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Owen217a
2007-Nov-05, 07:50 AM
FREE FALL OF BODIES
3-body problem (special case)

According to the Newton's Mechanics:
Let us consider a mass M of radius Ro. At a distance h from the center of this mass M, we place a spherical shell (e.g a spherical elevator) of mass m1 and radius R.
Moreover, at the center of the spherical shell we place another point mass m2, (m1 ≠ m2).
Now t= 0 (Phase I), we allow these three masses m1, m2, M to move freely under the influence of the force of universal attraction.
Let us also assume that after a time dt (Phase ΙΙ), υ1, υ2 and V are respectively the velocities of masses m1, m2 and Μ, relative to an inertial observer Ο.

Note:Masses m1, m2, and M are considered to be homogeneous and absolutely solid bodies.
In addition, velocities υ1, υ2 and V are considered to be positive numbers (that is, only the meters of their magnitudes are taken into account), while mass m2 is considered to be found always within the spherical shell (spherical elevator) m1.

The basic question that is being raised is the following:

QUESTION :
At what velocities do masses m1 (spherical elevator) and m2 (point mass) fall in the gravitational field of Mass M once simultaneously dropped in free fall from a height h, namely at the time dt>0 we have u1 = u2 or u1 ≠ u2 ?

Thanks,

Tony

Celestial Mechanic
2007-Nov-05, 03:29 PM
You have not stated the initial conditions completely. The velocities of the various objects at t=0 need to be stated. I am guessing that you mean that the velocities of m1 and m2 relative to M are zero, that is, all three bodies have the same velocity relative to some observer O. This should be made clear before we can proceed.

I suspect that this has been thrashed out here before, however. I'll look for the thread(s) over lunch break, unless someone finds them before I do.

grav
2007-Nov-05, 10:37 PM
Well, if instead of a spherical shell, we we first take that of a solid sphere at distance h, with radius R and a uniform density D, then the force toward M is F = GMD[(4pi/3)R^3]/h^2. If we then take that of another solid sphere with the same density but with just a slightly smaller radius, then the force becomes F = GMD[(4pi/3)(R-dR)^3]/h^2. So if we subtract the volume of the slightly smaller sphere from the larger one, then we end up with a spherical shell of radius R and the forces will subtract likewise, so that F = [GMD(4pi/3)/h^2][R^3 - (R-dR)^3] for the spherical shell. Now, (4pi/3)[R^3 - (R-dR)^3] is just the volume of the shell itself, so M1 = DV = D(4pi/3)[R^3 - (R - dR)^3], and then with simple substitution, we get F = G M M1 / h^2, whereby a = GM/h^2, the same as with the point mass. Some might not agree with my using a force for gravity, by the way, but I figure was the simplest means by which to show this. This is also using Newtonian gravity. Anyway, since both the point mass and the spherical shell will accelerate in the same way at any distance h from M, and the point mass is not affected by M1 since it is at its center, then they always accelerate together regardless of how M moves toward them in the meantime, so we have u1 = u2 at all times, as far as I can tell.

Gsquare
2007-Nov-06, 02:03 AM
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I suspect Owen217 (Tony) is testing your knowledge of an experiment ...described in the most recent volume of 'Advances in Space Research',,,

Or maybe he's trying to get the contents of the full report without having to pay the price. .....

Gsquare :D

Hornblower
2007-Nov-06, 01:00 PM
Why is this in ATM? So far it looks like a mainstream question about gravitational dynamics. The experiment in question looks like an analogy to Galileo's reputed experiment on the Leaning Tower, but used as a test of Einstein's theory.

Jim
2007-Nov-06, 01:55 PM
Thread moved to Q&A. Owen217a, if you decide to present an ATM concept, the thread can/will be moved back there.

Ken G
2007-Nov-06, 04:54 PM
Using an ideal treatment in Newtonian physics they would fall together, as spherical shapes act identically under gravity as if all the mass were at the center. But stability in the presence of tidal effects are another matter, and could cause the behavior of the central mass to be akin to a pencil balanced on its point. Probably not over the time it would take for the whole business to fall, however.