another ball - it's difficult. any off-centered-ness of the ball will increase and increase and increase exponentially. i mean, as the ball starts to roll off it will roll faster and faster until it falls off the edge.
the only way to do it is to apply constant corrections. In the ball-on-ball example this means you are constantly moving your hands around to balance it.
In the center of the earth example, it'd need somethind else like adjustable electromagnets or rockets.
basically, it wouldn't occilate. just as a ball balancing on another ball doesn't occilate - but just rolls off.
what'd happen is you'd put it into the center of the room and it'd slowly, then faster and faster, drift towards the closest wall.
my nerd-fu is all HAI-YA!
but trying to balance a ball on another ball - it's difficult. any off-centered-ness of the ball will increase and increase and increase exponentially. i mean, as the ball starts to roll off it will roll faster and faster until it falls off the edge.
the only way to do it is to apply constant corrections. In the ball-on-ball example this means you are constantly moving your hands around to balance it.
In the center of the earth example, it'd need somethind else like adjustable electromagnets or rockets.
basically, it wouldn't occilate. just as a ball balancing on another ball doesn't occilate - but just rolls off.
what'd happen is you'd put it into the center of the room and it'd slowly, then faster and faster, drift towards the closest wall.
my nerd-fu is all HAI-YA!
hmm. i was thinking of it like a ringworld - like in the Halo game universe. Earth with a hollow sphere in the center of it is the same as the Halo - just make the hole in the center of the Halo smaller and add an extra dimension, creating a sphere from the Halo.
When you start to talk about tunnels though it changes things as you are traveling THROUGH the mass. When you are floating around in the empty sphere you are not traveling through the mass - on ALL sides of that sphere there is an equal amount of mass on all sides. No matter where you are in the sphere the mass is the same - only your proximity to that fixed mass has changed and that's the force pulling you towards the nearest wall (if you are between 2 identical masses - you will fall towards the closer one - and one will ALWAYS be closer, equilibrium will be impossible). But when you start tunneling through the ground you are changing the ratio of mass, as you go further through the tunnel you leave more mass behind you and less mass infront.
In Halo it's like comparing flying to digging.
If you dug a tunnel all the way through the Halo, you'd end up floating on the up/down axis halfway before you tunneled out the other side - but you'd be standing on the edge of the tunnel surface, you'd be turned 90degrees and could walk around the tunnel like a hamster wheel
.
So... Ghengis, if you were in the spherical room in the center of the Earth and you fell into a tunnel like you said. What would happen is you'd be suspended in the doorway of that tunnel. Since the further you travel up that tunnel the more mass you would be taking from infront and putting behind, gravity would be increasingly pulling you back towards the sphere - but anywhere INSIDE that sphere gravity is not dictated by mass, but only by proximity to it. The sphere pulls you towards it's sides, and the tunnel pulls you towards the sphere. You'd be suspended between the 2.
(okok, actually you'd be on the side of the tunnel coz the tunnel's edges would pull at you the same as the sphere's sides would.
The tunnel would be a cylender intersecting the sphere, you'd likely end up somewhere on the ring intersecting the 2. But it could be possible to also be anywhere on the inside surface of that sphere.
This so far ignores the earth spinning... so lets add that to the mix!
Inside the sphere: Centrifical force and your promixity to Mass would BOTH work the same to pull you towards the edge of that sphere. The centrifical force would also allow you to enter the tunnel and travel a certain distance up it. However, you'd eventually get to a point in that tunnel where the difference in mass infront vs behind you is creating enough gravity to pull you back into the sphere to counter the centrifical force. You would find yourself stuck in equilibrium between gravity and centrifical force. Also, like the sphere but in 2D - you woudl be pulled towards the edges of the tunnel itself, so you would have a single RING of equilibrium you could walk around on the side of teh tunnel.
I wish i could show you an illustration. it's so interesting, it was fun to work out.
When you start to talk about tunnels though it changes things as you are traveling THROUGH the mass. When you are floating around in the empty sphere you are not traveling through the mass - on ALL sides of that sphere there is an equal amount of mass on all sides. No matter where you are in the sphere the mass is the same - only your proximity to that fixed mass has changed and that's the force pulling you towards the nearest wall (if you are between 2 identical masses - you will fall towards the closer one - and one will ALWAYS be closer, equilibrium will be impossible). But when you start tunneling through the ground you are changing the ratio of mass, as you go further through the tunnel you leave more mass behind you and less mass infront.
In Halo it's like comparing flying to digging.
If you dug a tunnel all the way through the Halo, you'd end up floating on the up/down axis halfway before you tunneled out the other side - but you'd be standing on the edge of the tunnel surface, you'd be turned 90degrees and could walk around the tunnel like a hamster wheel
.So... Ghengis, if you were in the spherical room in the center of the Earth and you fell into a tunnel like you said. What would happen is you'd be suspended in the doorway of that tunnel. Since the further you travel up that tunnel the more mass you would be taking from infront and putting behind, gravity would be increasingly pulling you back towards the sphere - but anywhere INSIDE that sphere gravity is not dictated by mass, but only by proximity to it. The sphere pulls you towards it's sides, and the tunnel pulls you towards the sphere. You'd be suspended between the 2.
(okok, actually you'd be on the side of the tunnel coz the tunnel's edges would pull at you the same as the sphere's sides would.
The tunnel would be a cylender intersecting the sphere, you'd likely end up somewhere on the ring intersecting the 2. But it could be possible to also be anywhere on the inside surface of that sphere.
This so far ignores the earth spinning... so lets add that to the mix!
Inside the sphere: Centrifical force and your promixity to Mass would BOTH work the same to pull you towards the edge of that sphere. The centrifical force would also allow you to enter the tunnel and travel a certain distance up it. However, you'd eventually get to a point in that tunnel where the difference in mass infront vs behind you is creating enough gravity to pull you back into the sphere to counter the centrifical force. You would find yourself stuck in equilibrium between gravity and centrifical force. Also, like the sphere but in 2D - you woudl be pulled towards the edges of the tunnel itself, so you would have a single RING of equilibrium you could walk around on the side of teh tunnel.
I wish i could show you an illustration. it's so interesting, it was fun to work out
.Don't forget about coriolis, Roid, so actually as you fell into the tunnel you would get smacked into the side of the tunnel, unless the tunnel where coaxial with the axis of rotation of the earth. You would also be accelerated somewhat due to coreolis by the sun. That's something that's interesting to think about- we actually weight differently in the morning and the evening due to coreolis- I'm not sure which is lighter and which is heavier- but it happens. (I think)
Roid, ignoring your Halo and centripetal force examples (which I wasn't quite able to fully wrap my mind around at the moment ), I'm just about positive that your \"sphere on a sphere\" analogy regarding a hollow Earth is incorrect. Remember, the gravitational force of a solid spherical body can ordinarily be treated as that of a point mass at the center of the body. Because of this, when using the standard gravitational formula (F = Gm1m2/(R^2)) to calculate the force of the Earth on an object at/near its surface, the R in question is simply the Earth's radius. The actual physical reality of the Earth's volume, the \"normal force\" of the ground pushing up against us at the molecular level, is what keeps all of us standing up here instead of smushed together down there.
Obviously, this situation changes somewhat if you're inside a hollow sphere in the center of the Earth, since now you're experiencing the pull of portions of the Earth's mass from all directions. And, like you mentioned, unless you're dead-center, you're always going to be closer to at least a portion of that mass. But there's one detail you've neglected: while you may get farther from the opposite side of the sphere as you move to one side, and while gravitational force does drop off by the inverse square, you also have more of the sphere's inner surface area pulling at you in one direction as you move to one side. And, since the surface area of a sphere is directly proportional to the square of its radius, this increasing directional pull from one side exactly counteracts the increasing pull from the side you're moving closer to; in the end, you're still left floating in there. The analogy doesn't extend to two dimensions (i.e. your Halo case), since the fact that the gravity law is an inverse square law means that the local mass will more than counteract the pull of the narrow area of mass across the \"sky.\" However, if you treat the gravitational force as being a simple inverse law, and draw a circle on paper, it becomes pretty plain that a point placed anywhere within the sphere will experience a net force of zero from the line of \"force\" all around it. The same holds true for the electrostatic force; given a spherical shell of charge, the total electric field inside the sphere works out to be zero.
After a bit of Googling, I was able to stumble across this link, which explains both the \"donut\" and hollow sphere problems in a pretty easy-to-understand context. It extends the example to someone standing in a cave deep within the Earth; that person would only experience a gravitational force generated by the mass inside his particular distance from the Earth's center, since the force generated by mass directly above the person would be counterbalanced by all the rest of the mass outside of that radius around the Earth's interior.
Roid, there is one case where your idea of \"oscillations\" does work out regarding the Earth .In the grand tradition of using approximations when setting up physics problems (one science site I stumbled across accurately stated that \"you know you're a physics major when you approximate a giraffe as a sphere\"), let's create a tunnel straight through the Earth's core and out the other side, the proverbial \"hole to China\" that every US kid dreams of digging at some point in their lives. Let's also neglect little things like the Earth's rotation, incredibly high temperatures and pressures, liquid rock, and air resistance. Given those conditions, if you jumped down into the hole, you'd experience positive acceleration until you reached the Earth's center, and then negative acceleration until you came out the other side. You'd wind up coming to a stop at the other end of the hole, and then you'd fall back down, in perpetual oscillatory motion. Considering the fact that the greatest acceleration you'd experience would be only one gee, if someone could actually set up those conditions, it'd be the safest and coolest ride around.
), I'm just about positive that your \"sphere on a sphere\" analogy regarding a hollow Earth is incorrect. Remember, the gravitational force of a solid spherical body can ordinarily be treated as that of a point mass at the center of the body. Because of this, when using the standard gravitational formula (F = Gm1m2/(R^2)) to calculate the force of the Earth on an object at/near its surface, the R in question is simply the Earth's radius. The actual physical reality of the Earth's volume, the \"normal force\" of the ground pushing up against us at the molecular level, is what keeps all of us standing up here instead of smushed together down there. Obviously, this situation changes somewhat if you're inside a hollow sphere in the center of the Earth, since now you're experiencing the pull of portions of the Earth's mass from all directions. And, like you mentioned, unless you're dead-center, you're always going to be closer to at least a portion of that mass. But there's one detail you've neglected: while you may get farther from the opposite side of the sphere as you move to one side, and while gravitational force does drop off by the inverse square, you also have more of the sphere's inner surface area pulling at you in one direction as you move to one side. And, since the surface area of a sphere is directly proportional to the square of its radius, this increasing directional pull from one side exactly counteracts the increasing pull from the side you're moving closer to; in the end, you're still left floating in there. The analogy doesn't extend to two dimensions (i.e. your Halo case), since the fact that the gravity law is an inverse square law means that the local mass will more than counteract the pull of the narrow area of mass across the \"sky.\" However, if you treat the gravitational force as being a simple inverse law, and draw a circle on paper, it becomes pretty plain that a point placed anywhere within the sphere will experience a net force of zero from the line of \"force\" all around it. The same holds true for the electrostatic force; given a spherical shell of charge, the total electric field inside the sphere works out to be zero.
After a bit of Googling, I was able to stumble across this link, which explains both the \"donut\" and hollow sphere problems in a pretty easy-to-understand context. It extends the example to someone standing in a cave deep within the Earth; that person would only experience a gravitational force generated by the mass inside his particular distance from the Earth's center, since the force generated by mass directly above the person would be counterbalanced by all the rest of the mass outside of that radius around the Earth's interior.
Roid, there is one case where your idea of \"oscillations\" does work out regarding the Earth .In the grand tradition of using approximations when setting up physics problems (one science site I stumbled across accurately stated that \"you know you're a physics major when you approximate a giraffe as a sphere\"
), let's create a tunnel straight through the Earth's core and out the other side, the proverbial \"hole to China\" that every US kid dreams of digging at some point in their lives. Let's also neglect little things like the Earth's rotation, incredibly high temperatures and pressures, liquid rock, and air resistance. Given those conditions, if you jumped down into the hole, you'd experience positive acceleration until you reached the Earth's center, and then negative acceleration until you came out the other side. You'd wind up coming to a stop at the other end of the hole, and then you'd fall back down, in perpetual oscillatory motion. Considering the fact that the greatest acceleration you'd experience would be only one gee, if someone could actually set up those conditions, it'd be the safest and coolest ride around. Even in the face of that link, i'm still not getting it. How does adding another dimention change it?
In my mind, if you suggest that a hollow sphere creates a uniform equilibrium inside it - then why does this not also suggest that you'd float into the center of the hole of a donut as well?
I don't understand how adding a dimention changes that.
Why do you stick to the inside surface of the donut, but float around anywhere inside the sphere?
Inside that hollow sphere in the center of the earth. The mass on all sides is constant, but your proximity to that mass changes as you float around. If you are to be at equilibrium, and the closest side to you is pulling you strongest merely via it's proximity - what is providing the counterforce to stop you from slamming into that closest wall? The only counterforce is the mass on the other side, which hasn't changed - but you are pretty far away from it now so it'd have almost no effect compared to the much closer wall right next too you. The mass in the closer wall would be pulling you strongest.
i know you tried to explain this, but can you explain it again better?
if you were online right now we could chat and draw diagrams using groupboards, but you are asleep damn you timezones!
In my mind, if you suggest that a hollow sphere creates a uniform equilibrium inside it - then why does this not also suggest that you'd float into the center of the hole of a donut as well?
I don't understand how adding a dimention changes that.
Why do you stick to the inside surface of the donut, but float around anywhere inside the sphere?
I'm not understanding the surface area bit. Is not surface area irrelevant? If we're talking gravity, are not the only relevant things MASS and your proximity to it?Top Gun wrote:But there's one detail you've neglected: while you may get farther from the opposite side of the sphere as you move to one side, and while gravitational force does drop off by the inverse square, you also have more of the sphere's inner surface area pulling at you in one direction as you move to one side. And, since the surface area of a sphere is directly proportional to the square of its radius, this increasing directional pull from one side exactly counteracts the increasing pull from the side you're moving closer to; in the end, you're still left floating in there.
Inside that hollow sphere in the center of the earth. The mass on all sides is constant, but your proximity to that mass changes as you float around. If you are to be at equilibrium, and the closest side to you is pulling you strongest merely via it's proximity - what is providing the counterforce to stop you from slamming into that closest wall? The only counterforce is the mass on the other side, which hasn't changed - but you are pretty far away from it now so it'd have almost no effect compared to the much closer wall right next too you. The mass in the closer wall would be pulling you strongest.
i know you tried to explain this, but can you explain it again better?
if you were online right now we could chat and draw diagrams using groupboards, but you are asleep damn you timezones!
I never sleep.
The difference between the concepts of the hollow sphere and donut ties into the consideration of surface area. You're absolutely right in saying that the mass of the body is the only thing that's creating the gravitational force. However, looking at things a certain way, the surface area of that hollowed-out sphere represents the distribution of that mass; since the physical bulk of the Earth is \"behind\" each unit of surface area in a radially outward direction, the amount of surface area you consider at any one time correlates with the amount of mass you're considering. So, in a very real sense, you can view the \"surface area\" of the sphere as representative of the mass outside of it.
This kind of leads me into the bulk of the argument. Again, you're right in saying that a body of mass closer to our test object is going to exert more gravitational force on it than an equivalent body of mass further away, due to the inverse square portion of the law. But, when you're looking at that closer wall pulling the object \"more strongly,\" you're missing how the effect of the rest of the mass, the majority of the Earth's \"bulk,\" is changing on it. To my utter surprise, I think I've come up with an analogy that manages to work semi-well. Let's treat the Earth as a round fruit (a cantaloupe or orange will do the job). First, slice the fruit in half directly down the middle and hollow both halves out; this represents our hollow cavity, with the remaining flesh/rind representing the Earth's \"bulk.\" In this case, you can see that an object directly in the center of the fruit would be experiencing an equal pull from all sides, since there's an equal amount of rind in any direction.
Now, take another fruit, and slice it not down the middle, but relatively close to one side; again, hollow out both pieces. This case represents our test mass being closer to one of the walls of the sphere; you can treat the mass as sitting at a point level with the \"rim\" of the smaller piece of fruit, right in the center. Now, in this case, the smaller portion of fruit represents the \"side\" of the sphere that you're closer to; like you said, area for area, this portion creates a greater gravitational force because of its closer proximity to the test mass. However, take a look at the larger portion of fruit. This represents the mass of the Earth that is pulling the test object away from the near wall. Obviously, there's a lot more of it than there is of the part closer to you. To take the case to the extreme, if you were right up against the wall of the sphere, looking out over the whole hollowed-out area, only the mass \"behind\" the area of wall right around you would be pulling you to that side; all the rest of the sphere you would see, your entire \"horizon,\" would be exerting a force on you in the opposite direction. While the mass closest to you does exert the greatest force proportionally toward that wall, the sheer amount of mass pulling you in the opposite direction balances it out.
Now, this is precisely where the surface area reference comes back into play. As you move away from the center of the sphere, the amount of surface area that's on the farther side, i.e. the bigger half of our fruit rind, increases by the square of the distance, by the simple formula of the area of a sphere. And since that surface area really represents the amount of mass behind it, the mass is likewise increasing as a square. At the same time, you're receiving an inverse square factor from the increasing distance from said mass. Square on top cancels square on bottom, and voila...you're left with a constant force in all directions, and so you get to float around to your heart's content.
This same exact argument explains why things don't work the same way in the case of the halo. Once again, in this case, the portion of the ring that you're closest to provides the greatest force proportional to the area that you're dealing with, and the same description of the mass of the opposite side of the ring falling off as a square also holds. However, there is one crucial difference. In this case, we don't have the surface area of the entire sphere, that complete and increasing square factor, to counter-balance the distance. Instead, we have only a tiny sliver of that sphere; as we get closer and closer to one side of the ring, the amount of surface area that we \"see\" in the \"sky,\" and thus the amount of mass that pulls us away from the near side, simply doesn't increase fast enough to cancel out the increasing force from said near side. Like the analogy in the article I linked to, when you stand on one side of the ring and look around, the \"near side\" stretches around you, filling your line of sight on the ground; however, the \"far side\" represents only a tiny sliver across the \"sky.\" This contrasts with our sphere picture, where the \"far side\" fills the entire \"dome\" of the \"sky.\" Due to that difference, the near side of the ring is able to more than overcome the pull of the far side, and so that's the side that you'd fall towards.
Hopefully, that helped clear things up a little bit; if not, I can always take one more crack at it. Trying to explain things like this is actually a great deal of fun for me.
P.S. There is a more mathematical way to look at the sphere problem, if you're familiar with the concept of vectors. (If not, then ignore this.) Since the concept of force is a vector concept, we can apply vector properties to it, and we can split the force provided by certain quantities of mass into different components. When you're close to one side of the hollow sphere, the mass represented by the smaller portion of our fruit produces a component of force in its own direction; since the portion we're looking at is perfectly round, there isn't any net force toward the edges of the \"fruit piece,\" since the component on one side is always balanced out by the component on the other. Likewise, all of the mass represented by the bigger portion of fruit has a component of force in the direction away from the near wall of the sphere. Now, the force vector components pointing toward the near wall and created by each mass unit of the \"smaller piece\" do have the greater magnitude, due to their closer distance. However, the components pointing away from this wall and created by the mass units of the \"larger piece\" are obviously much greater in number, since there's a lot more \"fruit,\" i.e. mass, on that side. Adding everything up, the vector sums cancel out, and you're once again left with your net force of zero.
The difference between the concepts of the hollow sphere and donut ties into the consideration of surface area. You're absolutely right in saying that the mass of the body is the only thing that's creating the gravitational force. However, looking at things a certain way, the surface area of that hollowed-out sphere represents the distribution of that mass; since the physical bulk of the Earth is \"behind\" each unit of surface area in a radially outward direction, the amount of surface area you consider at any one time correlates with the amount of mass you're considering. So, in a very real sense, you can view the \"surface area\" of the sphere as representative of the mass outside of it.
This kind of leads me into the bulk of the argument. Again, you're right in saying that a body of mass closer to our test object is going to exert more gravitational force on it than an equivalent body of mass further away, due to the inverse square portion of the law. But, when you're looking at that closer wall pulling the object \"more strongly,\" you're missing how the effect of the rest of the mass, the majority of the Earth's \"bulk,\" is changing on it. To my utter surprise, I think I've come up with an analogy that manages to work semi-well.
Let's treat the Earth as a round fruit (a cantaloupe or orange will do the job). First, slice the fruit in half directly down the middle and hollow both halves out; this represents our hollow cavity, with the remaining flesh/rind representing the Earth's \"bulk.\" In this case, you can see that an object directly in the center of the fruit would be experiencing an equal pull from all sides, since there's an equal amount of rind in any direction.Now, take another fruit, and slice it not down the middle, but relatively close to one side; again, hollow out both pieces. This case represents our test mass being closer to one of the walls of the sphere; you can treat the mass as sitting at a point level with the \"rim\" of the smaller piece of fruit, right in the center. Now, in this case, the smaller portion of fruit represents the \"side\" of the sphere that you're closer to; like you said, area for area, this portion creates a greater gravitational force because of its closer proximity to the test mass. However, take a look at the larger portion of fruit. This represents the mass of the Earth that is pulling the test object away from the near wall. Obviously, there's a lot more of it than there is of the part closer to you. To take the case to the extreme, if you were right up against the wall of the sphere, looking out over the whole hollowed-out area, only the mass \"behind\" the area of wall right around you would be pulling you to that side; all the rest of the sphere you would see, your entire \"horizon,\" would be exerting a force on you in the opposite direction. While the mass closest to you does exert the greatest force proportionally toward that wall, the sheer amount of mass pulling you in the opposite direction balances it out.
Now, this is precisely where the surface area reference comes back into play. As you move away from the center of the sphere, the amount of surface area that's on the farther side, i.e. the bigger half of our fruit rind, increases by the square of the distance, by the simple formula of the area of a sphere. And since that surface area really represents the amount of mass behind it, the mass is likewise increasing as a square. At the same time, you're receiving an inverse square factor from the increasing distance from said mass. Square on top cancels square on bottom, and voila...you're left with a constant force in all directions, and so you get to float around to your heart's content.
This same exact argument explains why things don't work the same way in the case of the halo. Once again, in this case, the portion of the ring that you're closest to provides the greatest force proportional to the area that you're dealing with, and the same description of the mass of the opposite side of the ring falling off as a square also holds. However, there is one crucial difference. In this case, we don't have the surface area of the entire sphere, that complete and increasing square factor, to counter-balance the distance. Instead, we have only a tiny sliver of that sphere; as we get closer and closer to one side of the ring, the amount of surface area that we \"see\" in the \"sky,\" and thus the amount of mass that pulls us away from the near side, simply doesn't increase fast enough to cancel out the increasing force from said near side. Like the analogy in the article I linked to, when you stand on one side of the ring and look around, the \"near side\" stretches around you, filling your line of sight on the ground; however, the \"far side\" represents only a tiny sliver across the \"sky.\" This contrasts with our sphere picture, where the \"far side\" fills the entire \"dome\" of the \"sky.\" Due to that difference, the near side of the ring is able to more than overcome the pull of the far side, and so that's the side that you'd fall towards.
Hopefully, that helped clear things up a little bit; if not, I can always take one more crack at it. Trying to explain things like this is actually a great deal of fun for me.
P.S. There is a more mathematical way to look at the sphere problem, if you're familiar with the concept of vectors. (If not, then ignore this.
) Since the concept of force is a vector concept, we can apply vector properties to it, and we can split the force provided by certain quantities of mass into different components. When you're close to one side of the hollow sphere, the mass represented by the smaller portion of our fruit produces a component of force in its own direction; since the portion we're looking at is perfectly round, there isn't any net force toward the edges of the \"fruit piece,\" since the component on one side is always balanced out by the component on the other. Likewise, all of the mass represented by the bigger portion of fruit has a component of force in the direction away from the near wall of the sphere. Now, the force vector components pointing toward the near wall and created by each mass unit of the \"smaller piece\" do have the greater magnitude, due to their closer distance. However, the components pointing away from this wall and created by the mass units of the \"larger piece\" are obviously much greater in number, since there's a lot more \"fruit,\" i.e. mass, on that side. Adding everything up, the vector sums cancel out, and you're once again left with your net force of zero.oh wow, so the increase in volume of the 3D semi-sphere far-from-you gives you the exact same increase in force as the increase in force from the proximity of the semisphere close-to-you.
(well, not increases... but you get what i mean)
i didn't know the maths were the same for both volume of a semi-sphere and proximity to mass. That's brilliant that is!
ok, so now with a tunnel intersecting the sphere, what would happen is: The hollow "missing" section of the side of the sphere would be a gap in the mass, it would upset the perfrect ballance of mass, so you would be repelled from it - you'd be attracted to the wall exactly opposite the hole.
So, you wouldn't drift into the tunnel, but assuming you were able to jump up it - you'd just fall back down.
adding another tunnel in the other side of the sphere, so the 2 tunnels together go all the way from one side of the earth to the other (with the hollow central sphere joining them). This would create the gravity of the Halo ringworld - you'd be repelled from both holes so you'd be attracted to the circular path running around the sphere between the holes. Increase the size of those holes and you have a ring. The larger the holes, the higher the gravity would be (until it gets to the horizon which would be your head level - horizons in ringworlds would be weird, opposite). Again... a diagram would explain it easier.
Now... add in SPINNING.
lol ★■◆● that, someone who knows coriolis forces can do it. Snoopy?
okok i can't help it... i'm gonna theorise that if it was a vacuum and there were no tunnels, and you wern't touching the sides, then it'd be exactly the same as if the earth wasn't spinning.
But if there were atmosphere (there'd be HEAPS of it being that deep*) then the air rotating with the earth would sweep you along and then you'd be at the mercy of coriolis forces - the same force probably explains why when i create a whirlpool in the pool the dirt congregates in the center. So hey - i bet it'd also make you spin around in the center of the sphere in the center of the earth. MIRITE?!
(*lol, oh i fergot since there'd be no tunnels - you'd probabaly be in a vacuum. but since we're ignoring that the center of the earth is LIQUID we'll also ignore that. So... tunnelless access to atmosphere ftw!)
, it's addictive fun.
(that is, when the gods gift my mind the cruelly temporary ability to concentrate)
(well, not increases... but you get what i mean)
i didn't know the maths were the same for both volume of a semi-sphere and proximity to mass. That's brilliant that is!
ok, so now with a tunnel intersecting the sphere, what would happen is: The hollow "missing" section of the side of the sphere would be a gap in the mass, it would upset the perfrect ballance of mass, so you would be repelled from it - you'd be attracted to the wall exactly opposite the hole.
So, you wouldn't drift into the tunnel, but assuming you were able to jump up it - you'd just fall back down.
adding another tunnel in the other side of the sphere, so the 2 tunnels together go all the way from one side of the earth to the other (with the hollow central sphere joining them). This would create the gravity of the Halo ringworld - you'd be repelled from both holes so you'd be attracted to the circular path running around the sphere between the holes. Increase the size of those holes and you have a ring. The larger the holes, the higher the gravity would be (until it gets to the horizon which would be your head level - horizons in ringworlds would be weird, opposite). Again... a diagram would explain it easier.
Now... add in SPINNING.
lol ★■◆● that, someone who knows coriolis forces can do it. Snoopy?
okok i can't help it... i'm gonna theorise that if it was a vacuum and there were no tunnels, and you wern't touching the sides, then it'd be exactly the same as if the earth wasn't spinning.
But if there were atmosphere (there'd be HEAPS of it being that deep*) then the air rotating with the earth would sweep you along and then you'd be at the mercy of coriolis forces - the same force probably explains why when i create a whirlpool in the pool the dirt congregates in the center. So hey - i bet it'd also make you spin around in the center of the sphere in the center of the earth. MIRITE?!
(*lol, oh i fergot since there'd be no tunnels - you'd probabaly be in a vacuum. but since we're ignoring that the center of the earth is LIQUID we'll also ignore that. So... tunnelless access to atmosphere ftw!)
no kiddingTop Gun wrote:Trying to explain things like this is actually a great deal of fun for me.
, it's addictive fun.(that is, when the gods gift my mind the cruelly temporary ability to concentrate
)
assuming the earth is a perfect sphere with a constant density, the gravitational effect at the center is 0
if you wanted to, I guess you could do a triple integration to prove it, but I don't have the time
the fact that the earth is spinning, orbiting the sun, the sky is blue, and it's tuesday doesnt change anything
if you wanted to, I guess you could do a triple integration to prove it, but I don't have the time
the fact that the earth is spinning, orbiting the sun, the sky is blue, and it's tuesday doesnt change anything
I haven't lost my mind, it's backed up on disk somewhere.
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Testiculese
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Roid and top wop are funny, tho top wop seems to be mistaking volume for mass. Mass and distance are the only governing factors for gravity, the shape of that mass only determines where the centre of gravity is. A mass of the moon as a hollow sphere of with a radius of 10,000km exerts the same gravitional force as your the moon(gravity either being the exchange of gravitrons or a dimple in fabric of space).
Depending on the construction of the ring world in question, drilling thru to the other side and falling thru would do one of two things:
1. if its constructed with gravity simulated in a manner like Babylon 5(ie the ring stops you from flying off due to acceleration), then you'd fly out once you exit the hole, but probably wouldn't feel any appricable affects going thru the tunnel
2. if its constructed with high-density mass, like in Halo, then you'd experience zero-g half-way thru the tunnel, and then face climbing out to the other side.
Gravity is a very weak force - you need something the mass of the moon in order to feel 1/6 of gravity, you need something more dense for a ringworld.
Depending on the construction of the ring world in question, drilling thru to the other side and falling thru would do one of two things:
1. if its constructed with gravity simulated in a manner like Babylon 5(ie the ring stops you from flying off due to acceleration), then you'd fly out once you exit the hole, but probably wouldn't feel any appricable affects going thru the tunnel
2. if its constructed with high-density mass, like in Halo, then you'd experience zero-g half-way thru the tunnel, and then face climbing out to the other side.
Gravity is a very weak force - you need something the mass of the moon in order to feel 1/6 of gravity, you need something more dense for a ringworld.
I'm not mistaking anything for anything, although you seem to be mistaking me for someone else.fliptw wrote:Roid and top wop are funny, tho top wop seems to be mistaking volume for mass.
Um, not really.snoopy wrote:Sorry TG, Roid's right. You will always be pulled toward the nearest wall by a slight amount.
Sorry to be sardonic about it, but what I'm arguing is a physical fact. That's it, plain and simple. You may not understand why it is so, and lord knows I've tried my darndest to explain it, but it's true whether you believe it or not. If you'd like some more proof, take a look at the Wiki page on the divergence theorem, a.k.a. Gauss' Theorem, of vector calculus. Putting it simply, it states that the net flow of "stuff" out of a region is equal to the sum of all "sources" and "sinks" of "stuff" within it. This law is used ubiquitously in electrostatics to calculate the electric field of symmetric charge distributions; I just correctly solved about six or seven problems using it two weeks ago. It applies equally well to the gravitational force. Like ccb (I believe) said, the proof is in the mathematics. The article about the shell theorem provides further reinforcement.
Heck, here's how Newton himself determined it. Need I go on?