Post by carnyx on Apr 4, 2011 17:35:02 GMT 1
Without a hole, the gravitational attraction of the whole shell integrates to cancel out to zero at any point within the central void.
If you drill a hole in the wall of the shell, the mass attraction from that bit of mass disappears, and in topological terms the shell now turns into a plate. The integration now produces a net very small gravitational attraction in the direction of the centroid of the shell wall. i.e. the ball is attracted to the point opposite the hole. But it will be a very small attraction, equivalent to the mass-attraction given by the amount of metal cut from the hole and it's distance from the ball (at 1/D^2 ) but away from the hole
What is interesting is that if the contact of the metal ball with the metal shell wall was perfectly elastic, then the ball will bounce back off the wall initially at the angle of reflection. .. and will continue to do so until it meets the hole!
So, there WILL be a huge explosion IF the ball was in the hemisphere of the sphere described from the centre of the hole ,.. at the time the hole was made... because as the ball bounces of the wall , it would be bound sooner or later to cross the centrepoint of the sphere where the trigger is.
(As an aside, it will eventually pass through the hole, but could it ever attain escape velocity?)
(And as a further aside, what kind of trajectory will the ball be making after the first few bounces?)
Other factors such as the gravity from large objects in the neighbourhood of the shell could add to the possibility of the ball reaching the centre, including the acceleration of the shell by rockets etc. In fact, there could even be the possibility of the escape rockets causing sufficent perturbation, which would be ironic.
And, I wonder if this puzzle works for non-spherical shell, and also nonsymmetrical ones. I suspect it will because I think the key insights in the topological transformation of the class of manifolds from a shell ( two surfaces ) to a plate ( one surface) ... by drilling that hole...
But to answer the puzzle, the explosion will happen if the Ball is less than half way across the sphere from where the hole is made, and the rebound height was greater than the radius.
If you drill a hole in the wall of the shell, the mass attraction from that bit of mass disappears, and in topological terms the shell now turns into a plate. The integration now produces a net very small gravitational attraction in the direction of the centroid of the shell wall. i.e. the ball is attracted to the point opposite the hole. But it will be a very small attraction, equivalent to the mass-attraction given by the amount of metal cut from the hole and it's distance from the ball (at 1/D^2 ) but away from the hole
What is interesting is that if the contact of the metal ball with the metal shell wall was perfectly elastic, then the ball will bounce back off the wall initially at the angle of reflection. .. and will continue to do so until it meets the hole!
So, there WILL be a huge explosion IF the ball was in the hemisphere of the sphere described from the centre of the hole ,.. at the time the hole was made... because as the ball bounces of the wall , it would be bound sooner or later to cross the centrepoint of the sphere where the trigger is.
(As an aside, it will eventually pass through the hole, but could it ever attain escape velocity?)
(And as a further aside, what kind of trajectory will the ball be making after the first few bounces?)
Other factors such as the gravity from large objects in the neighbourhood of the shell could add to the possibility of the ball reaching the centre, including the acceleration of the shell by rockets etc. In fact, there could even be the possibility of the escape rockets causing sufficent perturbation, which would be ironic.
And, I wonder if this puzzle works for non-spherical shell, and also nonsymmetrical ones. I suspect it will because I think the key insights in the topological transformation of the class of manifolds from a shell ( two surfaces ) to a plate ( one surface) ... by drilling that hole...
But to answer the puzzle, the explosion will happen if the Ball is less than half way across the sphere from where the hole is made, and the rebound height was greater than the radius.