Stoner Science: Pooping on the Equator

[Wicked Pilot]

Well, yes, although such a structure might still be a problem for the person in it, because it will still collapse under its own gravity.

Good observation Einstein, I'd never would have thought of that.

[Hardy]

The direction of the swirl in toilets, where the water comes down rather forcefully, yes, but the myth does have some truth in it, since there really is a slight preference in rotation.

Right. Large storms, for example, fall victim to this.

No amount of gravity can damage the person unless there is some other force trying to prevent the freefall.

Thank you. I was trying to confirm that before I answered question three, but I simply supposed the opposite.

[Kuroneko]

Well, actually your example is somewhat interesting, since it is less than an order of magnitude away from structural integrity. For high-quality steels, if the thickness was 1/5 of the given figure or less, which is still a very impressive 200mi, the structure should hold, although only just.

[LordShaithis]

Hrm. Well if there's a hollow spherical chamber inside this high-gravity mass, and you're clinging to one side, and you let go... You would naturally float back towards the center, correct? Is there a point where gravity could be powerful enough that this "everything to the center" impulse would leave you a symmetrical ball of goo floating dead-center?

[Hardy]

Hrm. Well if there's a hollow spherical chamber inside this high-gravity mass, and you're clinging to one side, and you let go... You would naturally float back towards the center, correct?

No. All graviational forces inside the hollow sphere cancel out. The only way you'd be able to float to the center is if you pushed, which I doubt you'd be able to stop, unless you had some means of propulsion (assuming that the interior is a vacuum).

Is there a point where gravity could be powerful enough that this "everything to the center" impulse would leave you a symmetrical ball of goo floating dead-center?

Gravity cancels out at all points in the sphere, so I don't think there would be. I may not understand what you mean, though.

[Jetfire]

For the second question, there was a chapter in Greg Benford's "Galactic Centre" series where aliens were using a Superstring to core out a planet. Among other things it created a vacuum along the axis of rotation throuhg the centre of the planet.

Anyways, one of the main characters was then sent throuhg that axis (he was in a space suit) and started bobbing from one end of the planet to the other. Air resistance wasn't a factor, but gravity pressure creating heat became an issue, with each trip throuhg the core, not to mention it was bleeding off a bit

I believe it was the 4th book that it occured in.

[Wicked Pilot]

Hrm. Well if there's a hollow spherical chamber inside this high-gravity mass, and you're clinging to one side, and you let go... You would naturally float back towards the center, correct? Is there a point where gravity could be powerful enough that this "everything to the center" impulse would leave you a symmetrical ball of goo floating dead-center?

No, regardless of the chamber's mass and regardless of your position gravity will be zero. You would not float to the center unless some other force pushed you there.

[Molyneux]

Hrm. Well if there's a hollow spherical chamber inside this high-gravity mass, and you're clinging to one side, and you let go... You would naturally float back towards the center, correct? Is there a point where gravity could be powerful enough that this "everything to the center" impulse would leave you a symmetrical ball of goo floating dead-center?

Think of it as if there were invisible rubber bands connecting you to every particle of matter in the hollow sphere (for the purposes of this, you're in a hamsterball, so your body acts as a point mass). Now, if you're in the center, every rubber band is pulling on you equally, so there's no net force. You don't go anywhere.

Here's where the rubber-band thing breaks down - imagine that as these rubber bands get shorter, they get *stronger*, too, such that the change in distance creates a change in strength such that a change from 2R to R causes a change in the force such that F (when you're 2R away from a given particle) becomes 4F, and you're exempted from the draft...er...never mind.

Basically, no matter where you go in the sphere (assuming you have a jetpack), the increase in strength of the gravity from one particular direction is exactly balanced by the increase in the number of particles exerting force on you in the opposite direction - at any point inside a hollow sphere, the net force of gravity is exactly the same, that is, zero.


Also, I may be mistaken on this, but I think that since the net force exerted on every particle of your body is zero, there is no force at all acting on you in any way (as in, crushing you into goo). No matter how massive the sphere is, if you're inside it you're fine.

[Lusankya]

Would there come a point when tidal forces squished your body (assuming the Earth was just getting denser, and not simply more voluminous at the same density)? (for case 3)

[Wicked Pilot]

Would there come a point when tidal forces squished your body (assuming the Earth was just getting denser, and not simply more voluminous at the same density)? (for case 3)

Nope, once you're in the center the net effect of gravity is zero for all points along your body.

[Lord Zentei]

Would there come a point when tidal forces squished your body (assuming the Earth was just getting denser, and not simply more voluminous at the same density)? (for case 3)

The density at a given point in/on an object and the distance to that point determine the contribution of that point to the total gravity field you experience. Inside a spherical shell, the contributions of all points on the shell cancel. If the density at point A is increased, the gravity due to that point goes up, but so does the density and hence gravity of the points that cancel A's gravity - and by the same relative amount. Thus the total gravity remains zero.

If one point on the shell is denser/less dense than the others, such that it's mass goes up/down by X kilograms, you will gravitate to/from that point exactly as though to a point mass of X/-X kilograms. Also, objects inside and outside the shell will still contribute gravity. It's the shell itself that can be disregarded for the purpose of constructing a mathematical model of the overall gravity inside the shell - and here I'm referring to the the "average" shell, so to speak, counting any imperfections as distinct objects.

[Molyneux]

Would there come a point when tidal forces squished your body (assuming the Earth was just getting denser, and not simply more voluminous at the same density)? (for case 3)

The density at a given point in/on an object and the distance to that point determine the contribution of that point to the total gravity field you experience. Inside a spherical shell, the contributions of all points on the shell cancel. If the density at point A is increased, the gravity due to that point goes up, but so does the density and hence gravity of the points that cancel A's gravity - and by the same relative amount. Thus the total gravity remains zero.

If one point on the shell is denser/less dense than the others, such that it's mass goes up/down by X kilograms, you will gravitate to/from that point exactly as though to a point mass of X/-X kilograms. Also, objects inside and outside the shell will still contribute gravity. It's the shell itself that can be disregarded for the purpose of constructing a mathematical model of the overall gravity inside the shell - and here I'm referring to the the "average" shell, so to speak, counting any imperfections as distinct objects.

Yeah, I was afraid of that...we've strayed into the realm of physical ideals, the place where perfect springs, massless ropes, and frictionless tabletops come from.

[Lord Zentei]

Yeah, I was afraid of that...we've strayed into the realm of physical ideals, the place where perfect springs, massless ropes, and frictionless tabletops come from.

The question was about a hole clean through the Earth. What the fuck were you expecting? A viable blueprint?

Anyway, considering imperfections as distinct objects and disregarding the shell itself is a perfectly valid mathematical tool that can be used to model reality. Disregarding the imperfections too that is physical ideal, and is still OK if the imperfections are small.

[LordShaithis]

Thank you friends, I have been enlightened. I wouldn't be surprised to see one of these posts again in the future, next time I'm asked a question that's just plain too weird to be within my layman's grasp.

[PigRotten]

for case 2, as i understand the physics of gravity, one is not actually attracted to mass in general, but rather the center of mass for a certain massive object, in this case the center of the earth. now because the absolute center of the earth is a coordinate, it has no area or volume, thus it is quite impossible to fit inside it. thus these forces would still be acting on you

now that being said the equation for the force of gravity between one object to another dictates that, as you get closer to this center of mass, the force that gravity exerts upon you only gets stronger, more specifically, at about 1 meter from the center of the earth the force gravity would be exerting on you would be around 4.5x10^13 times greater than at the surface, only to get much greater as you get closer, now if you were at the absolute center of the earth, because of what was stated in the above paragraph, this force would be pulling from all sides of you, thus crushing you into a stony mass of what was once your body.

now im no physics professor, so i may very well be wrong, but this is how i understand it, so if im just crazy please feel free to ridicule the crap outa me.

[Molyneux]

for case 2, as i understand the physics of gravity, one is not actually attracted to mass in general, but rather the center of mass for a certain massive object, in this case the center of the earth. now because the absolute center of the earth is a coordinate, it has no area or volume, thus it is quite impossible to fit inside it. thus these forces would still be acting on you

now that being said the equation for the force of gravity between one object to another dictates that, as you get closer to this center of mass, the force that gravity exerts upon you only gets stronger, more specifically, at about 1 meter from the center of the earth the force gravity would be exerting on you would be around 4.5x10^13 times greater than at the surface, only to get much greater as you get closer, now if you were at the absolute center of the earth, because of what was stated in the above paragraph, this force would be pulling from all sides of you, thus crushing you into a stony mass of what was once your body.

now im no physics professor, so i may very well be wrong, but this is how i understand it, so if im just crazy please feel free to ridicule the crap outa me.

Ridicule? Naah...making fun of people never accomplishes anything, except for creating a supply of annoyed people.

Gravity does indeed work from every particle with mass; however, the center of gravity of an object is a point such that the total sum of all mass of that object can be approximated by a single point mass at the center of gravity. For a uniform sphere or most other regular geometric objects, the center of gravity is the center of the object; with weirder shapes, the center can be anywhere, depending on the distribution of mass.

For someone outside of a large, hollow sphere,(and a good distance away) the gravity does indeed seem to be acting from the center of mass; however, once you're inside the object, the gravity no longer affects you the same way, because the mass of the shell is now acting on you in different directions. Inside of a hollow sphere of uniform density, there is NO gravitational force at all (barring outside factors).