A gravity conundrum involving a solid cylinder

In summary: The force of gravitation is calculated between the centers of mass, which for the cylinder is at a distance k from the point mass, giving a total force of GmM/k^2. Now consider the force between the top half of the cylinder and the point mass. The center of gravity of the top half is now k/2 from the point mass, and the mass of the top half is M/2. So the force is now Gm(M/2)/(k/2)^2= 2GmM/k^2, or twice the force with the whole cylinder. That the whole gravitational force should be less than that from half of the cylinder seems to me wrong, but obviously I am some elementary
  • #1

nomadreid

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Given a cylinder with constant density, and a mass at one of its ends, it would seem via GmM/r^2) that the gravitational force between the cylinder and the mass would be less than that between the mass and the half of the cylinder closest to the mass, which is absurd, so what is wrong?
Given a cylinder of height 2k with constant density and total mass M, and another object (for simplicity, a point mass) with mass m on the top of the cylinder; the force of gravitation is calculated between the centers of mass, which for the cylinder is at a distance k from the point mass, giving a total force of GmM/k^2. Now however consider the force between the top half of the cylinder and the point mass. The center of gravity of the top half is now k/2 from the point mass, and the mass of the top half is M/2. So the force is now Gm(M/2)/(k/2)^2= 2GmM/k^2, or twice the force with the whole cylinder. That the whole gravitational force should be less than that from half of the cylinder seems to me wrong, but obviously I am some elementary piont. I would be grateful for someone to point out my mistake. Thanks.
 
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  • #2
nomadreid said:
the force of gravitation is calculated between the centers of mass,
This is only true for spherically symmetric masses. As a counter example consider the solar system. Its center of mass is somewhere inside the Sun, yet here we stand on the Earth.

If you can work out or find the gravitational field of a disc on axis, you can integrate over the length of the cylinder.
 
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  • #3
The nearby mass is a point on axis at the end of the cylinder.
Consider the two ends of the cylinder are spheres, at centre of mass.
The near end is at r = k/2; with mass M/2.
The far end is at r = 3k/2; with mass M/2.
The force due to the near end is; F = Gm(M/2) / ( k² / 4 ) = 2GmM / k²
The force due to the far end is; F = Gm(M/2) / ( 9k² / 4 ) = 2GmM / 9k²
When you add the two ends you get an increase.
 
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Thanks, Ibix and Baluncore!
 

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