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 Proof  forthe advanceof Mercury'sperihelion
 TECHNOLOGYFusion:The “ZEUS”machine
 PROOF FOR THE ADVANCE OF MERCURY’S PERIHELION A. CALCULATING THE CENTER OF MASS OF THE SOLAR SYSTEM
 As it is well known, all celestial bodies of the Solar system, planets, asteroids, comets, etc, as well as the Sun, revolve around the center of mass of our Solar system. Given, however, that every planet Pi(i = 1,2,3,…9), i.e. Mercury, Venus, Earth, … etc, has a period of revolution Ti around the Sun, then theoretically at a time t all planets will be found on the same semi-straight line oχ´ (or extremely close to it), that is, they will be found on a total general planetary synod, Fig. 1.
 Let us assume in this case that C is the center of mass of our Solar system.As itis well known, the center of mass of our Solar system remains stable at all times at position C on the straight line χχ´ and this position C is independent of any planetary motion around it. Taking now the Sun as the measurement principle, distance d between the Sun and the center of mass C of the Solar system is expressed (as it is commonly known) by the following formula:
 (1)
 where M1, M2, M3, … M9 are the masses of the planets and R1, R2, R3… R9 their respective distances from the Sun.M is the mass of the Sun.By substituting into formula (1) the values of M1, M2, M3, … M9 and R1, R2, R3… R9, such as thoseare given in Table 1 (NASA Solar System data), formula (1) yields the following:
 or
 which is distance d, separating the Sun from the center of mass C of our Solar system. Moreover, because the Sun’s radius r0 is:
 this signifies that the center of mass of our Solar system is beyond the Sun and in fact it lies at a distance d0, that is:
 or
 m
 or
 Inotherwords, the center of mass of our Solar system lies at a distance greater than one radius of the Sun, outside its surface. Note: In the above calculation of the center of mass of our Solar system, the satellites of the planets were not taken into account, since their mass is extremely small in relation to the total mass of our Solar system.
 Consequently, as it has been stated above, because all celestial bodies of our Solar system revolve around the center of mass of the Solar system, then the Sun, also, will revolve around the center of mass of the Solar system in an approximate circular orbit of radius d = 1,505 109 m. This fact, as we shall see later in this text, plays a decisive role especially in the advance of Mercury’s perihelion, as well as of the perihelia of other planets of our Solar system. .NOTE: Radius d=1,505.109m is approximately considered to be average value of the radiuses of the sun`s rotation around the center of mass of our solar system, pic.1(see, http://astro.berkeley.edu/~eliot/Astro7A/Gravity.pdf).
 Pic.1 B   CALCULATING THE ADVANCE OF MERCURY’S PERIHELIONLet us assume that (Fig. 2) we have a frame of reference xoy whose starting point o is the center of mass of our Solar system
 In this frame of reference xoy, the Sun moves in a circular orbit C0, with the center being point o and radius d = 1,505 109 m.   Furthermore, for the sake of simplicity, we consider that in this frame of reference xoy the orbits Ci of all planets Pi are circular, where i = 1,2,3,…9 are in sequence Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and Pluto.    In this frame of reference xoy, we consider that distance Ri between a planet Pi and the Sun is the distance Ri between two concentric circumferences, i.e. circumference Ci of planet Pi and circumference C0 of the Sun.   These distances Ri between planets Pi and the Sun are those listed in Table 1.  A. PLANET MERCURY     In frame of reference xoy, with the Sun revolving in its orbitC0 having a period T0 and with Mercury revolving in its orbit C1 having a period T1 (T0<
 or
 (1)
 or
 B. THE OTHER PLANETS   As regards the other planets Pi(i = 2,3,…9), i.e. Venus, Earth, Mars,… etc, because distance d=1,505 109m is extremely small relative to their distanceRi from the Sun (d<< Ri), their orbits Ciare not perceptibly affected by the revolution of the Sun around the center of mass of the Solar system, as is the case with planet Mercury.   Therefore, for planets Piwe accept the accuracy of Kepler’s first law according to which (as it is well known) the Sun coincides with the center of mass of the Solar system.   Consequently, in this case, distances Ri between planets Pi(i = 2,3,…9) and the Sun are those listed in Table 1.    After everything explained above, the calculation of the advance of Mercury’s perihelion assumes the form of the following equivalent problem:  THE EQUIVALENT PROBLEM   Consider a heliocentric frame of reference xoy, with the Sun being motionless in position o, where o is the center of mass of our Solar system.     Planet Mercury moves in a circular orbit of radius a with the center being point o, according to relation (1), i.e.:
 or
 (2)
 All the other planets Pi(i = 2,3,…9), i.e. Venus, Earth, Mars,… etc, move in circular orbits Ci of radiuses Ri(distances from the Sun) stated in Table 1, withthe center being point o.    Note: In relation (2), the exact distance R1 between the Sun and Mercury will be calculated here below.   PROOF1.      The classical case   As it is commonly known, since the era of Le Verriere to this day, we have been working on the advance of Mercury’s perihelion by considering the Sun as motionless and its center of mass as coinciding with the center of mass of the Solar system, without taking into account its revolution around the center of mass of our Solar system.   Therefore, in this classical case, attractive force F0 exerted by the Sun on Mercury will be:
 N (3)
 (See Chris Pollock http://www.math.toronto.edu/~colliand/426_03/Papers03/C_Pollock.pdf)   This attractive force F´0 corresponds to a distance R1 between the Sun and Mercury, which is given by Newton’s Law:
 (4)
 where, G = 6,67  10-11the universal gravitational constant            m = 3,302  1023  Kg the mass of planet Mercury           M = 1, 989  1030 Kg the mass of the Sun, and           F΄0 = 1,318  1022Ν. } (5)
 By substituting the above values (5) into relation (4), we will have:R1 = 57,651  109m.      (6)    This is the exact distance between the Sun and Mercury, in a heliocentric system where the Sun is motionless at the center of mass of the Solar system, without taking into account the Sun’s revolution around it. (It is the classical case, known as such to this day).   Note: In this classical case, the advance of Mercury’s perihelion equals 531,9´´/century (See Chris Pollock’s paper referred to above). 2 . The case based on the new data and the calculation of the advance of Mercury’s perihelion.   On the contrary, according to the new data, i.e. by taking into account the Sun’s revolution around the center of mass of our Solar system, then on the basis of the above-mentioned equivalent problem resulting from relations (2) and (6), distance a between the Sun and Mercury will be:α = R1 + 1,505  109 m    orα = 57,651  109 m + 1,505  109 m    orα = 59,156  109m  (7)   After everything explained above, we can now work on the calculation of the advance of Mercury’s perihelion.   According to the equivalent problem and on the basis of value (7), force F0exerted by the Sun M on Mercury m will be:
 or
 (8)
 or
 Moreover, as it is well known, on the basis of the equivalent problem, the advance δφ of Mercury’s perihelion, per Mercury’s period T (See Chris Pollock’s paper mentioned above), is given by relation (9):
 where
 (9)
 (10)
 Ψ is the angle of the line of Mercury’s apses and Fa is the force exerted by all planets Pi(i = 2,3,…9) together on Mercury, which [force] is expressed by the following relation:
 (11)
 Relations (10) and (11), yield the following final relation:
 (12)
 where λiin relation (12) is the linear masses of planets Pi(i = 2,3,…9), i.e. Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and Pluto.    Since the linear mass λi of a planet is given by:
 (13)
 where Mi is the mass of planet Pi, and Riis its distance from the Sun, then based on relation (13) and Table 1, we obtain the following values for the linear masses λi of these planets, namely:  LINEAR MASSES OF THE PLANETS
 Venus  : λ2 = 7,159  1012Earth    :λ3 = 6,354  1012Mars    :λ4 = 4,481  1011Jupiter : λ5 = 3,880  1014Saturn:  λ6 = 6,344  1013Uranus : λ7 = 4,815  1012Neptune: λ8 = 3,623  1012Pluto :     λ9 = 350.324.121 (14)
 Moreover, in relation (12):
 G = 6,67 10-11, the universal gravitational constantm = 3,302 1023, the mass of planet Mercuryλi = the linear masses of planets Pi(i = 2,3,…9),         as expressed by relations (14)Ri = the distance of planets Pi(i = 2,3,…9) from the Sun,        as listed in Table 1.α = the distance between the Sun and Mercury as given       by relation (7) F0 = – 1,252 1022 N  the force exerted by the Sun          on Mercury as expressed by relation (8) (15)
 Note: The minus sign of force F0signifies that force F0 exerted by the Sun on Mercury is opposite to Force Faexerted by all other planets together on Mercury.   From relation (11), based on values given by (14) and (15), it results that force Fa is Fα = 7,752 1015 Ν   By substituting now into the final relation (12) the values provided by (14) and (15), we obtain:
 or
 or (16)
 or
 Also,
 (17)
 From relations (16) and (17), it results that
 (18)
 Thus, relations (18) and (9) yield:
 per period T of Mercury. (T = 87,969 days) or per period T of Mercury (2π radians = 1.296.000´´) or (1 year = 365,24 days), or finally:
 This, therefore, is the requested value of advanceδφof Mercury’s perihelion, once we take into account the Sun’s revolution around the center of mass of our Solar system.The resulting value of 576,7´´/century greatly converges with the value of 574,8´´/century yielded by astronomic observations about the advanceof Mercury’s perihelion with error εwhich is:
 that is, ε= 3%0 (per thousand) / centuryor with difference δ:
 that is,
 As it can be observed, this difference δ = 1,9´´/century is extremely small and reveals the true causes of the advance of Mercury’s perihelion, which are mainly the following:    a)    The perturbative forces exerted by other planets on Mercury, and   b)    The Sun’s revolution around the center of mass of the Solar system, as demonstrated above.     After everything described above, the following basic conclusion can be drawn: CONCLUSION   The difference of 43´´/century with astronomic observations as regards the advance of Mercury’s perihelion is not attributed to the curvature of space-time around the Sun, as the Theory of Relativity erroneously maintains.     The 43´´/century of the advance of Mercury’s perihelion (as demonstrated above) are due to the revolution of the Sun around the center of mass of our Solar system, a fact that until today has never been taken into account when calculating the advance of Mercury’s perihelion. Finally, after everything discussed in this paper, the Theory of Relativity should be unquestionably deemed erroneous.IMPORTANT REMARK   In this paper the advance of Mercury’s perihelion has been calculated with error ε = 3%o (per thousand)/century or with a difference δ = 1,9´´/century.    Certainly, there are other theoretical methods resting on a different rationale according to which the advance of Mercury’s perihelion can be calculated probably with a smaller error ε and a smaller difference δ.      Should the calculations of these theoretical methods yield an error ε < 3%o (per thousand)/century and a difference δ < 1,9´´/century (as compared to this paper), it would be very interesting to see them published.  Copyright 2006: Christos A. Tsolkas                                  Christos A. Tsolkas                                                                                                   May 2006 ©  Copyright 2001 Tsolkas Christos.  Web design by Wirenet Communications