1. We change the surdic formula {√Box}=√(2ax/(2a-x) into a form of {1/√Box}, because the derivative of d{√Box}/dx=(d(Box)/dx)/(2√Box)
√(2ax/2a-x)=√{(2ax)/2a-x)}.√2ax/√2ax=(2ax)/√(4a
2x-2ax
2)
Because the surdic quadratic form f(x)=4a
2x-2ax
2 has derivative f'(x)=4a
2-4ax , we rewrite the numerator (2ax) = -½(4a
2-4ax) + 2a
2 Since d{√(4a
2x-2ax
2)}/dx =½(4a
2-4ax)/√(4a
2x-2ax
2), the integral equation then reads:
t(r) = ∫dr = -√(4a2r-2ar2)/√(2G[m+M]) (1/√(2G[m+M])∫2a2/√(4a2r-2ar2)dr
2. The second part of the integral engages the inverse trigonometric functions as integral solutions and we complete the square of the denominator to obtain a usable integrable form:
d{arcsin[x/a]}/dx=1/√(a
2-x
2)
from x=asiny with inverse y=arcsin(x/a) and dx/dy=acosy and (acosy)
2 (asiny)
2=a
2 for dy/dx=1/acosy=1/√(a
2-(asiny)
2)=1/√(a
2-x
2)
2a
2/√(4a
2x-2ax
2) = 2a
2/√{2a(a
2-[x-a]
2)} = √(2a
3)/√{(a
2-[x-a]
2}
The integral equation is now evaluated as a function of the arclength L of the orbit from Aphelion at L=½π radian or 90° (1 radian=180°/π) and where arcsin(90°)=arcsin(½π radians)= 1 to Perihelion at L=-½π radian or -90° and where arcsin(-90°)=arcsin(-½π radians)=-1 with respect to the arclength of |½π|+|-½π|=π radians as the semiorbit of Nibiru.
The integration limits from r=r
maximum=r
aphelion=2a-a(1-e)=a(e+1)=2a-1AU to r=r
minimum=r
perihelion=|-a+ae}=|a(e-1)|=a(1-e)=|-1AU|=1AU, for a |positive arclength 2π| now redefine the integral equation in:
t(r) = ∫dr = √(4a
2r-2ar
2)/√(2G[m M]) + {√(2a
3)/√(2G[m+M])}{arcsin[(r-a)/a]} + Integration Constant C
Constant C is then taken at Perihelion as time t=0, say at December 21st, 2012 for r=a(e-1)=-1AU and defines:
C= √(4a
3(1-e)-2a
3(1-e)
2)/√(2G[m+M]) - {√(2a
3)/√(2G[m+M])}{arcsin[(a-ae-a)/a]}= {√(2a
3[1-e
2]/2G[m+M]) - √(2a
3/(2G[m+M]){arcsin[-e]}=(1.808080x10
10){0.092220+1.478445}=2.83989x10
10C=√(a3/G[m+M]).{√(1-e2)+{arcsin[e]} then calculates as {2.84x10
10 seconds or 899.925 civil years} for the 'flat line' L=π pathlength approximation of Nibiru in 1800 civil years, stated as eccentricity e=0.99573864 and for G(m+M)~GM~1.3348x10
20m
3/s
2and a
3~4.36367x10
40 meters and with arcsin[e]~84°52'25" or 0.471π~ 1.478445 radians and as the semi arclength joining Perihelion to Aphelion in 900 civil years as the integration constant yielding a: Total Time = Time at coordinate r + Initial Time with period T=2π√(a
3/GM) from a
3=GMT
2/4π
2for √(a
3/GM)=T/2π~1.808080x10
10 seconds for Nibiru's 3600 civil year period.
The completed integral equation so becomes:
t(r) = -√(4a2r-2ar2)/√(2G[m+M]) + √(a3/G[m+M]).{√(1-e2) + arcsin[(r-a)/a] + arcsin[e]}For any given distance r of Nibiru from the Sun, the time t for this displacement can then be calculated.
We indicate the domain for x from -90° at Perihelion to +90° at Aphelion for a counterclockwise semiorbit from maximum Aphelion to minimum Perihelion with an intermediate value for r=a and so when Nibiru would be at a distance of its own orbital semimajor axis from the sun. This path then takes a combined semiperiod of 1800 years from Apehelion to Perihelion for the arcsin values of 90°=π/2 radians and -90°=-π/2 radians respectively. The value at 0°=0 radians then defines a timeframe within this 1800 year half cycle measured from Perihelion say, as the minimum time coordinate at t=0 for Perihelion at displacement r
perihelion=a(1-e), which requires the elliptical orbit to become circular for e=0 and a linear approximation as e approaches 1 in relation to the arclength travelled.
The mass m of Nibiru can be neglected in this formulation, because M+m=10
30+m with m much smaller than M.
a) r=a t(r=a) = -√(4a
3-2a
3)/√(2GM]) + √(a
3/GM){√(1-e
2)+arcsin[0]+arcsin[e]}=-√(a
3/GM){1-√(1-e
2)-0-arcsin[e]}=-1.8081x10
10(1-√0.007984-1.4813..)=-1.8081x10
10{-0.5707}~1.032x10
10seconds or so 327.0 civil years into the future or past from the Perihelion. This would then describe a timeperiod of 2012-327=1685 and the 'Age of Newton, Kepler and Galileo' from the 15th century and following the classical period of the renaissance.
The semi latus rectum coordinates (±ae,b√(1-e
2))=(3.5056x10
13,2.9939x10
11) then give the ordinate (y-value) for the distance of Nibiru vertically and orthogonally to the central force focal center of the Sun. The hypotenuse of the formed right triangle then approximates the distance D of Nibiru from the Perihelion in D
2=(ae)
2 b
2(1-e
2)=for D=√(1.2289x10
27 8.963410
22)~3.5057x10
13 and a value near r=a as the semimajor axis of Nibiru.
For D=v
Nibiruaveraget
Nibiru, t
Nibiru=D/v
Nibiruaverage]=3.5057x10
13/1947~1.800x10
10seconds or 570.4 civil years for a timemarker of the year in 2012-570=1442 and when the Renaissance began in the European 'Old World' and when the 'Age of Science' 'enlightened' the 'Dark Middle Ages' of superstition and ignorance with names like Michelangelo, Boticelli, Leonardo da Vinci and Christopher Columbus.
This marker also indicates the beginning of the 13th Mayan Longcount baktun ending on 13.0.0.0.0=4Ahau3Kankin as December 21st, 2012 - 144,000 days = September 18th, 1618 in 12.0.0.0.0=5Ahau13Zotz.
For a value of
r=a/2 one calculates t(r=a/2) =-√{3a
3/4GM} + √(a
3/GM).{0.08935 arcsin[-½]+arcsin[e]}=-1.8081x10
10(√0.75-0.08935+0.5236-1.4813..)=-1.8081x10
10{-0.1810}~3.273x10
9 seconds or so 103.72 'civil' years into the future or past from Perihelion for an approximate dateline 2012-104=1908 and the start of World War 1 and the Industrial-Military conglomerations.
If Nibiru is destined to intersect the 1 AU scale in 2012, then Nibiru was 234.7/2=117.4 AU or so 17.6 billion kilometers from the Sun's position about 104 years ago.
b) r=rperihelion=a(1-e) t
perihelion(r=1AU)=t(r=a-ae) = -√(a
3/GM){√(1-e
2)} + √(a
3/GM){√(1-e
2)+arcsin[-e]+arcsin[e]}=-√(a
3/GM){√(1-e
2)-√(1-e
2)-arcsin[e]-arcsin[e]}=0 as the zeropoint of calibration, chosen at the Perihelion and as required for the boundary-initial condition for the integral equation.
Remember, that the velocity of Nibiru changes in slowing down in the transversing of Aphelion and speeds up in the transversion of the Perihelion. So it takes longer to travel from Aphelion to the midpoint of the elliptical orbit as a quarter of the total path, then it does to travel from this midpoint to the Perihelion, namely at the r=a time coordinate of 327 civil years being shorter and as measured from Perihelion, than the 900-327=573 year interval from the Aphelion to the r=a time coordinate and nexus point.
c) raphelion=a(1+e)t
aphelion(r=2a-1AU)=t(r=a+ae)=-√(a
3/GM){√(1-e
2)}-√(1-e
2)-arcsin[e]-arcsin[e]}=-√(a
3/GM){2arcsin[e]}=-1.8081x10
10(2.9626..)~-5.3567x10
10 seconds or so 1697.5 civil years into the future or past from the Perihelion for a time setting of about 2012-1697.5=314.5 AD.
Should 2arcsin[e]=2[½π]=π for e=1; the arclength between Perihelion and Aphelion would be 2x½π radians for -1.8081x10
10(π)~-5.680x10
10seconds or 1800 years in the past or the future from Perihelion.
t(r=2AU)= -√(4a
2r-2ar
2)/√(2GM) √(a
3/GM).{√(1-e
2) arcsin[(r-a)/a]+arcsin[e]}=-1/√(2GM){√{4(1.2395x10
27)(3.0x10
11)-2(3.5206x10
13)(9.0x10
22)}-(2.0889x10
20).(0.09222 arcsin[-0.991479] arcsin[e])}~-1/√(2GM){√(1.4874x10
39-6.3371x10
36) - (2.0889x10
20)(0.09222-1.44016 1.47845)}~-1/√(2GM){√(1.4811x10
39) - (2.0889x10
20)(0.1305)}~(3.8485x10
19-2.7260x10
19)/(1.6339x10
10)~6.870x10
8 seconds or 21.8 civil years.
This then shows, that Nibiru would have been at a distance approaching the orbit of Mars for a semimajor axis of 1.524 AU and at the distance of the Asteroid Belt (about 2.2-3.3 AU) about 22 years ago in the year 2012-22=1990 and would have approached ever closer to its Perihelion position from 1990 to 2012 in the displacement interval for Nibiru from 2 AU around 1990 and at 1 AU in 2012.
d) rMars=1.5AU or 2.25x1011meters Intersection in approximately the year 1993t(r=1.5 AU)= -√(4a
2r-2ar
2)/√(2GM) √(a
3/GM).{√(1-e
2) arcsin[(r-a)/a] arcsin[e]}=-1/√(2GM){√{4(1.2395x10
27)(2.25x10
11)-2(3.5206x10
13)(5.0625x10
22)}-(2.0889x10
20).(0.09222 arcsin[-0.99361] arcsin[e])}~-1/√(2GM){√(1.1156x10
39-3.5646x10
36) - (2.0889x10
20)(0.09222-1.45769+1.47845)}~-1/√(2GM){√(1.1120x10
39) - (2.0889x10
20)(0.1130)}~(3.3347x10
19-2.360x10
19)/(1.6339x10
10)~5.965x10
8seconds or 18.9 civil years. So Nibiru would have intersected the orbit of Mars anno 1993, should it be at Perihelion in 2012.
Nibiru's Maximum speed near the Perihelion is so v
perihelion =42,138 m/s or 1.33x10
12meters per year or 1.33x10
12/1.5x10
11P=8.86 AU/year and Nibiru's Minimum speed at Aphelion is about v
aphelion = 90 m/s or 2.84x10
9m/year or 2.84x10
9/1.5x10
11=0.019 AU/year for a geometric mean speed of v
Nibiruaverage=√GM/a)=√{GM√(1-e
2)/b}~1947 m/s as (√(8.86x0.019)~√(0.168)~0.41 AU/year or so 168,400 kilometres per day or 7016 kilometers per hour in the semiminor axis b=√{r
aphelion.r
perihelion)=a√(1-e
2)=3.246430x10
12 meters.
As the speed near Perihelion would exceed the average speed of Nibiru, in a time from December 2011 to January 2012, Nibiru so would have been no farther away from Perihelion, than 0.41 AU, which is about half the orbit of Venus at 0.72 AU and near the orbit of Mercury at 0.39 AU. And for the June Solstice 2012 (with the Venus Transit 2012), Nibiru would have then been located about 0.2 AU from Perihelion, intersecting its highly unstable path in the orbit of the innermosts planet in the solar system in Mercury and Venus in gravitational perturbations. One could then mythologize Nibiru as the planet nearest the Sun, commonly called Vulcan.