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To find the attractions between planets and stuff like that, you use the center of gravity/mass to apply to Newton's equation. So even if those planets collided into each other, you could separate them if you give enough force, because $r$ (distance between the center of gravity/mass of each planet) in the gravitation equation is not $0$ therefore $r^2$ is not $0$. But the problem comes when you put the centers of gravity/mass of two objects on each other. Then $r$ is $0$, $r^2$ is $0$ and when you divide by $r^2$ (in the gravitation equation), you’re dividing by $0$ which means the gravity is infinite; i.e you'll never be able to separate them. Now you might say that there will never be such an instance where the two centers of gravity/mass will never be on each other, but consider this-

Two hoops, one 1/2 in radius of the other, placed on a table such that the circumference of those 2 hoops are parallel (like a train track that goes in perfect circles). The center of mass of the bigger hoop will be at the very center of the area (circle) enclosed by the bigger hoop. The same goes for the second, smaller hoop. The center of mass of each hoop will lie on the same point. So does that mean no matter how much you tried, you'll never be able to separate them? This question has been puzzling me for ages so help would be great.

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The idea that the force between two spherical bodies goes as $1/r^2$ is only valid outside of the bodies.

Once you're inside the bodies, things are different. If the bodies are of uniform density, the "shell theorem" applies, and the force goes to zero as $r$ goes to zero. (It may not be obvious, but if you work out the math, there is no net gravitational force from any of the mass outside of your $r$, so as $r$ gets smaller so does the mass, and the mass gets smaller faster than $1/r^2$.)

The argument is similar for the hoops, but the math will be more complicated.

But, in the end, being very close to the center of mass without actually being very close to an actual mass will never result in very large forces.

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  • $\begingroup$ seems to me quite obvious that you need zero force at the center of spherical symmetry...Shell theorem is nontrivial only for $r\neq 0$ $\endgroup$ – Umaxo Oct 20 '20 at 4:48
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No. When centered, your two concentric hoops don’t exert any net force at all on each other. There is a net force in other positions, but never an infinite force. You can separate them.

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I think your question has 2 parts: i) would the center of masses of the hoops exert gravitational force upon each other and ii) would there be a gravitational interaction between the parts of the hoops when trying to separate them. i) has been answered above and I'll only mention that the centers of the hoops are also where the force vectors from every infinitesimal part of the hoop exactly cancel out. I do not think, however, that ii) has been fully answered, in part because focus on the center of mass ignores the force that may be exerted upon two infinitesimal parts of the two hoops that are being pulled apart. Now, when pulling two hoops apart with 2 hands, you are expected to focus upon 1 finite element of one hoop and another finite element of the other hoop. As such, you must consider the force exerted by, say the outer hoop, upon the finite element (where your other hand is) of the inner hoop. Furthermore, you must consider how this force changes as the hoops start to move apart. This second part is tricky because, as the hoop is a shell in 2-d space but not in 3-d space, the instant you start to pull the hoops apart a gravitational force will be exerted... While any 2 hoops in reality will be of small line density, it is theoretically possible for the line density to approach infinity. In this impractical case, and assuming that your hands are of finite breadth, the force could approach infinity. Of course, in reality, I think GR predicts that such a hoop (with sufficiently large line density) would collapse to form a (potentially unusual) black hole.

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