MASS REMAIN THE SAME. ADDITIONAL ENERGY and FORCE THAT
WOULD BE PROVIDED TO “ADDITIONAL TANGIBLE MASS” IS ADDED TO “ADDITIONAL
GRAVITATIONAL INERTIA, WHICH GROWTH WITH VELOCITY RATIO V/c”
Eo
= mo . c2 (internal energy at rest, at Vo=0) ----(1)
m1 =mo. (1 – V12/c2
)-1/2 ----(2)
Increase of mass = m1-o = mo. (1 – V12/c2
)-1/2 - mo = mo . { (1 – V12/c2 )-1/2 - 1} ---(3)
E1
= mo . c2 . (1 – V12/c2 )-1/2
= m1 . c2
DIFERENTIATING first left pair:-
---(4)
dE
= mo . c2 . ( -1/2) . (1 – V2/c2
)-1/2 –1 . ( - 2 . V/c2 )
. dV -------(5)
dE
= mo . V . (1 – V2/c2 )-3/2
. dV -------(6)
F
. V = P (Power = Force x Velocity) Multiplying both sides by "dt"
------(7)
Notice
that E = Energy = Power x time dE = P
. dt and replace with -----(8)
F. V. dt = P . dt = dE = mo . V . (1 – V2/c2
)-3/2 . dV -----(9) Divide all by V . dt
F
= mo . V . (1 – V2/C2 )-3/2
. dV / (V . dt) -----(10)
dV /dt = a = acceleration
F = mo
. a . (1 – V2/c2
)-3/2 = -----(11); adding and
subtracting mo
. a
= mo
. a . 1 + mo . a . {(1
– V2/c2
)-3/2 -
1} = -----(11)
= F1
(Newton’s
traditional force) + F2nd
(additional inertia
force) -----(12)
F2nd = Additional inertia force = mo . a . { ( 1 – V2/c2 )-3/2 - 1 }
-----(13)
E1 =
Eo + Increse E1-o (= “increase of energy stored” into tangible
mass “mo”) -----(14)
E1 = mo . c2 .
(1 – V12/c2 )-1/2 -----(15)
E1 = Eo + Increase E1-o = mo . c2 + Increase E1-o -----(16) Mass not change.
Increase E1-o = E1 - Eo = mo
. c2 . ( 1 – V12/c2 )-1/2 - mo . c2 = -----(17)
Increase E1-o = mo . c2 . {
(
1 – V12/c2 )-1/2 – 1} -----(18)
Increase of mass = m1-o = mo . { ( 1 – V12/c2 )-1/2 - 1} -----(19)
F2nd = Additional inertia force
= mo
. a . { (1 – V2/c2 )-3/2 - 1} -----(20)
E1
= mo . c2 . { (1 – V12/c2 )-1/2
–1}= m1 . c2 DIFERENTIATING first left pair:----(21)
dE
= mo . c2 . ( -1/2) . (1 – V2/c2
)-1/2 –1 . ( - 2 . V/c2) .
dV ------(22)
dE
= mo . V . (1 – V2/c2 )-3/2
. dV -------(23)
F
. V = P (Power = Force x Velocity) Multiplying both sides by "dt"
-----(24)
Notice
that E = Energy = Power x time dE = P
. dt and replace with -----(25)
F. V. dt = P . dt = dE = mo . V . (1 – V2/c2
)-3/2 . dV ------(26) Divide all by V . dt
F
= mo . V . (1 – V2/c2 )-3/2
. dV / (V . dt) -----(27)
dV /dt = a = acceleration
F = mo . a . (1 – V2/c2 )-3/2 ------(28)
At V=0 (at rest, when
starts Force for acceleration) Ftotal at “o” = mo .
a . 1 = F1 –--(29)
At final V=V1 ) Ftotal at “1” = mo
. a . (1 – V12/c2
)-3/2 = F1 (=Fo) + F2nd at “1” –--(30)
F2nd at “1” = Ftotal
at “1” – F1 = mo . a
. (1 – V12/c2
)-3/2 - mo
. a . 1 = –--(31)
F2nd at “1” = mo . a . { (1 – V2/c2 )-3/2 - 1} F2nd =
Additional inertia force –---(32)
IT WAS TESTED THE EQUATION for the F2nd, additional
force of INERTIAL UNIVERSAL GRAVITATION FIELD, deducted in this site: http://br.geocities.com/celsosavelligomes/SECOND_LAW_GRAVITATIONexactly.htm
Ftotal = mo. a
+ mo. a . (V/c)2 [{3/2 + 15/8.(V/c)2 +
35/16.(V/c)4 + 315/128 . (V/c) 6 + 693/256 . (V/c) 8 ) +
...] = ----(33)
(See in http://www.anti-relativity.com/forum/viewtopic.php?t=239&postdays=0&postorder=asc&start=30
Posted Tue Apr 11, 2006
11:47 pm Post subject: How 2nd GRAVITATIONAL LAW affects
gravitational acceleration)
= F1 (Newton’s inertial
force = mo.
a ) +
+ F2nd (gravitational
inertial force of 2nd type Gravitation = mo. a . (V/c)2
[ { 3/2 + 15/8.(V/c)2 + 35/16.(V/c)4 + 315/128 . (V/c)
6 + 693/256 . (V/c) 8 ) + ...] ) ---(34)
We were able to provide support for
F1 remaining constant, as mo . a,
thus in fact
there is none mass increase, as “m1 –mo” with increasing
ratio “V/c”. In fact what increases is a potential gravitational energy,
generating the second type of gravitational force “opposing” to any change of
velocity or direction:
F2nd equals to “mo . a2nd“, with “acceleration” as a2nd being the second type (inertial
gravitational force, of 2nd type of Universal Gravitation, further
than Newton’s Law) of gravity acceleration (quite like our “g” on Earth) but
that applies only for very huge velocities “V”, as if when the ratio V/c
approaches 1 (from under or above “c”, just a transition barrier):
a2nd = a . (V/c)2 . [
{ 3/2 + 15/8.(V/c)2 +
35/16.(V/c)4 + 315/128 . (V/c) 6 + 693/256 . (V/c) 8 ) + ...] ) ----(35)
a2nd depends on (V/c) as direct controlling parameter for acceleration. Thus a2nd is an
expression that depends
directly on (V/c) ratio, being “zero” when we are at
rest (V=0) when it applies the classic gravity acceleration and Newton’s Law of
inertia: Ftotal = mo .
a only.
Only
in large Velocities, approaching “c”, is that a2nd becomes much
greater than “a” (usual
acceleration making to increase or decrease kinetics energy of mo). And also a2nd can be a NEGATIVE acceleration, acting as a “break”,
as when is opposing the free fall of a
particle (of tangible mass or of light energy) into
a very massive and dense star, to collide with it or just to orbit it with perihelion nearby the surface of
such star (as that was astronomic experiments on deflection of
light, one century ago, to provide support for Einstein’s theory). The problem is that the 2nd type of
gravitation, through F2nd forces (formulas
“32” and “35”, above), will avoid particles coming too close to star at before
computed perihelion. Today it is
believed that such measurements were not good tests to validate relativity
theory (http://www.geocities.com/autotheist/Physics/gr.htm?200615
)