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Walking on a cube-shaped planet

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Re: Walking on a cube-shaped planet

#41
post #7

"Ask a Physician" has something to say about this: http://www.askamathematician.com/?p=6657

[There should be a standard disclaimer for "I'm correcting you because I think you would benefit from knowing, not because I mean to be an ass"] A physician is a medical doctor. A physicist studies fundamental laws of our universe.

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.

Re: Walking on a cube-shaped planet

#42

Earlier quoted context omitted.

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 equilibri…

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.

Re: Walking on a cube-shaped planet

#43
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.

Re: Walking on a cube-shaped planet

#44

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.

Pretty cool story, at least thats what I thought.

Re: Walking on a cube-shaped planet

#45
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).

Re: Walking on a cube-shaped planet

#46

Earlier quoted context omitted.

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 tha…

It's still the case that if you're sufficiently far away from a nonspherical object, it'll effectively act like a point source.

I'd guess this is what the OP was recalling.

Re: Walking on a cube-shaped planet

#47

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…

I think people find the sudden transition into horror the silly or unexpected part.

Re: Walking on a cube-shaped planet

#49
post #45

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 experi…

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.

Re: Walking on a cube-shaped planet

#50
post #45

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 experi…

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!

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