Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: John Chapman on 02/05/2010 00:04:18

Could you help me solve a problem I've been pondering over. If you lived on a doughnut shaped planet would you be able to walk around it's entire surface?
Yesterday I thought ‘no’. I reckoned that because a doughnut is radially symmetrical it’s centre of gravity is in the centre of the hole and so all the planets inhabitants will fall to that point.
Today I believe that the ‘me’ of yesterday was clearly a moron. I now believe you could stand on any point of it’s surface, even within the hole, because the ground which is closest will exert the greatest gravitational pull and would stop you from floating away. But would this make the centre of gravity not a single point but a line running internally through the centreline of the doughnut? That can’t be right.
What do you think?

Could you help me solve a problem I've been pondering over. If you lived on a doughnut shaped planet would you be able to walk around it's entire surface?
Yesterday I thought ‘no’. I reckoned that because a doughnut is radially symmetrical it’s centre of gravity is in the centre of the hole and so all the planets inhabitants will fall to that point.
Today I believe that the ‘me’ of yesterday was clearly a moron. I now believe you could stand on any point of it’s surface, even within the hole, because the ground which is closest will exert the greatest gravitational pull and would stop you from floating away. But would this make the centre of gravity not a single point but a line running internally through the centreline of the doughnut? That can’t be right.
What do you think?
Obviously, you need to get another hobby.
I think you are right in thinking that you'd be able to walk all over it without falling off, although you would be a bit lighter on your feet when you were on the inside. Its centre of gravity as a whole (haha) would still be in the middle of the hole though.
If you fell from space towards the hole you'd keep on going then eventually stop, and fall back through the hole again  ad nauseam. It's similar to the hole that goes right through the centre of the Earth question.

If it were spinning (and most things do) and the axis was perpendicular to the hole; could you not get a nice balance of gravity towards centre and 'artificial gravity' because you are on the inside surface of a rotating object?

If it were spinning (and most things do) and the axis was perpendicular to the hole; could you not get a nice balance of gravity towards centre and 'artificial gravity' because you are on the inside surface of a rotating object?
Good point!

An interesting thought about imat's question. If his doughnut had the mass of an Earth and was spinning like a merrygoround, wouldn't the overall time dilation of the space surrounding his object exhibit a sort of north and south pole?

With real "back of envelope" calcs  I reckon the torus (of rought same vol/mass as earth) would have to be spinning 10100 times quicker than the earth to make any difference to the force felt on inside surface . Might sit down and crunch numbers later  but glued to the shennanigans on the tv election broadcast. Would need really nice balancing.
ps hope i havent missed out a vital factor

Imat, I think you missed my point. If we look at the time dilation of your object as a field, analogous to the field surrounding a magnet, would it have a sort of time dilation north and south pole?

Hi Geezer
Yes, that’s what I think. (About walking on Planet Doughnut, not about needing a hobby).
But if you think the centre of gravity is in the hole then what about the centre of mass? That can’t be in the hole, surely. Are the two concepts effectively the same thing?

Hi Imatfaal
If it were spinning (and most things do) and the axis was perpendicular to the hole; could you not get a nice balance of gravity towards centre and 'artificial gravity' because you are on the inside surface of a rotating object?
Wouldn’t the gravity from the mass of the doughnut and the gravity from the spin both be acting in the same direction if you were standing on the ‘hole’ (inner) surface of the doughnut but cancel out if you were standing on the outer surface. This is the opposite to the effect that would exist if the doughnut was not spinning – ie the gravity near the hole would be less due to the pull of the mass of the doughnut from the other side of the hole. As Geezer said “you would be a bit lighter on your feet when you were on the inside”. If you could make Planet Doughnut spin at just the right speed then presumably you could get equal gravity at all points on the surface.
Just a thought ... If the centrifugal force of a spinning planet affects the net pull of gravity, does that mean that if the planet Earth stopped spinning we would all get heavier?

Hi Ron
A? How’s that work, then?

Hi Geezer
Yes, that’s what I think. (About walking on Planet Doughnut, not about needing a hobby).
But if you think the centre of gravity is in the hole then what about the centre of mass? That can’t be in the hole, surely. Are the two concepts effectively the same thing?
Hi John,
Yes. I think the centre of mass would be in the middle of the doughnut, even though there is nothing there.
According to Wikipedia  Center of Mass
"The term center of mass is often used interchangeably with center of gravity, but they are physically different concepts. They happen to coincide in a uniform gravitational field, but where gravity is not uniform, center of gravity refers to the mean location of the gravitational force acting on a body. This results in small but measurable gravitational torque that must be accounted for in the operation of artificial satellites."

Ron,
I kinda got your point but it is beyond me/my physics. My response was meant to indicate that (in my best guess) the speeds involved are so small to be insignificant in relativistic calculations.
If we look at the, albeit small, time dilation due to relative speeds and distances from centres of gravity it all gets beyind my ken. However I am not sure that side of the calcution to do with speeds vary on a torus as compared to a sphere  there is greater variation of speed "through the ether" on a sphere (pole vs equator) than on a torus (inside vs outside). On the gravity side (assuming perfect sphere) everyone of similar mass experiences similar force (1/r^2 is equal); on a torus (assuming size of hole is same as crosssection of ring) those on outside feel force towards centre of gravity nine times less than those inside ring (1/R^2 vs 1/(3R)^2)
How these difference positions in the gravity well and varying speeds would affect time perception are beyond me but I am pretty certain that they would be unnoticiable on earthlike size and spin
Matthew

Wouldn’t the gravity from the mass of the doughnut and the gravity from the spin both be acting in the same direction if you were standing on the ‘hole’ (inner) surface of the doughnut but cancel out if you were standing on the outer surface. This is the opposite to the effect that would exist if the doughnut was not spinning – ie the gravity near the hole would be less due to the pull of the mass of the doughnut from the other side of the hole. As Geezer said “you would be a bit lighter on your feet when you were on the inside”. If you could make Planet Doughnut spin at just the right speed then presumably you could get equal gravity at all points on the surface.
Just a thought ... If the centrifugal force of a spinning planet affects the net pull of gravity, does that mean that if the planet Earth stopped spinning we would all get heavier?
Hello John,
I don't think it would. I agree with Nick  sorry I agree with Geezer. I think the centre of mass would be at the centre of the hole of the torus and any calculation can proceed as if there is a single force acting on your centre of mass and that of the torus/planet.
Standing on the inside surface of a nonspinning torus would, I believe, be impossible. Although the ground would be beneath your feet  the aggregate force would be towards the centre of mass; ie straight up in the air. I guess at this point I disagree with Geezer too.
I am unsure that I could work out the balance when spinning or even if it is possible  the force from gravity varies inversely with the square of the radius; however the force from rotation vary linearly with radius. Two points of equality (ie inside and out) would be doable  but I think anything more is not possible.
Great question by the way
Matthew

A hollow sphere is a slightly simpler but similar model. Shell Theorem says that if you are outside the sphere, you are attracted towards its centre.
I'm not sure what it says if you are inside the sphere, but I have a suspicion that the dead center is not a stable position, so as soon as you move in any direction you may be drawn towards the interior surface of the sphere.

I think that inside a uniform hollow sphere there is no net gravitational attraction.
Imagine you are in the middle there's no net force (by symmetry which way would it go?).
Imagine that you move a bit to the right. There's now more stuff on the left of you so the attraction is bigger, but the stuff on the right is nearer and attracts more; I think the two effects cancel.
Anyone better at calculus than me who can prove this?
(come to think of it, isn't the maths the same as the "ice pail" experiment?)

Ok, What I'm asking, there is time dilation as an object falls into a gravity well. There is also time dilation due to the acceleration of the mass spinning in a circle. If we look at the doughnut shape edge on wouldn't the top dilation be different from the bottom dilation. If the direction of spin, as viewed from the top is clockwise then viewed from the bottom it is counter clockwise. It just looks to me like there would be a difference in the two.

There's now more stuff on the left of you so the attraction is bigger, but the stuff on the right is nearer and attracts more; I think the two effects cancel.
Anyone better at calculus than me who can prove this?
(come to think of it, isn't the maths the same as the "ice pail" experiment?)
Looks like you are correct. According to Wikipedia  Shell Theorem, Newton proved that:
"2.If the body is a spherically symmetric shell (i.e. a hollow ball), no gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell."
Unfortunately Newton never got around to publishing his Theorem Doughnuticus, so someone fluent in The Calculus (and it's the dreaded integration!) is going to have to figure this one out the hard way. Where's JP?

Unfortunately Newton never got around to publishing his Theorem Doughnuticus, so someone fluent in The Calculus (and it's the dreaded integration!) is going to have to figure this one out the hard way.
Consider a slender cone whose vertex is at a point within the hollow shell, and which extends in both directions through the surface of the shell, cutting two small sections. The volume of each section is its area times the thickness of the shell. Because the cone subtends a solid angle, the area is equal to the solid angle times the square of the distance, divided by the cosine of the angle of incidence (which is the same for both sections). Therefore, the mass of each subtended section is proportional to the square of the distance to the point in question. But the gravitational effect at the said point is inversely proportional to the square of the distance. Therefore the gravitational attraction exerted by the two sections are equal, and of course, in opposite directions. therefore they cancel. Inasmuch as the entire sphere can be cut up into paired sections in this manner, each of which has zero net pull, therefore the entire net pull at any point within the sphere is zero.

Just a thought ... If the centrifugal force of a spinning planet affects the net pull of gravity, does that mean that if the planet Earth stopped spinning we would all get heavier?
Yes. The amount (at the equator) by which one's weight would increase can be calculated by the formula for the acceleration associated with circular motion. Or one can consider that the orbital period at the surface of the earth is 85 minutes. The rotation of the earth takes 1440 minutes. Inasmuch as the acceleration associated with circular motion varies as the square of the angular speed. the ratio of such forces in this case would be (85/1440)^{2}, which is 0.003484 . That is the ratio between the levitation one would feel if the earth were rotating once ever 85 minutes, and what one actually does feel. But because 85 minutes is the orbital period, the levitation would equal 1 times one's weignt, so that multiplying one's weight by 0.003484 will give the amount of one's weight which would be "regained" if the earth stopped rotating. That works out to be about half a pound for a typical person by my calculations. (More than I would have expected).

An interesting thought about imat's question. If his doughnut had the mass of an Earth and was spinning like a merrygoround, wouldn't the overall time dilation of the space surrounding his object exhibit a sort of north and south pole?
For ordinary masses and speeds, there would be no significant effect. For much greater masses and speeds, we would expect that there would be. As to just what this would be like, I don't know; however it is my understanding that the kind of gravity we are accustomed to experiencing is not the only kind there is, and that there is a second kind of gravity associated with momentum. The kind of effects you speak of are likely associated with this. (As to anisotropies in the time dialation, which may be a separate phenomenon, I think that the symmetries of the situation indicate that these would be symmetric with respect to a plane passing through the equator of the doughnut.)

Unfortunately Newton never got around to publishing his Theorem Doughnuticus, so someone fluent in The Calculus (and it's the dreaded integration!) is going to have to figure this one out the hard way.
Consider a slender cone whose vertex is at a point within the hollow shell, and which extends in both directions through the surface of the shell, cutting two small sections. The volume of each section is its area times the thickness of the shell. Because the cone subtends a solid angle, the area is equal to the solid angle times the square of the distance, divided by the cosine of the angle of incidence (which is the same for both sections). Therefore, the mass of each subtended section is proportional to the square of the distance to the point in question. But the gravitational effect at the said point is inversely proportional to the square of the distance. Therefore the gravitational attraction exerted by the two sections are equal, and of course, in opposite directions. therefore they cancel. Inasmuch as the entire sphere can be cut up into paired sections in this manner, each of which has zero net pull, therefore the entire net pull at any point within the sphere is zero.
Er, I'm not sure if you are referring to a shell or a doughnut here. We're trying to figure out what happens on the Welsh guy's Planet Doughnut.
BTW, if you are talking about a shell, many people would like to examine your proof, but I think they'll need a couple of diagrams or some calculus to be convinced.

Oh my God. Now I remember why I dropped out of physics A Level! [???]
.... because the cone subtends a solid angle, the area is equal to the solid angle times the square of the distance, divided by the cosine of the angle of incidence (which is the same for both sections). Therefore, the mass of each subtended section is proportional to the square of the distance to the point in question. But the gravitational effect at the said point is inversely proportional to the square of the distance, etc, ..... Zzzzzzzz
I'm going to try that as a chatup line in the pub on Friday night. I wonder if any of the naughty Welsh girls will find it irresistible.
That works out to be about half a pound for a typical person by my calculations.
I didn't fully understand the calculation but thank you for the answer. Does that mean that if I drive my van in the direction of the rotation of the Earth I can legally carry a greater payload than if travel in the opposite direction? And my wife will lose extra weight? [:)]

Oh my God. Now I remember why I dropped out of physics A Level! [???]
Don't fall for it.
Some take their time so that others can understand while others take no time so that many cannot understand.

Atomic, I suspect the calculations required to give us a three dimensional picture of what that dilation looks like would be labor intensive. I know I don't have the skills. I can imagine a rudimentary sort of image.

Atomic, I suspect the calculations required to give us a three dimensional picture of what that dilation looks like would be labor intensive. I know I don't have the skills. I can imagine a rudimentary sort of image.
Ron,
It's not clear (to me at least) that John C was tremendously interested in resolving questions about time dilation in this topic (although I understand that John does find this a rather intriguing subject).
Perhaps you could start a new topic on time dilation.

Today I believe that the ‘me’ of yesterday was clearly a moron. I now believe you could stand on any point of it’s surface, even within the hole, because the ground which is closest will exert the greatest gravitational pull and would stop you from floating away. But would this make the centre of gravity not a single point but a line running internally through the centreline of the doughnut? That can’t be right.
Condidering the nonrotating case: If we imagine such a planet to be a very thin ring, like a wire hoop, then it is clear, by considering the attraction as it applies to a long linear mass (which falls off inversely as the distance), that the dominant force will be near the surface, whether inside or outside. On that basis, we may assert that for a planet of proper dimensions, a person would be able to stand on any part of its surface without falling off, even if the planet is not rotating.

On that basis, we may assert that for a planet of proper dimensions, a person would be able to stand on any part of its surface without falling off, even if the planet is not rotating.
I'm pretty good at asserting things too, but a proof is much more convincing (unfortunately).

There is more detail of the proof of the thin ring and also of the donut here; it is beyond me to give a precis, so here is a link
http://www.mathpages.com/home/kmath402/kmath402.htm

There is more detail of the proof of the thin ring and also of the donut here; it is beyond me to give a precis, so here is a link
http://www.mathpages.com/home/kmath402/kmath402.htm
Thanks Imat!
It looks as if my initial WAG was about right [;D]

Gravitational attraction/acceleration, as you know, decays with distance from the surface of any massive object in accordance with the inverse square law. If we imagine this decay from the surface of the doughnut (at any position), let`s say on the internal "equator", then the attraction becomes much smaller very quickly as you take off towards the centre of the system, but the effects from the opposite parts of the doughnut are even less significant at any position, until we get to the centre of the hole where all attractions are equal and opposite. If we go back to the Earth/Moon system where there is a neutral point where the Earth`s field counteracts the Moon`s, then we have the same situation here and the neutral point is a point in the centre of the hole in the equatorial plane. At this position, there is no net effect in any direction as the vector sum of the fields is zero in all directions, and one would simply float at this position. One could, however, walk freely all over the surface of the doughnut without noticing the small decrease in weight on the internal equator and the slight weight gain on the outside equator.
At the outside equatorial plane, we have a thin "disc" of space, spreading towards infinity, where the gravitational effects are the same as if the doughnut where spherical.
As we move up, over and around from this imaginary disc and curl towards the neutral position, then through an axial section cutting the doughnut, we see two effects superimposed. The first is the "normal" gravitation from the immediate part of the doughnut, but superimposed on this is the gradually increasing effect from the opposite section of the doughnut. I ignore the effects to the sides of our considered section as these are equal and opposite and therefore cancel.
The distribution of the gravitational acceleration around this curve is one of a normal spherical planetary attraction, but skewed slightly towards the other side of the doughnut. The equivalent gravitational force will therefore vary slightly, as previously described, with the maximum on the external equator and the minimum on the internal equator.
I trust this helps.