The seas wouldn't even be flat to the "surface" of the cube, right? They'd deform into a spheroid. So looking at it from a peak (or from space) it would appear to be a liquid sphere intersected with a solid cube.
Interesting point, but I'm not so sure. We take the shortcut of assuming that the earth's gravity is drawing everything toward a point at the center of the earth. But that's not really correct; every bit of matter is drawing everything else toward it.
So those giant mountains that comprise the vertexes of the cube would be pulling the oceans toward them. That works against your point. I don't know where the equilibrium between those forces sits, nor do I know how to calculate it.
A sphere with some gravitational influence from the points, but with most of the mass towards the center of the cube, it would still be mostly sphere-like, I think.
For a cube, what portion of its volume lies within an inscribed sphere? In other words, given a cube of edge=X and a sphere of diameter=X, what's the difference in volume?
Of course, this doesn't begin to account for the vectors, and it assumes that the density is uniform. But it should give us a sense of how much of the gravitational pull differs from our simplistic model.
Easier to figure using the radius rather than the diameter. Volume of the sphere is 4pi/3 * r^3, volume of the cube is (2r)^3. Factor out r^3 and you get (4pi/3)/8, or 52% of the cube's volume is within the sphere.
The math on that is easy. The volume of a cube is X^3, the volume of a sphere is 1/6 * pi * X^3, so the difference is a little under 1/2 X^3. Put another way, the cube is (roughly) twice the volume of the sphere.
That's right. If you are in the center of a face, the center of an edge, or on one of the vertices the gravity vectors all point towards the core.
An imagined sphere inside the cube holds 52% of the mass and so the remaining mass is divided among eight vertices, about 6% for each.
If you are off a ways from one of the points listed above, the sphere part (52%) will always pull towards the center. A lot of the force from the corners will be mitigated/cancelled by symmetry, and whatever is left over, again, it's only 6% maximum per corner.
For all we know, anyone making a square planet would be able to put all the mass in the corners, with only 0.00001% for the super-material used to make the faces of the cube.
The math is relatively easy with uniform density. Otherwise we have to start conjecturing on the density distribution, and without real-world feedback it degenerates to wondering if an infinite number of angels can dance on the head of a pin.
One place to start would be the Schiehallion Experiment of 1774 where a symmetric and isolated mountain in Scotland was used as the basis of an experiment to measure how much the gravitational attraction of the mountain pulled a pendulum away from the vertical. The goal being to estimate the mass of the Earth - with what turns out to have been a reasonably accurate result.
During the course of the experiment they had to measure the shape of the mountain quite carefully and they had the idea to join points of similar height together with a line on the map - inventing contour lines.
Sorry I don't have a reference, but as long as you are outside of an object, its gravitational force on you is the same as if it were a point-mass (at least for classical physics).
This is only true for objects that have a spherically symmetrical mass distribution and for a point test object. That is, the field of a spherically symmetric object is the same as the field of a point mass of its mass located at its center.
As soon as your test object is non-point you get tidal forces due to the field typically being nonuniform. As soon as your object is not spherically symmetric you get a field that looks nothing like that of a point mass.
Why not? The most important thing about a round planet in this case (and, in fact, the reason planets tend towards roundness) is that it means gravity is (roughly) equal at every point on the surface (and, in fact, at every point on any planet-shaped shell inside it, as well). On a cube, this is entirely gone. The centers of the faces would have the strongest gravities; the corners would have the weakest. For localised walks we wouldn't notice the gravitational changes (just as we don't notice them when we walk up a staircase), but if you made the trek from the center of a face to the edge, you will definitely notice.
I took the question (which I thought was rather absurd) and rationalized (and imagined it) this way: this cubic planet will not rotate, or it will not be cubic then. I took out gravity as we know it, and imagined I could just "walk" in it.
We walk on a sphere shaped planet, yet we do not "slide down" if we stand on the North Pole, yes? Same with a cubic one, as I imagine, the 90 degrees angle will be crossed without even realizing is there. I imagine we would perceive it as a flat planet.
As you can see, I did not take the whole thing too seriously. And that, of course, diminishes karma. So it goes. Cheers!
Think about the corners like giant mountains on an otherwise sphere shaped planet. If you were at the top of a huge mountain and lost your footing, you would 'slide down.'
Most likely you're losing karma because you come across as trolling. In case you're not, picture "down" as towards the center of mass. On the surface of a sphere, that means perpendicular to the surface. At the edge of two faces of a cube, "down" is through the edge (between the faces), with the faces 45 degrees off-vertical. Like on a mountaintop.
"The Highest Possible Level of Development civilization. A gravely injured hermit comes to Trurl's house and tells Trurl of Klapaucius's adventure: Klapaucius wanders across an old robot, who tells him that he has logically deduced the existence of a civilization that reached the highest possible level of development (hence "HPLD"). He has inferred the existence of such a civilization by figuring that if there are different stages of development, there will be one that is the highest. He was then faced with a problem of identifying that one; as he noted, everyone claimed that theirs was the HPLD. Upon much research and thought, he decided that the only way to find it is by looking for a "wonder", i.e. something that has no rational explanation. Eventually Klapaucius discovers one such wonder: a star in the shape of a cube, orbited by a planet also shaped like a cube with the huge letters HPLD written on it."
Sigh, I should have simply copy&pasted the "Ask a Physicist" from the title. And I actually know the difference between physician and physicist (when I consciously think about it), but when not careful I do sometimes mix them up.
> If you were standing on the edge of a face, and looked back toward the center, you’d be able to clearly see the round bubble of air and water extending above the flat surface. I strongly suspect that it would be pretty.
Anyone here willing to take on the task of rendering an image of what this would look like?
> On spherical earth the horizon on average is a little over three miles away.
I read (long time ago) that the horizon is 29km on a seashore. In other words if you see a ship disappear on a horizon, it was 29km away. And so his 3mi vs my 29km is a bit of a discrepancy. Can anyone set things straight here?
The "distance to the horizon" is strongly dependant on how high the viewer is and how tall the object being observed is. If you put your eyes pretty much down to the ground the horizon will be pretty darn close. If you are standing 6 feet tall the horizon will be 3 miles away. If you're standing a bit higher up on the shore and looking at the top of a tall ship then 29 kilometers is easy to achieve. I'm providing a link to a handy calculator. You have to add the results from both the height of the viewer and the viewed object.
http://www.ringbell.co.uk/info/hdist.htm
As a point of reference - the cliffs of Dover can be seen from Calais on a good day, and that's a distance of 34 km (21 miles). Wikipedia gives the max. elevation at Calais as 18m, and 107m at the White Cliffs, so that works out nicely (though the calculations would be a bit more involved than what the calculator you link to allows).
Regarding the perceived gravity on a cube planet, the movie Sunshine tries to portray accurate physics for a similar situation. It has great sci-fi visuals, though the end gets a bit... silly.
I don't understand why everyone thinks the horror aspect of the film was silly. The captain of the other ship was driven mad by pure light and retreats into the darkness. He then tries to sabatoge the attempt to restart the sun so he (and all of the human race) can be forever encased in the night. These effects were illustrated in some of the characters on the second ship as well.
And the final moment of the film was merely playing with the idea that the intense gravity of the sun would alter time to the point where Capa would experience each nanosecond before his death, observing the nuclear reaction as it occured.
Thanks for this. The weird ending had me mystified and not thinking well of an otherwise good movie. It finally all makes sense. I'm going to have to rewatch it this weekend.
It was working up until about five minutes ago. I suspect I'm not the only one who'd never heard of the The Straight Dope and started clicking through the archives after reading this article.
The Straight Dope was where I first learn of the Monty Hall problem. It's also where I found out exactly how stubbornly wrong some people can be in the face of absolute proof.
"Earth-scale gravity is so strong that a cube made of the strongest rock would soon be deformed into a ball."
Not necessarily true. If the cube-shaped world has a cube-shaped moon orbiting at the same period as our moon, it will be sooner transformed into a ball by creepers.
If no sci-fi-3d-renderer makes a complete physically based rendering of this I will be surprised. Some nice Mie-scattering-atmosphere simulation with awesome distances and displacements maps from a Borg cube.. Yumm..
Wow, this illustration might be the first one ever to help me understand how space can be 'curved'. If you altered the space around the cube to reflect a linear gravitational pull, you'd have a cube with streteched out points and a flat ocean within the curved bason of the cube face.
Correct me if I'm completely imagining things here, but is this how space is 'curved' by gravity? The disorted shape of the ocean and cube would reflect how the cube-planet felt to an observer within its gravitational pull.
I had the same thought. you see yourself at the bottom of a bowl with ridges above you yet you know the surface is flat. If you take away the gravity then there is no wicked climb, ergo no mountain: the world suddenly appears flat. I know little physics but I think you're right! If true then your insight is greater than the authors.
Very fun thought experiment. It reminded me that we did a problem in calculus to compute the effect of gravity inside a hollow sphere. It turns out to be ZERO.
So, if the Earth's mass, was all densely concentrated in a, say, 1 mile thick shell, you could drill a hole through the shell and experience total weightlessness when you popped out on the "inside" (assuming a total vacuum on the inside - if not - you'd experience a very small gravity toward the center based on the mass of the contained atmosphere).
If the inside was a total vacuum, the moment you finished drilling through the 1 mile shell, the vast majority of the planet's atmosphere would be sucked through the open hole at an insanely fast rate.
So on the surface of the shell planet, the effects of gravity would be the same as normal, but when you're on the inside there is no net gravitational effect? ie. you'd just float near the inside surface? That doesn't seem intuitive. Wouldn't the force from gravity be acting towards the centre of mass in both cases?
Edit: Actually never mind, I think the density of the shell is what threw me off! On the outside the net gravitational effect is the local shell (RHS) plus the rest of the shell (RHS), but just on the inside - the net effect is the remaining shell surface (RHS) minus the local shell (LHS). Because the effect of gravity is 1/r^2 everything works out!
Another planetary thought experiment (edit: sorry, credit where it's due: excerpt from The Algebraist by Iain M. Banks):
I was born in a water moon. Some people, especially its inhabitants, called it a planet, but as it was only a little over two hundred kilometres in diameter, 'moon' seems the more accurate term. The moon was made entirely of water, by which I mean it was a globe that not only had no land, but no rock either, a sphere with no solid core at all, just liquid water, all the way down to the very centre of the globe.
If it had been much bigger the moon would have had a core of ice, for water, though supposedly incompressible, is not entirely so, and will change under extremes of pressure to become ice. (If you are used to living on a planet where ice floats on the surface of water, this seems odd and even wrong, but nevertheless it is the case.) The moon was not quite of a size for an ice core to form, and therefore one could, if one was sufficiently hardy, and adequately proof against the water pressure, make one's way down, through the increasing weight of water above, to the very centre of the moon.
Where a strange thing happened.
For here, at the very centre of this watery globe, there seemed to be no gravity. There was colossal pressure, certainly, pressing in from every side, but one was in effect weightless (on the outside of a planet, moon or other body, watery or not, one is always being pulled towards its centre; once at its centre one is being pulled equally in all directions), and indeed the pressure around one was, for the same reason, not quite as great as one might have expected it to be, given the mass of water that the moon was made up from.
I used to think that the ice core would work that way, but water is most dense at 3.7C (or something like that). So, under enough pressure, it would remain liquid.
Water has a very complicated phase diagram. I'm not sure how high a pressure water at a core of a planetoid of that size would be, but if it's above 100GPa then it would certainly be solid (Ice X or XI). Below that pressure, it would depend on the temperature.
pressure = Integral[h=x to infinity] of rho g(h) dh where x is the current distance from the center. "h" here means "height." Since we're at the center, x=0. Since rho(vacuum) = 0, this reduces to Integral[h=0:200km] rho g(h) dh. I simplify and say that water is incompressible, with density of rho = 1000 kg/mmm. (At 1 giga pascals the density is 1.2 that of its normal density.)
g(r) = G * M(r) / r * r where M(r) is the mass inside a sphere of radius r. That's G (rho 4/3 pi r * r * r) / r * r or 4/3 rho G pi r.
Therefore, pressure = Integral[h=0:200km] rho * (4/3 rho G pi h) dh. This is 2/3 rho * rho G pi h * h, giving 5.6 mega pascals.
Looking at the phase diagram you pointed out, 6MPa is well within the liquid range. (Presumably the temperature would be around 4C). Even with a higher density it's at most about 7MPa, and you need to get to 1 GPa for water to always be a solid.
In other words, unless I did my math wrong or left something out of the calculations, this water sphere isn't close to one of the ice phases of water.
BTW, for more fun, since there's no heat source for the water (radioactive decay, heat from the phase change to ice, etc), then how does it stay liquid? If exposed to vacuum it would turn to vapor until it cooled down enough (about -60C) to freeze. There isn't enough mass to hold down much atmosphere, so I presume it's covered.
How also does it get enough energy to stay liquid? Most of the energy will be dumped into the top layer, so there will be a warm zone on top of a thermocline, like with Earth's oceans. If the energy comes from the sun, then do the poles freeze? If it doesn't rotate fast enough then the backside will freeze.
If ice does form, it reflects more heat than water so that region will stay ice.
Actually, I haven't read The Algebraist yet, but if I had to guess, it's a vaguely magical Culture explanation, like artificial suns surrounding the moon, or industrial processes within giving off enough waste heat to keep it warm, like the Puppeteer homeworld.
A waterdrop world with industrial processes would be a Culture-like bauble since there's nothing to base the processes upon but water. In any case, they can control gravity and therefore pay little heed to my calculations.
By the way, the pressure in the Mariana Trench is about 108 megapascals. The 7MPa I calculated is therefore about the same as 700 meters of depth.
Humans now, with the Atmospheric Diving System, have dived to 610 meters, and free divers have made it to 265m. We could probably make a (nuclear) sub that could cruise through the entire thing. The max depth for a steel sub is 250–400 meters and titanium to 1,000 meters.
Naturally, the large airless portions of the surface would be pocked by craters, with enormous scree fields washing center-ward from from many of those craters, as the debris fell back to earth to bounce, slide, and roll "downhill".
The Integral Trees (Larry Niven): the habitable space is a torus.
It's a torus of air in orbit. The trees look like integral signs because they align pointed toward the star, but have constant wind in opposite directions at either end, because the air there has different orbital speeds, being closer or further from the star.
It seems that both the straight dope and ask-a-physicist links describe the effects as something similar to having eight giant mountains on Earth whose summits make up the points of a cube. This led me to wonder, what if we consider the case where the Earth is exactly the same (geologically) as it is now, but in the future we put up eight giant towers whose tops describe those summit points? Assuming a supermaterial has been discovered whose properties allow for an extremely light structure with enormous compression strength relative to its weight, like aerogel ( http://bashinginminds.com/2010/01/23/playing-with-nasas-soli... ) but even aerogel-ier. What would that be like?
Is it then too far-fetched to imagine that some freak process of nature or other might conceivably allow for the creation of a bizarre planet with an immensely dense spherical core and a lighter mantle and crust that take on the shape of a cube externally? Nature is not averse to giving us cubes, after all: http://www.gemstoneslist.com/pyrite.html
There are all sorts of rules for drama and character development. What are the rules for incorporating neat scientific/sci-fi ideas into a story? In lots of Larry Niven's stories, the ideas are necessary to solve the central mystery, but not always.
EDIT: In a mashup of The Culture and Dilbert, a godlike nanotech Dogbert forces the hapless Homo Sapiens Dilbert to work on a "cubical".
78 comments
[ 6.9 ms ] story [ 143 ms ] threadSo those giant mountains that comprise the vertexes of the cube would be pulling the oceans toward them. That works against your point. I don't know where the equilibrium between those forces sits, nor do I know how to calculate it.
Of course, this doesn't begin to account for the vectors, and it assumes that the density is uniform. But it should give us a sense of how much of the gravitational pull differs from our simplistic model.
An imagined sphere inside the cube holds 52% of the mass and so the remaining mass is divided among eight vertices, about 6% for each.
If you are off a ways from one of the points listed above, the sphere part (52%) will always pull towards the center. A lot of the force from the corners will be mitigated/cancelled by symmetry, and whatever is left over, again, it's only 6% maximum per corner.
The math is relatively easy with uniform density. Otherwise we have to start conjecturing on the density distribution, and without real-world feedback it degenerates to wondering if an infinite number of angels can dance on the head of a pin.
During the course of the experiment they had to measure the shape of the mountain quite carefully and they had the idea to join points of similar height together with a line on the map - inventing contour lines.
http://en.wikipedia.org/wiki/Schiehallion_experiment
As soon as your test object is non-point you get tidal forces due to the field typically being nonuniform. As soon as your object is not spherically symmetric you get a field that looks nothing like that of a point mass.
I'd guess this is what the OP was recalling.
That doesn't apply to being on the face of a cube, though; that's not nearly far enough away, obviously. :)
We walk on a sphere shaped planet, yet we do not "slide down" if we stand on the North Pole, yes? Same with a cubic one, as I imagine, the 90 degrees angle will be crossed without even realizing is there. I imagine we would perceive it as a flat planet.
As you can see, I did not take the whole thing too seriously. And that, of course, diminishes karma. So it goes. Cheers!
Yes, I can picture it as you describe it.
Well, if you throw out ALL the rules....
"The Highest Possible Level of Development civilization. A gravely injured hermit comes to Trurl's house and tells Trurl of Klapaucius's adventure: Klapaucius wanders across an old robot, who tells him that he has logically deduced the existence of a civilization that reached the highest possible level of development (hence "HPLD"). He has inferred the existence of such a civilization by figuring that if there are different stages of development, there will be one that is the highest. He was then faced with a problem of identifying that one; as he noted, everyone claimed that theirs was the HPLD. Upon much research and thought, he decided that the only way to find it is by looking for a "wonder", i.e. something that has no rational explanation. Eventually Klapaucius discovers one such wonder: a star in the shape of a cube, orbited by a planet also shaped like a cube with the huge letters HPLD written on it."
Now I know how it is like, to be one of HPLD.
You probably know about this already but for the benefit of others I will post this:
http://en.wikipedia.org/wiki/List_of_ships_of_the_Culture_se...
Trurl, attempting to build a machine to write poetry:
http://books.google.com/books?id=kWElP9YZkzQC&lpg=PA44&#...
A physician is a medical doctor. A physicist studies fundamental laws of our universe.
I think it would quickly lose it's tone of respect, and take on one of sarcasm.
Anyone here willing to take on the task of rendering an image of what this would look like?
I read (long time ago) that the horizon is 29km on a seashore. In other words if you see a ship disappear on a horizon, it was 29km away. And so his 3mi vs my 29km is a bit of a discrepancy. Can anyone set things straight here?
Wikipedia says 3.1 miles.
And the final moment of the film was merely playing with the idea that the intense gravity of the sun would alter time to the point where Capa would experience each nanosecond before his death, observing the nuclear reaction as it occured.
Pretty cool story, at least thats what I thought.
One family of orbits is parallel to a face of the cube, if I understand the paper even a little.
Not necessarily true. If the cube-shaped world has a cube-shaped moon orbiting at the same period as our moon, it will be sooner transformed into a ball by creepers.
Correct me if I'm completely imagining things here, but is this how space is 'curved' by gravity? The disorted shape of the ocean and cube would reflect how the cube-planet felt to an observer within its gravitational pull.
So, if the Earth's mass, was all densely concentrated in a, say, 1 mile thick shell, you could drill a hole through the shell and experience total weightlessness when you popped out on the "inside" (assuming a total vacuum on the inside - if not - you'd experience a very small gravity toward the center based on the mass of the contained atmosphere).
Well, at the same rate as air at 1 atmosphere pressure will fill any vacuum container through a hole of that size.
You seem to suggest that fact that that speed somehow depends on the size of the hollow earth and the quantity of the atmosphere.
The pressure depends only on the height of the air column.
Edit: Actually never mind, I think the density of the shell is what threw me off! On the outside the net gravitational effect is the local shell (RHS) plus the rest of the shell (RHS), but just on the inside - the net effect is the remaining shell surface (RHS) minus the local shell (LHS). Because the effect of gravity is 1/r^2 everything works out!
I was born in a water moon. Some people, especially its inhabitants, called it a planet, but as it was only a little over two hundred kilometres in diameter, 'moon' seems the more accurate term. The moon was made entirely of water, by which I mean it was a globe that not only had no land, but no rock either, a sphere with no solid core at all, just liquid water, all the way down to the very centre of the globe.
If it had been much bigger the moon would have had a core of ice, for water, though supposedly incompressible, is not entirely so, and will change under extremes of pressure to become ice. (If you are used to living on a planet where ice floats on the surface of water, this seems odd and even wrong, but nevertheless it is the case.) The moon was not quite of a size for an ice core to form, and therefore one could, if one was sufficiently hardy, and adequately proof against the water pressure, make one's way down, through the increasing weight of water above, to the very centre of the moon.
Where a strange thing happened.
For here, at the very centre of this watery globe, there seemed to be no gravity. There was colossal pressure, certainly, pressing in from every side, but one was in effect weightless (on the outside of a planet, moon or other body, watery or not, one is always being pulled towards its centre; once at its centre one is being pulled equally in all directions), and indeed the pressure around one was, for the same reason, not quite as great as one might have expected it to be, given the mass of water that the moon was made up from.
This was, of course,—
http://upload.wikimedia.org/wikipedia/commons/0/08/Phase_dia...
g(r) = G * M(r) / r * r where M(r) is the mass inside a sphere of radius r. That's G (rho 4/3 pi r * r * r) / r * r or 4/3 rho G pi r.
Therefore, pressure = Integral[h=0:200km] rho * (4/3 rho G pi h) dh. This is 2/3 rho * rho G pi h * h, giving 5.6 mega pascals.
Looking at the phase diagram you pointed out, 6MPa is well within the liquid range. (Presumably the temperature would be around 4C). Even with a higher density it's at most about 7MPa, and you need to get to 1 GPa for water to always be a solid.
In other words, unless I did my math wrong or left something out of the calculations, this water sphere isn't close to one of the ice phases of water.
BTW, for more fun, since there's no heat source for the water (radioactive decay, heat from the phase change to ice, etc), then how does it stay liquid? If exposed to vacuum it would turn to vapor until it cooled down enough (about -60C) to freeze. There isn't enough mass to hold down much atmosphere, so I presume it's covered.
How also does it get enough energy to stay liquid? Most of the energy will be dumped into the top layer, so there will be a warm zone on top of a thermocline, like with Earth's oceans. If the energy comes from the sun, then do the poles freeze? If it doesn't rotate fast enough then the backside will freeze.
If ice does form, it reflects more heat than water so that region will stay ice.
Actually, I haven't read The Algebraist yet, but if I had to guess, it's a vaguely magical Culture explanation, like artificial suns surrounding the moon, or industrial processes within giving off enough waste heat to keep it warm, like the Puppeteer homeworld.
By the way, the pressure in the Mariana Trench is about 108 megapascals. The 7MPa I calculated is therefore about the same as 700 meters of depth.
Humans now, with the Atmospheric Diving System, have dived to 610 meters, and free divers have made it to 265m. We could probably make a (nuclear) sub that could cruise through the entire thing. The max depth for a steel sub is 250–400 meters and titanium to 1,000 meters.
http://books.google.com/books/about/Mathenauts.html?id=tHQsA...
Great book, BTW.
It's a torus of air in orbit. The trees look like integral signs because they align pointed toward the star, but have constant wind in opposite directions at either end, because the air there has different orbital speeds, being closer or further from the star.
Is it then too far-fetched to imagine that some freak process of nature or other might conceivably allow for the creation of a bizarre planet with an immensely dense spherical core and a lighter mantle and crust that take on the shape of a cube externally? Nature is not averse to giving us cubes, after all: http://www.gemstoneslist.com/pyrite.html
EDIT: In a mashup of The Culture and Dilbert, a godlike nanotech Dogbert forces the hapless Homo Sapiens Dilbert to work on a "cubical".