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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 centre-line of the doughnut? That can’t be right.What do you think?

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?

Hi GeezerYes, 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?

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?

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?)

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.

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?

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

QuoteUnfortunately 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.

.... 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

That works out to be about half a pound for a typical person by my calculations.

Oh my God. Now I remember why I dropped out of physics A Level! []

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.

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 centre-line of the doughnut? That can’t be right.

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.

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...sorry, you cannot view external links. To see them, please REGISTER or LOGIN