             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 m_{1 }and a radius of R_{o}.        Fig. 1 At the centre of the spherical shell m_{1}, we place a point mass m_{2}. At a distance h from the centre of the spherical shell m_{1}, is a body of mass M. Note: Each one of masses m_{1}, m_{2}, 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 m_{1} and m_{2 }(i.e. of the spherical shell m_{1} and the point mass m_{2}) does not behave as a totally solid body of mass m_{1} + m_{2}. 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 m_{2 }is at the centre of the spherical shell m_{1 }and the centre of the spherical shell m_{1}_{ }is at a distance h from the mass M. Also, during this phase (t_{o} = 0), the spherical shell m_{1}, 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 õ_{1} = 0, õ_{2} = 0 and V = 0
, where õ_{1} is the velocity of mass m_{1}, õ_{2} is the velocity of point mass m_{2} and V is the velocity of mass Ì. In order words, Velocities õ_{1} = 0, õ_{2} = 0, V = 0 obviously refer to the inertial reference system S. PHASE ÉÉ: During this phase, we allow, from a height h, the spherical shell m_{1} (with the point mass m_{2}, which is at its
centre) to move freely under the influence of the force of universal attraction exercised between masses m_{1}, m_{2} and Ì. Let us assume now that after a time t_{1} > 0 of free fall, the velocity of mass m_{1} is õ_{1} > 0, the velocity of point mass m_{2} is õ_{2} > 0 and the velocity of mass Ì 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 t_{1} > 0 (Phase ÉÉ) of free fall, will the point mass m_{2} remain at the centre of the spherical shell m_{1} where it was originally (t_{o} = 0) placed, or will it move from the centre of the spherical shell m_{1}?
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 m_{1}_{ }and m_{2} are equal (m_{1} = m_{2}), then, at the moment t_{1} > 0 the point mass m_{2} will be at the centre of the spherical shell m_{1}. 2.Conversely, if the two masses m_{1} and m_{2} are not equal (
m_{1} ≠ m_{2}), then, at the moment t_{1} > 0 the point mass m_{2} will move from the centre of the spherical shell m_{1}. PROOF Let us assume, Fig.1, that at the moment t_{1} > 0 the point mass m_{2} is with the spherical shell m_{1} and the distance of mass m_{1} (i.e. the centre of the spherical shell m_{1}
) from the centre of mass Ì is h_{1} and the distance of point mass m_{2} from the centre of mass Ì is h_{2}. If we now apply to the system of the three masses m_{1}, m_{2}, M at the moment t_{1} > 0: 1. The principle of conservation of energy, and 2. The principle of conservation of momentum, we have relations:        
In relations (1) the numbers õ_{1}, õ_{2}, V, h_{1}, h_{2}, h are all considered positive numbers. Also, relations (1), can also be expressed as follows:      
   Posing now: 
   and Â = ÌV relations (2) can be expressed as: 
   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), Ê 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 É In relation (8): When Ê = 0, then õ_{1} = õ_{2} and, conversely, when õ_{1} = õ_{2},
then Ê = 0. When Ê > 0, then õ_{1} > õ_{2} and, conversely, when õ_{1} > õ_{2}, then Ê > 0.

    Law É, 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). É. EQUAL MASSES In the spherical shell problem, Fig. 1, and in the event that the masses m_{1} and m_{2} are equal (m_{1} = m_{2}), we will prove that: In relations (4), when the masses m_{1} and m_{2 }are given, (m_{1} = m_{2}), then õ_{1} ≠ õ_{2} can never apply.
Proof According to the problem, since when masses m_{1} and m_{2} are given (m_{1} = m_{2}) õ_{1} ≠ õ_{2} can never apply, it means (correspondingly), that: When masses m_{1} and m_{2} are given m_{1} = m_{2}_{ } (9) then the only relation that applies is:
õ_{1} = õ_{2} (10) because õ_{1} ≠ õ_{2} and õ_{1} = õ_{2} 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: m_{1}õ_{1} = m_{2}õ_{2} (11) m_{1}õ_{1} – m_{2}õ_{2} = 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, m_{1} = m_{2}, then, based on relation (9), relation (15) yields:   
     Based on relation (16), relation (8) yields: 
      
Relation (17) corroborates our hypothesis (õ_{1} = õ_{2}). Subsequently, our hypothesis is correct. Our problem, therefore, has been proven. We have proven that: In relations (4), when masses m_{1} and m_{2} are given (m_{1} = m_{2}), then õ_{1} ≠ õ_{2} can never apply, because we have proven that relation õ_{1} = õ_{2 }always
applies. Therefore, the following relation will always apply:         where, in relation (18) m_{1} and m_{2}, (m_{1} = m_{2}) are the data of the problem, and õ_{1}, õ_{2} (õ_{1} = õ_{2}) are the conclusion. b) Conversely: In relations (4), when the velocities õ_{1} and õ_{2} are given (õ_{1} = õ_{2}), then m_{1}
≠m_{2} can never apply. Proof According to the problem, since when velocities õ_{1} and õ_{2}, (õ_{1} = õ_{2}) are given, m_{1} ≠ m_{2} can never apply, this means (correspondingly) that: When velocities õ_{1} and õ_{2} are given: õ_{1} = õ_{2} (18.1)
then the only relation that applies is: m_{1} = m_{2} (18.2) because m_{1} ≠ m_{2} and m_{1} = m_{2} 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: m_{1}õ_{1} = m_{2}õ_{2}
(18.3) m_{1}õ_{1} – m_{2}õ_{2} = 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 õ_{1} = õ_{2}
, relation (8) yields: Ê = 0 (21) Based on relation (21), relation (20) yields:        Relation (22) corroborates our hypothesis (m_{1} = m_{2}). Therefore our hypothesis is correct. Which means that our problem has been proven. We have proven that: In relations (4), when velocities õ_{1} and õ_{2} (õ_{1} = õ_{2}) are given, then m_{1}
≠ m_{2 }can never apply, because we have proven that relation m_{1} = m_{2} always applies. Therefore, the following relation will always apply:         where, in relation (23) õ_{1} and õ_{2}, (õ_{1} = õ_{2}) are the data of the problem and m_{1}, m_{2} (m_{1}=m_{2}) are the conclusion.
After what we discussed above, we can draw the following basic conclusion.     CONCLUSION É In the spherical shell problem, Fig. 1, and in relations (4): a. When the masses m_{1} and m_{2} are given (m_{1} = m_{2}), then the velocities õ_{1} and õ_{2} are always equal (õ_{1} = õ_{2}) and, conversely, b. When the velocities õ_{1} and õ _{2} are given (õ_{1} = õ_{2}) then masses m_{1} and m_{2} are always equal (m _{1} = m _{2}).
From conclusions (a) and (b) results the relation:
when, in relation (24), m_{1} = m_{2} and õ_{1} = õ_{2}. Therefore, in accordance with relation (23.1), it emerges that: c. When the masses m_{1} and m_{2} are given (m_{1} ≠ m_{2}) then the velocities õ_{1} and õ_{2} are always õ_{1} ≠ õ_{2}
and, conversely, d. When the velocities õ_{1} and õ _{2} are given (õ_{1} ≠ õ_{2}), then masses m_{1} and m _{2} are always m_{1} ≠ m_{2}.

 
   Based on the above conclusion, we can now formulate the following law: 
                  LAW ÉÉ
In the spherical shell problem, when the two masses m_{1} and m_{2} are equal (m_{1} = m_{2}), then: During the fall for of the equal masses m_{1} and m_{2} within the field of gravity of mass
Ì, then both masses m_{1} and m_{2} always fall with the same velocities õ_{1} and õ_{2}, (õ_{1} = õ_{2}) as per the inertial reference system S. m_{1} = m_{2 , }õ_{1} = õ_{2} 
    ÉÉ. UNEQUAL MASSES According to conclusion I (paragraph (c)) as above, when m_{1} ≠ m_{2}, then õ_{1} ≠ õ_{2.}In the spherical shell problem, Fig. 1, we will prove that: When masses m_{1} and m_{2} are given (m_{1} < m_{2}), then õ_{1} < õ_{2} can never apply.
Proof According to the problem, since when masses m_{1} and m_{2} are given (m_{1} < m_{2}) õ_{1} < õ_{2} can never apply, this means (correspondingly) that: When masses m_{1} and m_{2 }are given,_{ } m_{1} < m_{2} (24.1)
then the only relation that applies is: õ_{1} > õ_{2 }(24.2) because õ_{1} < õ_{2 }and õ_{1} > õ_{2 }are 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 m_{1} and m_{2}, we have:
       
Since, according to the data of the problem: m_{1} < m_{2} (26) then relations (25) and (26) yield:        In addition, since according to conclusion É, when m_{1} ≠ m_{2} then õ_{1} ≠ õ_{2}, this means that: In relations (25), when m_{1} < m_{2} then:
õ_{1} ≠ õ_{2 } (27.1) Therefore, relation (27), (which results from relations (25)) will always be õ_{1} ≠ õ_{2}, i.e.: õ_{1}  õ_{2 }> 0 (27.2) according to the hypothesis. Based, therefore, on relation (27.2), relation (27) yields: (Âõ_{1} – Á) õ_{1} > (Á – Âõ_{2})õ_{1
} Â(õ_{1}^{2} + õ_{2}^{2}) > Á(õ_{1} + õ_{2}) (27.3) In relation (27.3), if we replace Á and Â as these result from relations (3), we have:         In relation (27.4), since, according to the data of the problem: m_{1} < m_{2}i.e.: m_{2 }– m_{1} > 0 (27.5)
relation (27.4) is verified only when: õ_{1} – õ_{2} > 0 i.e.: õ_{1} > õ_{2} (27.6) Relation (27.6) corroborates our hypothesis õ_{1} > õ_{2 }. 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 m_{1} and m_{2} are given (m_{1} < m_{2}) then õ_{1} < õ_{2} can never apply, because we have proven that relation õ_{1} > õ_{2} always applies.
Therefore, the following relations will always apply:      
   where, in relations (27.7) m_{1} < m_{2 } are the data of the problem and õ_{1}>õ_{2 }are the conclusion.
Conversely: Based on relations (21) and in the same way we described above, we can prove that: When the data is õ_{1} > õ_{2}, then relation m_{1} > m_{2} can never apply. Therefore, relation m_{1} < m_{2} must apply, because m_{1} > m_{2} and m_{1} < m_{2} are mutually exclusive. In that case, we are lead back to the same relation: (õ_{1} – õ_{2}) (m_{2} – m_{1})
> 0 (a) where the data in relation (a) are õ_{1} > õ_{2}, i.e. õ_{1} – õ_{2} > 0. SUMMARY 1. In accordance with the above proof: When the data is m_{1} < m_{2}, then õ_{1} > õ_{2} and conversely, when the data is õ_{1} > õ_{2}, then m_{1} < m_{2}, i.e. the following relation applies:

                  
    2. Therefore, according to relation (b): When the data given is m_{1} > m_{2}, then õ_{1} < õ_{2} will apply and, conversely, when the data given is õ_{1} < õ_{2}, then m_{1} > m_{2}, i.e. the following relation applies:
       
CONCLUSION ÉÉ In the spherical shell problem, Fig. 1: a. When m_{1} < m_{2}, then õ_{1} > õ_{2} (in accordance with relation (b), as above). b. When m_{1} > m_{2}, then õ_{1} < õ_{2} (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 ÉÉÉ In the spherical shell problem, Fig. 1, when the two masses m_{1} and m_{2} are unequal
(m_{1} < m_{2}), then: During the free fall of these unequal masses m_{1} and m_{2} within the field of gravity of mass Ì, the smaller mass m_{1} falls with a higher velocity õ_{1}, while, conversely, the larger mass m_{2}, falls with a lower velocity õ_{2}, (õ_{1} > õ_{2}), as per the inertial reference system S. m_{1} < m_{2 } ,_{ }õ_{1} > õ_{2}

    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 É, ÉÉ, ÉÉÉ apply for every moment t_{1} (t_{1} > t_{0} = 0) of the free fall of masses m_{1} and m_{2},
within the field of gravity of mass Ì. 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 m_{1} and m_{2} are m_{1} < m_{2}_{,} then relations (4) apply and relations (4.1) are rejected. In this case, in relations (4), m_{1} < m_{2}, õ_{1} > õ_{2} and Ê > 0.
b. When masses m_{1} and m_{2} are m_{1} > m_{2}, then relations (4.1) apply and relations (4), are rejected. In this case, in relations (4.1), m_{1} > m_{2}, õ_{1} < õ_{2} and Ê > 0. c. When masses m_{1} and m_{2 }are m_{1} = m_{2}_{,} then either relations (4) or relations (4.1)
apply. In this case, in relations (4) or relations (4.1), m_{1} = m_{2}, õ_{1} = õ_{2} and Ê = 0. d. Finally, we need to underline that: We are led to these same laws É, ÉÉ, ÉÉÉ and the same conclusion É 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 m_{1} < m_{2} based on relation (27.4) the velocities õ_{1} and õ_{2} result from relations (4), where (in relations (4)) m_{1} < m_{2}, õ_{1} > õ_{2 }and Ê > 0. Obviously, for m_{1} < m_{2} based on relation (27.4) relations (4.1) are rejected,
because relations (4.1) for Ê > 0, yield:
        i.e.: õ_{1} < õ_{2} 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: É. EQUAL MASSES
Step 1: In accordance with the problem, if the data given is m_{1} = m_{2} then õ_{1}≠õ_{2} can never apply. Proof Therefore, if the data given is m_{1} = m_{2}, then õ_{1} = õ_{2} must always apply, as we have proven, i.e.: Step 2: Conversely: In accordance with the problem, if the data given is õ_{1} = õ_{2} then m_{1}≠m_{2} can never apply. Proof Therefore, if the data given is õ_{1} = õ_{2} then m_{1} = m_{2} must always apply, as we have proven, i.e.:
Conclusion Therefore, from relations (Á.1) and (Á.2), we have: ÉÉ. UNEQUAL MASSES Step 3: Therefore, based on relation (Á.3) it emerges that: Step 4: In accordance with the problem and based on relation (Á.4), when m_{1}< m_{2}, then õ_{1} < õ_{2} can never apply. Proof Therefore, if the data given is m_{1}< m_{2}, then õ_{1} > õ_{2} must always apply, as we have proven, i.e.: Step 5: Conversely: In accordance with the problem and based on relation (Á.4), when õ_{1} > õ_{2}, then m_{1} > m_{2} can never apply.
Proof Therefore, if the data given is õ_{1} > õ_{2}, then m_{1} < m_{2} must always apply, as we have proven, i.e.: Conclusion Therefore, from relations (Á.5) and (Á.6) we have:
Step 6: Therefore, based on relation (Á.7) it emerges that: Proof
Relation (Á.8) can be proven very easily, by the same method we used to prove relation (Á.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 m_{1}, m_{2}, Ì (m_{1 }< m_{2)} and the distance h work out the three functions:        which give us functions of the time t, the velocities õ_{1}, õ_{2}, V, the masses m_{1}, m_{2}
, Ì in terms of the inertial reference system S, Fig. 1. Note: We assume that ratio k, i.e.:      
   åis 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 m_{1}) we place an aluminium sphere with mass m_{2} (similar to the size of a football). We now allow the Moon m_{1} and the aluminium sphere m_{2} 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 t_{1} (from the beginning t_{o} = 0 of the free fall of masses and , within the field of
gravity of the Earth Ì) will the distance h_{o} (between the centre c΄ of the Moon and c the Earth Ì) be ; 2. At that moment t_{1}, what is the value of velocities , , V masses , , and Ì in terms of the inertial reference system S? Given: The mass of the Moon,
m_{1}= 7,347^{.}10^{22} kg The mass of the aluminium sphere, m_{2 }= 10 kg The mass of the Earth, M = 5,973^{.}10^{24} 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 m_{1} and m_{2}Based 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 m_{1}, (m_{1} < m_{2}) falls with higher velocity õ_{1} (compared to the larger mass m_{2}), which falls with lower velocity õ_{2}, (
õ_{1}>õ_{2}), within the field of gravity of mass Ì. In other words, during the free fall of the unequal masses m_{1} and m_{2}, (m_{1}< m_{2}) the velocity õ_{1} of mass m_{1} is always higher than the velocity õ_{2} of mass m_{1}, (õ_{1}> õ_{2}). Conversely, with the momentum J_{1}=m_{1 }õ_{1} of mass m_{1}
and J_{2}=m_{2 }õ_{2} of mass m_{2} that is not the case. Specifically, with regard to momentums, J_{1} and J_{2}, the following happens: From the beginning t_{0} = 0 of the free fall of the unequal masses m_{1} and m_{2}, (m_{1} < m_{2}), until a moment t_{p}, (t_{p} > t_{0} = 0) the momentum J_{1}
is lesser that the momentum J_{2}, i.e.: 
                          At the moment t_{p} momentums J_{1} and J_{2} are equal, i.e.:         After the moment t_{p} momentum J_{1}
is always greater than momentum J_{2}, i.e.:      
 Note: The fact that, in relation (28) m_{1}õ_{1 }< m_{2}õ_{2} and not, e.g. m_{1}õ_{1 }> m_{2}õ_{2}, (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 t_{0} = 0 of the free fall of masses m_{1} and m_{2}, (m_{1}< m_{2}) 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 m_{1} and m_{2}, (m_{1}< m_{2}). 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 m_{1} and m_{2}, (m_{1}< m_{2}) during their free fall within the field of gravity of mass M is obviously accelerated and, specifically, nonuniformly accelerated, in accordance with relations (4). Finally, since throughout the free fall the velocity õ_{1} of the smaller mass m_{1},
(m_{1}< m_{2}) is always higher than the velocity õ_{2} of the larger mass m_{2}, this means that: In Fig. 2, the velocity õ_{1} of mass m_{1}, corresponds to curve (õ΄_{1}) and the velocity õ_{2} of mass m_{2}, corresponds to curve (õ΄_{2}).         Fig. 2
When, during the free fall of masses m_{1} and m_{2}, (m_{1}< m_{2}) within the field of gravity of mass Ì, 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 J_{1}=m_{1}õ_{1} of mass m_{1} and J_{2}=m_{2}õ_{2} of mass m_{2} will be equal J_{1}= J_{2}, (m_{1}õ_{1}=m_{2}õ_{2}) 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 m_{1} and m_{2}In the same way that we approached the momentums J_{1} and J_{2} of unequal masses m_{1}
and m_{2}, (m_{1} < m_{2}), we will now approach the kinetic energies Ê_{1} and Ê_{2} of those masses. Thus, the kinetic energy Ê_{1} of mass m_{1}, will be:        and the kinetic energy Ê_{2} of mass m_{2}, is:         In the spherical shell problem, Fig. 1, during the free fall of the unequal masses m_{1} and m_{2}
, (m_{1} < m_{2}) at a moment t_{k}, Fig. 3.
     
 Fig. 3 the kinetic energy Ê_{1} of mass m_{1} will be equal to the kinetic energy Ê_{2} of mass m_{2}, 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 m_{1} and m_{2}. In addition, the moment t_{k} will be known as moment t_{k} of equal kinetic energies of masses m_{1} and m_{2}. 2. The time period, Fig. 3, from the beginning t_{0 }= 0 of the free fall of masses m_{1} and m_{2} until the moment t_{k} will be: 
   Relations (48) and (49) yield: 
   
  
Relation (50) will be known as the second constant relation of unequal kinetic energies of masses m_{1} and m_{2}. 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 m_{1} and m_{2}, Fig. 3. Following what we discussed above, a (difficult) problems arises: PROBLEM In the spherical shell problem, Fig. 1, what are the minimum masses m_{1}, m_{2}, Ì and the minimum height h from which the two unequal masses m_{1} and m_{2}, (m_{1} < m_{2}) can fall freely
within the field of gravity of mass Ì, so that we have a moment t_{p} of equal momentums and a moment t_{k}_{ }of 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 Ì. 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 m_{1} , m_{2 , }<< Ì, (m_{1} < m_{2} ) In the spherical shell problem, let us now examine the special case where mass m_{1} of the spherical shell and the point mass m_{2}, are much smaller compared to mass Ì, 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 Ì, 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 m_{1}
and the point mass m_{2} (which is at the centre of the spherical shell m_{1}) fall freely and together (simultaneously) from a height h within the field of gravity of the motionless (V = 0) mass Ì. 
   Fig. 4 Let us also assume that, at the moment t_{1} > 0 (of the free fall of masses m_{1} and m_{2}), in terms of the inertial reference system S, we have: 1. V_{cm} = the velocity of the centre of mass of masses m_{1} and m_{2}
õ_{1} = the velocity of mass m_{1} õ_{2} = the velocity of mass m_{2} 2. h΄_{c} = the distance of the centre of mass of masses m_{1}_{ }and m_{2} from the upper limit of height h. h΄_{1} = the distance of mass m_{1}h΄_{2} = the distance of mass m_{2}

   where:

 
   If we now apply the principle of conservation of energy at the moment t_{1} > 0
of the free fall, we have: 
   Relation (56) yields: 
  
              
    As we can see, in relation (57) the velocity V_{cm} of the centre of mass of masses m_{1} and m_{2}
, is independent of masses m_{1} and m_{2}, i.e. valid for all values of masses m_{1} and m_{2}. b. Similarly, for mass m_{1}, we have:        Relation (58) yields:    
    As we can see, in relation (59) the velocity õ_{1} of mass m_{1} at the moment t_{1} > 0 of its free fall, is independent of mass m_{1}
, i.e. valid for all values of mass m_{1}. c. Similarly, for mass m_{2}, we have:         Relation (60) yields:         As we can see, in relation (61) the velocity õ_{2} of mass m_{2} at the moment t_{1} > 0 of its free fall, is independent of mass m_{2}, i.e. valid for all values of mass m_{2}
. Note: In cases (a), (b), (c) above, we assume that at the moment t_{1} > 0, the point mass m_{2} is always within the spherical shell m_{1}. Also, as we know, the velocity V_{cm} of the centre of mass of masses m_{1 }and m_{2} at the moment t_{1} > 0 is:        Relation (62) yields:  
     Substituting now in relation (63) V_{cm}
, õ_{1}, õ_{2} that result from relations (57), (59), (61) we have:  
