Spherical Shell Problem


		SPHERICAL SHELL PROBLEM (SSP) SUMMARY The spherical shall problem (SSP) is a very simple and very important experiment of Classical Physics. The problem (SSP) is a special (and simple) case of the “three bodies” problem. By solving this problem, which is done exclusively on the basis of Newtonian Mechanics, and assuming Newton’s axiom on the equality of the inertial mass m_i and the gravitational mass m_g, (m_i = m_g), we formulate the fundamental law of the free fall of bodies. Based on the fundamental law of the free fall of bodies, we can prove that: Gravitational fields never have the remarkable property of imparting all bodies with the same acceleration, regardless of their mass and their material constitution, as Galileo and Einstein mistakenly claim. See: http://www.tsolkas.gr/forums/tga6.jpg (Book: THE PRINCIPLE OF RELATIVITY, by A. EINSTEIN, H.A. LORENTZ, H. WEYL, H. MINKOWSKI). Simply put, based on the fundamental law of the free fall of bodies, we can prove that: 1) The results of Galileo’s experiment (the tower of Pisa experiment) are not theoretically correct, but empirically and approximately. 2) The “equivalence principle” (strong and weak) of the General Theory of Relativity is a completely false principle of Physics. SOLVING THE SPHERICAL SHELL PROBLEM (SSP) Let us assume, Fig. 1, that we have a homogenous spherical shell, with a mass of m₁and a radius of R_o.



	Fig. 1 At the centre of the spherical shell m₁, we place a point mass m₂. At a distance h from the centre of the spherical shell m₁, is a body of mass M. Note: Each one of masses m₁, m₂, M is considered a totally solid body (i.e. not altered by the gravitational forces exercised between them). Obviously, the system of the two masses m₁ and m₂(i.e. of the spherical shell m₁ and the point mass m₂) does not behave as a totally solid body of mass m₁ + m₂. We now perform, Fig. 1, our experiment in two phases (Phase I and Phase II), as follows: PHASE I: During this phase, and at the moment t_o = 0 , the point mass m₂is at the centre of the spherical shell m₁and the centre of the spherical shell m₁is at a distance h from the mass M. Also, during this phase (t_o = 0), the spherical shell m₁, the point mass m_{2 and} the mass M are motionless as to the inertial reference system S. In order words, at t_o = 0, we have υ₁ = 0, υ₂ = 0 and V = 0 , where υ₁ is the velocity of mass m₁, υ₂ is the velocity of point mass m₂ and V is the velocity of mass Μ. In order words, Velocities υ₁ = 0, υ₂ = 0, V = 0 obviously refer to the inertial reference system S. PHASE ΙΙ: During this phase, we allow, from a height h, the spherical shell m₁ (with the point mass m₂, which is at its centre) to move freely under the influence of the force of universal attraction exercised between masses m₁, m₂ and Μ. Let us assume now that after a time t₁ > 0 of free fall, the velocity of mass m₁ is υ₁ > 0, the velocity of point mass m₂ is υ₂ > 0 and the velocity of mass Μ is V > 0. After what we discussed above, the critical question that emerges is this: QUESTION: During the performance of the spherical shell experiment, Fig. 1, and at the moment t₁ > 0 (Phase ΙΙ) of free fall, will the point mass m₂ remain at the centre of the spherical shell m₁ where it was originally (t_o = 0) placed, or will it move from the centre of the spherical shell m₁? In accordance with the proof (which follows), the answer to that question is: In the spherical shell problem, Fig. 1: 1. When the two masses m₁and m₂ are equal (m₁ = m₂), then, at the moment t₁ > 0 the point mass m₂ will be at the centre of the spherical shell m₁. 2.Conversely, if the two masses m₁ and m₂ are not equal ( m₁ ≠ m₂), then, at the moment t₁ > 0 the point mass m₂ will move from the centre of the spherical shell m₁. PROOF Let us assume, Fig.1, that at the moment t₁ > 0 the point mass m₂ is with the spherical shell m₁ and the distance of mass m₁ (i.e. the centre of the spherical shell m₁ ) from the centre of mass Μ is h₁ and the distance of point mass m₂ from the centre of mass Μ is h₂. If we now apply to the system of the three masses m₁, m₂, M at the moment t₁ > 0: 1. The principle of conservation of energy, and 2. The principle of conservation of momentum, we have relations:



In relations (1) the numbers υ₁, υ₂, V, h₁, h₂, h are all considered positive numbers. Also, relations (1), can also be expressed as follows:


	Posing now:


	and Β = ΜV relations (2) can be expressed as:




	In relations (3) Α and Β are obviously positive numbers, i.e. Α > 0 and Β > 0. Now, if we solve the system of equations (3), in terms of υ₁ and υ₂, we have:


	In relations (4) and (4.1), it is obviously:


	where,


	From now on, relations (4) will be referred to as the basic relations of the spherical shell problem (SSP). As mentioned above, in relations (4), Κ can only have one of two values, i.e.:




	Note: As we will see below: For m1<m2, relations (4) are acceptable and relations (4.1) are rejected. For m1>m2, relations (4.1) are acceptable and relations (4) are rejected. Where, in cases (a) and (b) in relations (4) and (4.1), K>0. For m1=m2, relations (4) or relations (4.1) are acceptable. Where, in case (c) in relations (4) and (4.1), K=0. RESEARCH ON THE SPHERICAL SHELL PROBLEM In the spherical shell problem, let us take, e.g. relations (4) of the basic relations (4) and (4.1). Relations (4) yield:

resulting in:


	Based, therefore, on relation (8), we can now formulate the following fundamental law:

LAW Ι

In relation (8):

When Κ = 0, then υ₁ = υ₂ and, conversely, when υ₁ = υ₂, then Κ = 0.
When Κ > 0, then υ₁ > υ₂ and, conversely, when υ₁ > υ₂, then Κ > 0.

Law Ι, as described above, plays a very important part in the research on the spherical shell problem, and especially in terms of basic relations (4) and (4.1).

Ι. EQUAL MASSES

In the spherical shell problem, Fig. 1, and in the event that the masses m₁ and m₂ are equal (m₁ = m₂), we will prove that:
In relations (4), when the masses m₁ and m₂are given, (m₁ = m₂), then υ₁ ≠ υ₂ can never apply.

Proof

According to the problem, since when masses m₁ and m₂ are given (m₁ = m₂) υ₁ ≠ υ₂ can never apply, it means (correspondingly), that:
When masses m₁ and m₂ are given

m₁ = m₂ (9)

then the only relation that applies is:

υ₁ = υ₂ (10)

because υ₁ ≠ υ₂ and υ₁ = υ₂ are mutually exclusive, (only one or the other can apply).
Hypothesis: Let us assume that relation (10) applies, then from relations (9) and (10), we have:

m₁υ₁ = m₂υ₂ (11)

m₁υ₁ – m₂υ₂ = 0 (12)

In addition, from relations (4), we have:

And relations (13) yield:

In addition, from relations (12) and (14) we have:

Because, according the data of the problem, m₁ = m₂, then, based on relation (9), relation (15) yields:

Based on relation (16), relation (8) yields:

Relation (17) corroborates our hypothesis (υ₁ = υ₂).
Subsequently, our hypothesis is correct.
Our problem, therefore, has been proven.
We have proven that:
In relations (4), when masses m₁ and m₂ are given (m₁ = m₂), then υ₁ ≠ υ₂ can never apply, because we have proven that relation υ₁ = υ₂always applies.
Therefore, the following relation will always apply:

where, in relation (18) m₁ and m₂, (m₁ = m₂) are the data of the problem, and υ₁, υ₂ (υ₁ = υ₂) are the conclusion.

b) Conversely: In relations (4), when the velocities υ₁ and υ₂ are given (υ₁ = υ₂), then m₁ ≠m₂ can never apply.

Proof

According to the problem, since when velocities υ₁ and υ₂, (υ₁ = υ₂) are given, m₁ ≠ m₂ can never apply, this means (correspondingly) that:
When velocities υ₁ and υ₂ are given:

υ₁ = υ₂ (18.1)

then the only relation that applies is:

m₁ = m₂ (18.2)

because m₁ ≠ m₂ and m₁ = m₂ are mutually exclusive, (only one or the other can apply).
Hypothesis: Let us assume that relation (18.2) applies, then from relations (18.1) and (18.2), we have:

m₁υ₁ = m₂υ₂ (18.3)
m₁υ₁ – m₂υ₂ = 0 (18.4)

But because relations (4) yield:

Then (as in case (a) above), relations (18.4) and (19), yield:

Ion addition, since according to the data of the problem υ₁ = υ₂ , relation (8) yields:

Κ = 0 (21)

Based on relation (21), relation (20) yields:

Relation (22) corroborates our hypothesis (m₁ = m₂).
Therefore our hypothesis is correct.
Which means that our problem has been proven.
We have proven that:
In relations (4), when velocities υ₁ and υ₂ (υ₁ = υ₂) are given, then m₁ ≠ m₂can never apply, because we have proven that relation m₁ = m₂ always applies.
Therefore, the following relation will always apply:

where, in relation (23) υ₁ and υ₂, (υ₁ = υ₂) are the data of the problem and m₁, m₂ (m₁=m₂) are the conclusion.

After what we discussed above, we can draw the following basic conclusion.

CONCLUSION Ι

In the spherical shell problem, Fig. 1, and in relations (4):

a. When the masses m₁ and m₂ are given (m₁ = m₂), then the velocities υ₁ and υ₂ are always equal (υ₁ = υ₂) and, conversely,

b. When the velocities υ₁ and υ₂ are given (υ₁ = υ₂) then masses m₁ and m₂ are always equal (m₁ = m₂).

From conclusions (a) and (b) results the relation:

when, in relation (24), m₁ = m₂ and υ₁ = υ₂.
Therefore, in accordance with relation (23.1), it emerges that:

c. When the masses m₁ and m₂ are given (m₁ ≠ m₂) then the velocities υ₁ and υ₂ are always υ₁ ≠ υ₂ and, conversely,

d. When the velocities υ₁ and υ₂ are given (υ₁ ≠ υ₂), then masses m₁ and m₂ are always m₁ ≠ m₂.


	Based on the above conclusion, we can now formulate the following law:

LAW ΙΙ

In the spherical shell problem, when the two masses m₁ and m₂ are equal (m₁ = m₂), then:
During the fall for of the equal masses m₁ and m₂ within the field of gravity of mass Μ, then both masses m₁ and m₂ always fall with the same velocities υ₁ and υ₂, (υ₁ = υ₂) as per the inertial reference system S.

m₁ = m_{2 ,}υ₁ = υ₂

ΙΙ. UNEQUAL MASSES

According to conclusion I (paragraph (c)) as above, when m₁ ≠ m₂, then υ₁ ≠ υ_2.In the spherical shell problem, Fig. 1, we will prove that:
When masses m₁ and m₂ are given (m₁ < m₂), then υ₁ < υ₂ can never apply.

Proof

According to the problem, since when masses m₁ and m₂ are given (m₁ < m₂) υ₁ < υ₂ can never apply, this means (correspondingly) that:
When masses m₁ and m₂are given,

m₁ < m₂ (24.1)

then the only relation that applies is:

υ₁ > υ₂(24.2)

because υ₁ < υ₂and υ₁ > υ₂are mutually exclusive, (only one or the other can apply).
Hypothesis: Let us assume that relation (24.2) applies.
If we solve the system of equations (3) for m₁ and m₂, we have:

Since, according to the data of the problem:

m₁ < m₂ (26)

then relations (25) and (26) yield:

In addition, since according to conclusion Ι, when m₁ ≠ m₂ then υ₁ ≠ υ₂, this means that:
In relations (25), when m₁ < m₂ then:

υ₁ ≠ υ₂ (27.1)

Therefore, relation (27), (which results from relations (25)) will always be υ₁ ≠ υ₂, i.e.:

υ₁ - υ₂> 0 (27.2)

according to the hypothesis.
Based, therefore, on relation (27.2), relation (27) yields:

(Βυ₁ – Α) υ₁ > (Α – Βυ₂)υ₁ Β(υ₁² + υ₂²) > Α(υ₁ + υ₂) (27.3)

In relation (27.3), if we replace Α and Β as these result from relations (3), we have:

In relation (27.4), since, according to the data of the problem:

m₁ < m₂i.e.: m₂– m₁ > 0 (27.5)

relation (27.4) is verified only when:

υ₁ – υ₂ > 0
i.e.: υ₁ > υ₂ (27.6)

Relation (27.6) corroborates our hypothesis υ₁ > υ₂.
Therefore, our hypothesis is correct.
Which means our problem has been proven.
We have proven that:
In the spherical shell problem, Fig. 1, when the masses m₁ and m₂ are given (m₁ < m₂) then υ₁ < υ₂ can never apply, because we have proven that relation υ₁ > υ₂ always applies.
Therefore, the following relations will always apply:

where, in relations (27.7) m₁ < m₂ are the data of the problem and υ₁>υ₂are the conclusion.
Conversely: Based on relations (21) and in the same way we described above, we can prove that:
When the data is υ₁ > υ₂, then relation m₁ > m₂ can never apply.
Therefore, relation m₁ < m₂ must apply, because m₁ > m₂ and m₁ < m₂ are mutually exclusive.
In that case, we are lead back to the same relation:

(υ₁ – υ₂) (m₂ – m₁) > 0 (a)

where the data in relation (a) are υ₁ > υ₂, i.e. υ₁ – υ₂ > 0.

SUMMARY

1. In accordance with the above proof:
When the data is m₁ < m₂, then υ₁ > υ₂ and conversely, when the data is υ₁ > υ₂, then m₁ < m₂, i.e. the following relation applies:

2. Therefore, according to relation (b):
When the data given is m₁ > m₂, then υ₁ < υ₂ will apply and, conversely, when the data given is υ₁ < υ₂, then m₁ > m₂, i.e. the following relation applies:

CONCLUSION ΙΙ

In the spherical shell problem, Fig. 1:

a. When m₁ < m₂, then υ₁ > υ₂ (in accordance with relation (b), as above).

b. When m₁ > m₂, then υ₁ < υ₂ (in accordance with relation (c), as above).

In other words, as demonstrated by the above cases (a) and (b), a higher velocity corresponds to the smaller mass, and a lower velocity corresponds to the larger mass.

After what we described above, we can now formulate the following basic law.

LAW ΙΙΙ

In the spherical shell problem, Fig. 1, when the two masses m₁ and m₂ are unequal (m₁ < m₂), then:
During the free fall of these unequal masses m₁ and m₂ within the field of gravity of mass Μ, the smaller mass m₁ falls with a higher velocity υ₁, while, conversely, the larger mass m₂, falls with a lower velocity υ₂, (υ₁ > υ₂), as per the inertial reference system S.

m₁ < m₂ ,υ₁ > υ₂

A notable observation

As we can see, laws I, II, II and conclusion I, cited above, are very important conclusions in the research on the spherical shell problem, Fig. 1.
Laws Ι, ΙΙ, ΙΙΙ apply for every moment t₁ (t₁ > t₀ = 0) of the free fall of masses m₁ and m₂, within the field of gravity of mass Μ.
Note: Following the above, in basic relations (4) and (4.1), is it plain to see that:
In the spherical shell problem, Fig. 1:

a. When masses m₁ and m₂ are m₁ < m₂_, then relations (4) apply and relations (4.1) are rejected.
In this case, in relations (4), m₁ < m₂, υ₁ > υ₂ and Κ > 0.

b. When masses m₁ and m₂ are m₁ > m₂, then relations (4.1) apply and relations (4), are rejected.
In this case, in relations (4.1), m₁ > m₂, υ₁ < υ₂ and Κ > 0.

c. When masses m₁ and m₂are m₁ = m₂_, then either relations (4) or relations (4.1) apply.
In this case, in relations (4) or relations (4.1), m₁ = m₂, υ₁ = υ₂ and Κ = 0.

d. Finally, we need to underline that:

We are led to these same laws Ι, ΙΙ, ΙΙΙ and the same conclusion Ι if, in basic relations (4) and (4.1) instead of relations (4) that we have used in this project, we use relations (4.1). The method is exactly the same, as applied above for relations (4).

Conclusion

In the spherical shell problem, Fig. 1, when m₁ < m₂ based on relation (27.4) the velocities υ₁ and υ₂ result from relations (4), where (in relations (4)) m₁ < m₂, υ₁ > υ₂and Κ > 0. Obviously, for m₁ < m₂ based on relation (27.4) relations (4.1) are rejected, because relations (4.1) for Κ > 0, yield:

i.e.: υ₁ < υ₂

which is in contrast to relation (27.4).

SUMMARY II

As described above, the reasoning (the steps) that we used to solve the spherical shell problem, Fig. 1, are:

Ι. EQUAL MASSES

Step 1: In accordance with the problem, if the data given is m₁ = m₂ then υ₁≠υ₂ can never apply.

Proof

Therefore, if the data given is m₁ = m₂, then υ₁ = υ₂ must always apply, as we have proven, i.e.:

Step 2: Conversely: In accordance with the problem, if the data given is υ₁ = υ₂ then m₁≠m₂ can never apply.

Proof

Therefore, if the data given is υ₁ = υ₂ then m₁ = m₂ must always apply, as we have proven, i.e.:

Conclusion

Therefore, from relations (Α.1) and (Α.2), we have:

ΙΙ. UNEQUAL MASSES

Step 3: Therefore, based on relation (Α.3) it emerges that:

Step 4: In accordance with the problem and based on relation (Α.4), when m₁< m₂, then υ₁ < υ₂ can never apply.

Proof

Therefore, if the data given is m₁< m₂, then υ₁ > υ₂ must always apply, as we have proven, i.e.:

Step 5: Conversely: In accordance with the problem and based on relation (Α.4), when υ₁ > υ₂, then m₁ > m₂ can never apply.

Proof

Therefore, if the data given is υ₁ > υ₂, then m₁ < m₂ must always apply, as we have proven, i.e.:

Conclusion

Therefore, from relations (Α.5) and (Α.6) we have:

Step 6: Therefore, based on relation (Α.7) it emerges that:

Proof

Relation (Α.8) can be proven very easily, by the same method we used to prove relation (Α.7), as above.

These are, therefore, the (6) steps we used to solve the spherical shell problem, Fig. 1, this very important problem of Physics.

A SECOND SOLUTION TO THE SPHERICAL SHELL PROBLEM

(An important physics problem, for University Professors, students, etc, who still think that the Theory of Relativity is correct!!!)

An interesting, second solution to the spherical shell problem, Fig. 1, is the following:
Given the masses m₁, m₂, Μ (m₁< m₂₎ and the distance h work out the three functions:

which give us functions of the time t, the velocities υ₁, υ₂, V, the masses m₁, m₂ , Μ in terms of the inertial reference system S, Fig. 1.
Note: We assume that ratio k, i.e.:

εis relatively small.

This second solution, i.e. working out the functions (27.1), is a very interesting solution to the spherical shell problem, Fig. 1.
Obviously, this second solution must be in agreement with law III, as above, and is also proof of this law.

Problem
Free fall of the Moon to Earth

According to Newtonian Mechanics:
Let us assume that the Earth and the Moon revolve around their axis and around the Sun and are motionless as per an inertial reference system S.
Let us also assume, Fig. 1, that the Moon has the form of a spherical shell with mass and radius R_o.
At the centre of the Moon (i.e. the spherical shell m₁) we place an aluminium sphere with mass m₂ (similar to the size of a football).
We now allow the Moon m₁ and the aluminium sphere m₂ to fall freely from a distance h, within the field of gravity of the Earth M (where h is the distance between the Earth and the Moon), Fig. 1. (see, fig. a).

We are looking for:
1. After how much time t₁ (from the beginning t_o = 0 of the free fall of masses and , within the field of gravity of the Earth Μ) will the distance h_o (between the centre c΄ of the Moon and c the Earth Μ) be ;
2. At that moment t₁, what is the value of velocities , , V masses , , and Μ in terms of the inertial reference system S?
Given: The mass of the Moon, m₁= 7,347^.10²² kg
   The mass of the aluminium sphere, m₂= 10 kg
   The mass of the Earth, M = 5,973^.10²⁴ kg
   The distance between Earth and Moon, h = 384.000 km
   The radius of the Moon, R_o = 1.738 km

Question

Which Physics Professor, student, etc, will give us the correct answer to this very important Physics problem?

A notable observation

Based on the first and second solution to the spherical shell problem described above, it is now proven, clearly and beyond any doubt, that:
a. Gravitational fields never have the remarkable property of imparting all bodies (regardless of their mass) with the same acceleration, as Einstein mistakenly claims, according to the “equivalence principle”.
b. The «equivalence principle» (Strong and weak) is a totally false theory of Physics, as demonstrated at http://www.tsolkas.gr/forums/tga5.jpg, http://www.tsolkas.gr/forums/tga4.jpg and http://www.tsolkas.gr/forums/tga6.jpg , mentioned above.

FREE FALL OF UNEQUAL MASSES

1. The momentum of masses m₁ and m₂Based on relations (4) and in accordance with law III as above, we have proven that:
In the spherical shell problem, Fig. 1, the smaller mass m₁, (m₁ < m₂) falls with higher velocity υ₁ (compared to the larger mass m₂), which falls with lower velocity υ₂, ( υ₁>υ₂), within the field of gravity of mass Μ.
In other words, during the free fall of the unequal masses m₁ and m₂, (m₁< m₂) the velocity υ₁ of mass m₁ is always higher than the velocity υ₂ of mass m₁, (υ₁> υ₂).
Conversely, with the momentum J₁=m₁υ₁ of mass m₁ and J₂=m₂υ₂ of mass m₂ that is not the case.
Specifically, with regard to momentums, J₁ and J₂, the following happens:
From the beginning t₀ = 0 of the free fall of the unequal masses m₁ and m₂, (m₁ < m₂), until a moment t_p, (t_p > t₀ = 0) the momentum J₁ is lesser that the momentum J₂, i.e.:

At the moment t_p momentums J₁ and J₂ are equal, i.e.:

After the moment t_p momentum J₁ is always greater than momentum J₂, i.e.:

Note: The fact that, in relation (28) m₁υ₁< m₂υ₂ and not, e.g. m₁υ₁> m₂υ₂, (see, in detail at the beginning of the link “Galileo and Einstein are wrong” at the site, www.tsolkas.gr.)

Therefore, the formulation of the various relations, which result from the above cases (a), (b), (c) is as follows:
1. In accordance with relation (28):

and since, in accordance with law III:

relations (31) and (32) yield:

Obviously, relation (33) is valid from the beginning t₀ = 0 of the free fall of masses m₁ and m₂, (m₁< m₂) until the moment t_p.

2. In accordance with relation (29):

and since, in accordance with law III:

relations (34) and (35) yield:

Obviously, relation (36) is valid only at the moment t_p.
3. In accordance with relation (30):

and since, in accordance with law III:

relations (37) and (38) yield:

Obviously, relation (39) is valid after the moment t_p.

The moment t_p will be known as the moment t_p of equal momentum. Relations (33), (36), (39), will henceforth be known as basic relations of momentum of the spherical shell problem for unequal masses m₁ and m₂, (m₁< m₂).
In particular, relation (33) will be known as the first constant relation of unequal momentum and relation (39) will be known as the second constant relation of unequal momentum. In contrast, relation (36) will be known as momentary relation of equal momentum.
In the spherical shell problem Fig.1, we also observe that the motion of the unequal masses m₁ and m₂, (m₁< m₂) during their free fall within the field of gravity of mass M is obviously accelerated and, specifically, non-uniformly accelerated, in accordance with relations (4).
Finally, since throughout the free fall the velocity υ₁ of the smaller mass m₁, (m₁< m₂) is always higher than the velocity υ₂ of the larger mass m₂, this means that:

In Fig. 2, the velocity υ₁ of mass m₁, corresponds to curve (υ΄₁) and the velocity υ₂ of mass m₂, corresponds to curve (υ΄₂).

Fig. 2

When, during the free fall of masses m₁ and m₂, (m₁< m₂) within the field of gravity of mass Μ, it occurs, at a moment t_p, that the ratio of velocities is equal to the ratio of masses , Fig. 2 in accordance with case (b) of relation (29), the momentums J₁=m₁υ₁ of mass m₁ and J₂=m₂υ₂ of mass m₂ will be equal J₁= J₂, (m₁υ₁=m₂υ₂) and the momentary relation (36 will apply.
Obviously, before the moment t_p the first constant relation (33) will apply, and after the moment t_p the second constant relation (39) will apply.
It is plain to see that everything we discussed above and which is illustrated in Fig. 2 is a very interesting conclusion of the spherical shell.

2. The kinetic energy of masses m₁ and m₂In the same way that we approached the momentums J₁ and J₂ of unequal masses m₁ and m₂, (m₁ < m₂), we will now approach the kinetic energies Κ₁ and Κ₂ of those masses.
Thus, the kinetic energy Κ₁ of mass m₁, will be:

and the kinetic energy Κ₂ of mass m₂, is:

In the spherical shell problem, Fig. 1, during the free fall of the unequal masses m₁ and m₂ , (m₁ < m₂) at a moment t_k, Fig. 3.

Fig. 3

the kinetic energy Κ₁ of mass m₁ will be equal to the kinetic energy Κ₂ of mass m₂, i.e.:

and since, in accordance with law III:

Relations (42) and (43) yield:

Relation (44) will be known as the momentary relation of equal kinetic energies of masses m₁ and m₂.
In addition, the moment t_k will be known as moment t_k of equal kinetic energies of masses m₁ and m₂.

2. The time period, Fig. 3, from the beginning t₀= 0 of the free fall of masses m₁ and m₂ until the moment t_k will be:

and since, in accordance with law III:

Relations (45) and (46) yield:

Relation (47) will be known as the first constant relation of unequal kinetic energies of masses m₁ and m₂.

3. The time period, Fig. 3, after the moment t_kwill be:

and since, in accordance with law II:


	Relations (48) and (49) yield:

Relation (50) will be known as the second constant relation of unequal kinetic energies of masses m₁ and m₂.

Everything that we discussed above is illustrated in Fig. 3.

NOTE: We need to stress, at this point, that in the spherical shell problem, Fig. 1, the moment t_p in general of equal momentums, Fig. 2, does not coincide with the moment t_k of equal kinetic energies of masses m₁ and m₂, Fig. 3.

Following what we discussed above, a (difficult) problems arises:

PROBLEM

In the spherical shell problem, Fig. 1, what are the minimum masses m₁, m₂, Μ and the minimum height h from which the two unequal masses m₁ and m₂, (m₁ < m₂) can fall freely within the field of gravity of mass Μ, so that we have a moment t_p of equal momentums and a moment t_kof equal kinetic energies?

Example

From what we discussed above, we observe that in free fall lighter bodies fall with higher velocity than heavier bodies, within the field of gravity of a mass Μ.
Thus, if (e.g. in a void) we simultaneously release from a height h over the surface of the Earth a feather and a tank, and allow them to fall freely, then the feather will reach the surface of the Earth before the tank!!!

Conversely (as is well known), Galileo, Newton, Einstein and many contemporary Physicists claim that the feather and the tank will reach the surface of the Earth at the same time.
However, what these Physicists claim is obviously a very big mistake, as proven by the solution of the spherical shell problem, as above.

SPECIAL CASE WHEN m₁ , m_{2 ,}<< Μ, (m₁ < m₂ )

In the spherical shell problem, let us now examine the special case where mass m₁ of the spherical shell and the point mass m₂, are much smaller compared to mass Μ, Fig. 1, i.e.:


	In this based, based on the second of relations (3), we have:


	or


	Therefore, relation (53) based on relations (51), yields:

In other words, in our case, based on relation (54) mass Μ, is considered, by close approximation, motionless, i.e. it’s velocity V is:

V = 0

in terms of the inertial reference system S, Fig. 1.

Let us assume now, Fig. 4, that we let (t_o = 0) the spherical shell m₁ and the point mass m₂ (which is at the centre of the spherical shell m₁) fall freely and together (simultaneously) from a height h within the field of gravity of the motionless (V = 0) mass Μ.

Fig. 4

Let us also assume that, at the moment t₁ > 0 (of the free fall of masses m₁ and m₂), in terms of the inertial reference system S, we have:

1. V_cm = the velocity of the centre of mass of masses m₁ and m₂ υ₁ = the velocity of mass m₁ υ₂ = the velocity of mass m₂

2. h΄_c = the distance of the centre of mass of masses m₁and m₂ from the upper limit of height h.
h΄₁ = the distance of mass m₁h΄₂ = the distance of mass m₂


	where:

If we now apply the principle of conservation of energy at the moment t₁ > 0 of the free fall, we have:

a. For the centre of mass of masses m₁ and m₂:


	Relation (56) yields:

As we can see, in relation (57) the velocity V_cm of the centre of mass of masses m₁ and m₂ , is independent of masses m₁ and m₂, i.e. valid for all values of masses m₁ and m₂.
b. Similarly, for mass m₁, we have:

Relation (58) yields:

As we can see, in relation (59) the velocity υ₁ of mass m₁ at the moment t₁ > 0 of its free fall, is independent of mass m₁ , i.e. valid for all values of mass m₁.

c. Similarly, for mass m₂, we have:

Relation (60) yields:

As we can see, in relation (61) the velocity υ₂ of mass m₂ at the moment t₁ > 0 of its free fall, is independent of mass m₂, i.e. valid for all values of mass m₂ .

Note: In cases (a), (b), (c) above, we assume that at the moment t₁ > 0, the point mass m₂ is always within the spherical shell m₁.

Also, as we know, the velocity V_cm of the centre of mass of masses m₁and m₂ at the moment t₁ > 0 is:

Relation (62) yields:

Substituting now in relation (63) V_cm , υ₁, υ₂ that result from relations (57), (59), (61) we have:

As we can see in relation (64) the coefficients of m₁ and m₂, are independent of masses m₁ and m₂. In other words, the coefficient (within the brackets) of m₁always returns the same value for all values of masses m₁and m₂ at the moment t₁ > 0 of the free fall of masses m₁ and m₂, with the field of gravity of the motionless (V = 0) mass Μ.

The same applies to the coefficient (within the brackets) of m₂.

Consequently, since (64) at the moment t₁ > 0 must be valid for all values of masses m1 and m2 (relation (64)) becomes an identity and we have:

Therefore, in order to verify the identity (65) for all values of m₁ and m₂ it must be:

Thus, from the first of relations (66), we have:

h_c = h₁ (67)

and from the second of relations (66), we have:

h_c = h₂ (68)

Relations (67) and (68) yield:

h_c = h₁ = h₂ (69)

According to relation (69), relations (57), (59), (61) yield:

From relation (70) it arises that: 1) The centre of mass of the spherical shell m₁ and the point mass m₂ 2) The spherical shell m₁ and 3) The point mass m₂, fall with the same velocity, within the field of gravity of the motionless (V = 0) mass Μ.

In addition, relations (69) and (55) yield:

Based on relation (71) we observe that, from the upper limit of height h: 1) The distance h΄_c of the centre of mass of masses m₁ (of the spherical shell) and the point mass m₂2) The distance h΄₁ of mass m₁ (of the centre of the spherical shell) and 3) The distance h΄₂ of the point mass m₂, are always equal at the moment t₁ > 0 of the free fall of the spherical shell m₁ and the point mass m₂, within the field of gravity the motionless (V = 0) mass Μ.
After, therefore, what we discussed above, and according to relations (70) and (71), we draw the following basic conclusion.

Conclusion

In the case of Fig. 4 and during t₁ > 0 of the free fall of mass m₁ (of the spherical shell) and the point mass m₂, from height h within the field of gravity of the motionless (V = 0) mass Μ, ( m₁, m₂ << M and m₁ < m₂) the point mass m₂ will always remain at the centre of the spherical shell m₁, where it was originally placed at the moment t_o=0.

As we can see, the above conclusion can be applied to Galileo’s experiment (the tower of Pisa experiment), fore the two masses m₁ and m₂ (m₁, m₂ << M and m₁ < m₂ ) that Galileo allowed to fall freely from the top of the tower of Pisa, where Μ, is the mass of the Earth.
NOTE: In theory, based on relation:

and relation (33):

relation (70), i.e.:

υ₁ = υ₂

is not, in reality, υ₁ = υ₂, but:

υ₁ ≈ υ₂

i.e.: υ₁ - υ₂= Δυ → 0 (73)
This fact (that Δυ → 0), was technically (experimentally) impossible for Galileo to identity with the technical resources he had.
That is the reason why Galileo was deceived and led to an incorrect formulation of the results of the experiment of the free fall of bodies (the tower of Pisa experiment).

Therefore, if the results of the Tower of Pisa experiment are examined in strict theoretical terms, it is proven that they are completely incorrect in accordance with relations (72) and (73). Simply put, the results of the Tower of Pisa experiment are empirical and approximate and are not valid in theoretical terms.

THE “TOWER OF PISA” EXPERIMENT
ON AN ASTEROID

Let us assume that the Tower of Pisa was built on an asteroid with a diameter, e.g. of D = 50 m and that Galileo released from the top of that tower e.g. a cotton wool sphere with a diameter of d₁=10 cm and a sphere with a diameter of d₂=10 cm, which consists of the material of a neutron star.
In this case, will the two spheres each the surface of the asteroid at the same time, as with the Tower of Pisa experiment that Galileo performed on Earth?

The answer to that question is negative.
In the case of the “asteroid experiment” , according to relation (72), i.e.:

the sphere made of cotton wool will be the first to reach the surface of the asteroid, followed by the sphere that consists of the material of a neutron star.
In the “asteroid experiment” described above, Galileo’s fallacy and Einstein’s mistake (regarding the “remarkable” property of gravitational fields), as cited in the beginning (in the project summary) and as we will demonstrate next, are completely clear.
Unfortunately, Galileo’s fallacy and Einstein’s mistake led contemporary Physics down the wrong path.

THE FUNDAMENTAL LAW OF THE FREE FALL OF BODIES

In recapitulating, the fundamental law of the free fall of bodies is formulated as follows:

Einstein’s big error!

In the General Theory of Relativity Einstein claims that:
“Gravitational fields have the remarkable property of imparting all bodies with the same acceleration (irrespective of their mass and material constitution) at all times”.

See:

1. http://www.tsolkas.gr/forums/tga4.jpg
(Book: Relativity, by Albert Einstein)

2. http://www.tsolkas.gr/forums/tga5.jpg
(Book: WAS EINSTEIN RIGHT?, by Clifford M. Will)

However, according to the law of the free fall of bodies, everything that Einstein maintains is utterly mistaken for the following reasons:
In nature, the above “remarkable” property (to which Einstein refers in the case of gravitational fields) is exhibited only by fields of inertial forces and by no other field of forces (e.g. gravitational, electrical, magnetic, etc.).
This is where Einstein greatly erred. In fact, this is one of his many serious errors in formulating the Theory of Relativity.

More specifically, looking at the text of the books (1) and (2) cited above, Einstein’s error is very much apparent --especially in book (2), where Einstein claims that the aluminum sphere and the Earth fall at the same velocity!!! In the gravitational field of a star having a mass M.
According to the fundamental law of the free fall of bodies and the example of the “asteroid experiment” mentioned above, Einstein was largely mistaken.

BASIC CONCLUSION

As demonstrated in the previous chapter and according to relation (70):

The special case (referred to above), where e.g. two bodies in free fall, with small masses m₁and m₂,happen coincidentally to fall at the same velocity υ₁ and υ₂ (υ₁=υ₂) inside the gravitational field of a large mass M (m₁,m₂<<M), does not signify in any case that:
Gravitational fields have the remarkable property of imparting all bodies unexceptionally with same velocity (regardless of their mass), as Einstein erroneously maintains in his equivalence principle (See the above two pictures (1) and (2)). Gravitational fields never have this “remarkable” property as Einstein’s asserts.
Reiteration: In nature, this “remarkable” property is exhibited only by fields of inertial forces and by no other field of forces (e.g. gravitational, electrical, magnetic, etc.).

Obviously, in formulating the General Theory of Relativity, Einstein made one of his greatest errors.

The principle of equivalence and the spherical shell problem (SSP)

Let us assume (Fig, 1) that spherical shell m₁ is a spherical elevator in which there is an observer (Ο). At time t_o=0 the distance between the spherical elevator m₁ and a mass Μ is h. Also, at time t_o=0 the velocity υ₁ of the spherical elevator m₁ and the velocity V of mass Μ, relative to the inertial reference frame S are υ₁=0 and V = 0 respectively.
The observer (Ο) at a moment t_o=0 places a point mass m₂, (m₁<m₂) in the centre of the spherical shell m₁.
At some point, we let the spherical elevator m₁ (in which point mass m₂ and observer (Ο) are found) fall freely from a height h, inside the gravitational field of mass Μ.
In this case (according to the fundamental law of the free fall of bodies referred to above) at the time t₁>0 of the free fall of the spherical elevator m₁and point mass m₂ inside the gravitational field of mass Μ relative to the inertial reference frame S, the following will be observed:
The velocity υ₁ of the spherical elevator will be υ₁>0 and the velocity υ₂ of point mass will be υ₂>0, where:

υ₂<υ₁ (75)

Consequently, on the basis of relation (75), at the time t₁>0 of the free fall, point mass m₂will move from the centre of spherical elevator m₁ where it was originally placed at a moment t_o=0.
Therefore, since from relation (75) it transpires that υ₂<υ₁, the observer (Ο) inside the spherical elevator m₁ (which is in free fall inside the gravitational field of mass M) will observe the following:
Point mass m₂ will move from the centre of the spherical elevator m₁and will continuously ascend towards the elevator’s top (“roof”) during the free fall of the spherical elevator m₁ and point mass m₂inside the gravitational field of mass M.
As is well known, the above conclusion is completely opposed to the “equivalence principle” of the General Theory of Relativity for the following reason:
According to the “equivalence principle”, Einstein claims that:
During the free fall of the spherical elevator m₁ inside the gravitationalfield of mass M, the observer (O) found inside the spherical elevator will observe that point mass m₂ remains at all times at the centre of the spherical elevator m_1..

However, according to the fundamental law of the free fall of bodies Einstein’s conclusion is wrong.
Therefore, after everything discussed above, we come to the following basic conclusion:

Conclusion

On the basis of the fundamental law of the free fall of bodies, it is demonstrated beyond any doubt that the “equivalence principle” (strong and weak) of the General Theory of Relativity is a completely false principle of Physics. Consequently, the Theory of Relativity should also be regarded as an utterly erroneous theory of Physics.

Two big mistakes many physicists make!

Mistake 1: One mistake still made by a great many physicists (Einstein included) is the following:
Let us assume that there is a body of mass m falling freely inside the gravitational field of a mass M, and at a moment t the distance between the two masses m and Μ is r.
In this case, according to the law of universal attraction and the fundamental law of Newtonian Mechanics, the following relation will apply:

where m_i and m_g are the inertial and gravitational mass of mass m respectively.
a, is the acceleration of the body during a moment t > 0 of its free fall inside the gravitational field of mass M, and
G is the universal gravitational constant.
In relation (76), since m_i=m_g (according to Newton’s axiom) we have:

a = g (77)

where g is the intensity of the gravitational field of mass M at a distance r, namely:

So, on the basis of relation (77), we observe that the acceleration a of the body falling freely inside the gravitational field of mass M is independent of that body’s mass m.
Consequently, physicists mistakenly claim that:
Based on relation (77), since the acceleration a of bodies falling freely inside the gravitational field of mass M is independent of these bodies’ mass, all bodies (heavy or light) will fall with the same acceleration a inside the gravitational field of mass M.
Moreover, all these physicists invoke Galileo’s experiment (the tower of Pisa experiment) for the experimental verification of relation (77). Thus, the following question now emerges:
On the basis of relation (77), which is the big mistake that all these physicists make?
The answer to this question is:
By using relation (77) alone, the problem of the free fall of bodies is solved in the wrong way, because according to this relation the solution of the problem is not complete and therefore is wrong.
Conversely, the solution of the problem of the free fall of bodies (e.g. two bodies m₁ and m₂) inside the gravitational field of a mass M is integral and accurate only when we apply to the closed system of the three bodies m₁ – m₂ – M:

The principle of conservation of energy,
The principle of conservation of momentum,
Newton’s axiom m_i=m_g, and
Taking obviously into account the free fall (movement) of mass M towards the two masses m₁ and m₂ (as was exactly the case with the solution of the spherical shell problem).

The above prerequisites (1), (2), (3), (4) must be taken into account for the correct solution of our problem. Conversely, these prerequisites were never considered by the various physicists (Einstein included) who dealt with this problem.
This is, therefore, the big mistake that many physicists made (and continue to make) in solving the spherical shell problem (SSP).

Mistake 2: Another big mistake that physicists have made and continue to do so is the following:
Newton’s axiom on the equality of the inertial mass m_i and gravitational mass m_g (m_i=m_g) is regarded as accurate and has been experimentally verified with great precision by means of various experiments. This axiom was used in the solution of the spherical shell problem (SSP) referred to above.
Unfortunately, many physicists confuse the equality m_i=m_gwith the equivalence between m_i and m_g ( as claimed by Einstein in the equivalence principle of the General Theory of Relativity).
This is a big mistake on the part of physicists for the following reason:
Under no circumstances does the equality m_i=m_g in Newton’s axiom signify an equivalence between m_i and m_g as physicists wrongly claim. This is therefore the big mistake of these physicists.
Simply put, the experimental verification of Newton’s axiom does not imply an experimental verification of an equivalence between m_i and m_g, as is (unfortunately) asserted by many scientists.
This is apparently a very delicate point, and caution is required when we refer to the equality m_i=m_g and the equivalence between m_i and m_g.
From the point of view of physics, the equality of and equivalence between m_i and m_g are two totally different concepts and should never be confused.
Specifically, whereas in classical Physics the equality m_i=m_g is an axiom (Newton’s axiom), in the theory of Relativity this particular axiom (m_i=m_g) is demonstrated on the basis of the General Theory of Relativity.
Yet, since (according to the fundamental law of the free fall of bodies) it is proven that the “equivalence principle” of the General theory of Relativity is an utterly erroneous principle of Physics, the following occurs:
A body’s inertial mass mi is at all times equal to its gravitational mass m_g (m_i=m_g) (as Newton accurately maintains), and these two masses m_i and m_g are never equivalent (as Einstein mistakenly claims) as per the “equivalence principle”.
So this is Einstein’s second big mistake –and of many other physicist’s also–, as regards the equality of and equivalence between a body’s inertial and gravitational mass m_i and m_g.

EPILOGUE

The Theory of Relativity (Special and General) is unquestionably a completely false Theory of Physics.
Therefore, a well-disposed physics researcher who studies carefully this site (www.tsolkas.gr) will wonder about the following:

Five Questions

Question 1: Why isn’t a simple experiment like the Cedarholm – Townes experiment (1959) carried out on a moving vehicle (e.g. train, aircraft, satellite, etc), so that it can be established once and for all whether ether exists in nature?
What is, therefore, the reason for not performing the Cedarholm – Townes experiment on a moving vehicle?

Question 2: Why isn’t experiment 21 (See www.tsolkas.gr) reperfomed, since it is a very simple and low-cost experiment which proves that the Special Theory of Relativity is completely false?
What is, therefore, the reason for not performing experiment 21?

Question 3: Why has experiment - 14 been kept quiet (See www.tsolkas.gr), since it demonstrates theoretically (in a very simple manner and without any cost) that the “principle of equivalence” of the General Theory of Relativity is an utterly erroneous principle of physics?
What is, therefore, the reason for keeping quiet about this very important experiment?

Question 4: Why is the real cause of the advance of Mercury’s perihelion (43΄΄per century), i.e. the Sun’s revolution around the centre of mass of the Solar system, being kept quiet?
Because this advance (of the planets’ perihelia) is not attributed to the curvature of time-space around the Sun as Einstein wrongly maintains (See στο www.tsolkas.gr).
Why is this very significant point (purposely) suppressed?

Question 5: Why is the spherical shell problem (discussed above) being kept quiet, since it proves in a very simple manner that : 1) the “principle of equivalence” of the General Theory of Relativity is a completely false principle of Physics, and 2) gravitational fields do not have the “remarkable” property of imparting the same acceleration to all bodies (irrespective of the latter’s mass), as Einstein erroneously claims?
What is, therefore, the reason for keeping quiet about the spherical shell problem?

What answers have physics professors given to Questions (1), (2), (3), (4), (5)?
Is there some “expediency” for not providing an answer to the above five questions? After everything discussed in this paper (and on www.tsolkas.gr in general), we come to the following important conclusion:

Conclusion

Modern physics should be rid once and for all of the “pseudo-science” of the Theory of Relativity.
This theory should no longer be taught in universities for the simple reason that it is false and does not reflect natural reality.
If however, the “pseudo-science” of the Theory of Relativity continues to be taught in universities, this will be the biggest shame in the history of Physics.

Lastly, I have a more general question to pose. In closing up.

QUESTION

Could it be that physics is currently controlled and directed by various “Centers” (universities, professors, journals, etc) towards an anti-scientific path that is taking us to a new Middle Ages?

I am posing this question because this inexplicable (and perhaps intentional) silence and non performance of the above-mentioned experiments probably lend truth to it.

I HAVE A QUESTION...

As is well-known, tremendous amounts of money have been spent on the Gravity Probe b experiment, and eventually this experiment failed!
This compels me to ask all these Physics “Centers” the following question:
Why not spend just a small amount of money to carry out two simple experiments, namely:

The Cedarholm – Townes (1959) experiment on a moving vehicle (e.g. train, aircraft, satellite, etc), and
Experiment – 21 (See www.tsolkas.gr), in order to establish once and for all the accuracy or fallacy of the Theory of Relativity.

I repeat, why don’t these Physics “Centers” explain to us the reason for not conducting these two experiments?
Is there some “expediency” behind this, --so that the foundations of contemporary physics won’t utterly shaken?

I believe this is a very rational question.
This is where my paper ends. Time will tell whether I am right or not.