Galileo's experiment and Equivalence Principle

[tsolkas]

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 ?

 

I await your anwer.

 

 

Also,As it is well-known, Einstein’s General Theory of Relativity postulates the following:

“Gravitational fields have the remarkable property of imparting the same acceleration to all bodies independent of their material composition and amount of mass”. (See THE PRINCIPLE OF RELATIVITY, BY A. EINSTEIN – H.A LORENTZ – H. MINKOWSKI – H. WEYL, Page 114 DOVER PUBLICATIONS, INC). Einstein based this conclusion on the all-too-familiar Galileo’s experiment (the experiment conducted from the Leaning Tower of Pisa).

 

Also, I have a question:

If, m1 = 10^6 tn (mass a white draft) , m2 = 1gr (mass a feather), and the mass M of Earth is e.g M = 10^6 tn , namely m1 = M, then according to the Galileo's experiment, (experiment of the Tower of Pisa) we have again u1 = u2 ?

 

 

 

Thanks,

 

 

Christos A. Tsolkas

[Country Boy]

In your first example, since the total force due to the spherical shell on the mass insde it is nil, yes, at a given instant and assuming non-zero relative speeds to begin with, u1= u2. More correctly, a1= a2

 

In your second example, under the same conditions as before, Yes, again, u1= u2.

 

As for your third question, I do not know what a "white mass" is but clearly u1 and v will be equal and opposite while massm2, the "feather mass", will have a much higher acceleration and will accelerate toward either m1 or m2 depending upon which is closer.

 

All that is pretty "Newton"- there is no reason to appeal to Relativity.

 

Since you kindly awaited my answer I will await your response!

[insane_alien]

hvae fun waiting. this guy is a hit an run spammer of the same crap over and over again. he has only replied to one of his threads once. ever.

[tsolkas]

In your first example, since the total force due to the spherical shell on the mass insde it is nil, yes, at a given instant and assuming non-zero relative speeds to begin with, u1= u2. More correctly, a1= a2

 

In your second example, under the same conditions as before, Yes, again, u1= u2.

 

As for your third question, I do not know what a "white mass" is but clearly u1 and v will be equal and opposite while massm2, the "feather mass", will have a much higher acceleration and will accelerate toward either m1 or m2 depending upon which is closer.

 

All that is pretty "Newton"- there is no reason to appeal to Relativity.

 

Since you kindly awaited my answer I will await your response!

 

 

 

Sorry, you are wrong!

 

A very good proof, see :

 

http://www.tsolkas.gr/english/document1/principle-equivalence/principle-equivalence.html

 

 

Christos A. Tsolkas

[martianxx]

whats the point in giving a question letting one person have an attempt and then post the answer when they get it wrong, at least let other people have a go and maybe the first person might want to correct themsleves.

[D H]

He didn't post the answer. Look at his site; it is so chock full of math errors and physics errors that I don't know where to start. Equation 3, I guess.

 

where [math]dv=(v_1-0)[/math] and [math]dv'=(v_2-0)\qquad(3)[/math]

 

That is not what [math]dv[/math] means. Christos, you need to learn how to integrate. Equation (6) happens to be correct. The derivation is not.

 

The problem with failing to integrate properly compounds in the case of different masses. Moreover, Christos, forces are additive. The force on the larger mass [math]M[/math] is the sum of the forces by the spherical shell and the little mass inside the spherical shell.

 

PS. This thread needs to be moved to pseudoscience, pronto.

[tsolkas]

GALILEO'S EXPERIMENT AND PRINCIPLE OF EQUIVALENCE

 

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?

As it is well-known, Einstein’s General Theory of Relativity postulates the following:

“Gravitational fields have the remarkable property of imparting the same acceleration to all bodies independent of their material composition and amount of mass”. (See THE PRINCIPLE OF RELATIVITY, BY A. EINSTEIN – H.A LORENTZ – H. MINKOWSKI – H. WEYL, Page 114 DOVER PUBLICATIONS, INC). Einstein based this conclusion on the all-too-familiar Galileo’s experiment (the experiment conducted from the Leaning Tower of Pisa).

Also, I have a very basic question:

 

QUESTION:

 

If, m1 = 10^6 tn (mass a white draft) , m2 = 1gr (mass a feather), and the mass M of Earth is e.g M = 10^6 tn , namely m1 = M, then according to the Galileo's experiment, (experiment of the Tower of Pisa) we have again u1 = u2 ?

I await your anwer on the above two questions.

 

 

Christos A. Tsolkas

[uncool]

Oy vey...you're just trying to confuse people because they haven't included the "equal and opposite force" of the falling objects on the "ground-like" object.

 

Therefore, your ideas of proving Einstein wrong are completely irrelevant.

 

Incidentally, you actually are getting incorrect answers. When m1 and m2 are released at the same time in the same place, then the force of each on the Earth is added - so the attraction of the Earth to the objects is the same, and the attraction of the objects to the earth (in terms of acceleration) is the same. Therefore, the earth would meet both objects at the same time.

=Uncool-

[swansont]

Note: 2nd Tsolkas post appended; moved from physics to speculations.

[brainfart94]

Therefore, the earth would meet both objects at the same time.

=Uncool-

 

Well if you assume that they have the same acceleration rate, then they would always have the same velocity. This means that since m2 is in the center of m1, then as the two masses are falling together, m2 will stay in the center of m1 until m1 meets the earth, at which time m2 will continue to fall, going from the center to the bottom of m1. So, m1 would stop and meet the earth a certain amount of time before m2 would. This means that for m2 to meet the earth at the same time as m1 does, then m2 must accelerate more than m1 so that it doesn't stay in the center of the shell, which is contraversial to what you say as the masses having equal attractions.