Areal velocity is constant: r² θ' =h Kepler's Law
h = 2π a b/T; b=a√ (1-ε²); a = mean distance value; ε = eccentricity
r² θ'= h = S² w'
Replace r with S = r exp (ỉ wt); h = [r² Exp (2iwt)] w'
w' = (h/r²) exp [-2(i wt)]
w'= (h/r²) [cosine 2(wt) - ỉ sine 2(wt)] = (h/r²) [1- 2sine² (wt) - ỉ sin 2(wt)]
w' = w'(x) + ỉ w'(y) ; w'(x) = (h/r²) [ 1- 2sine² (wt)]
w'(x) – (h/r²) = - 2(h/r²)sine²(wt) = - 2(h/r²)(v/c)² v/c=sine wt
(h/ r²)(Perihelion/Periastron)= [2πa.a√ (1-ε²)]/Ta² (1-ε) ²= [2π√ (1-ε²)]/T (1-ε) ²

Δ w' = (d w/d t – h/r²] = -4π {[√ (1-ε²)]/T (1-ε) ²} (v/c) ² radian per second
Δ w' = (- 4π /T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² radians
Δ w' = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² degrees; Multiplication by 180/π
Δ w' = (-720x36526/T) {[√ (1-ε²)]/(1-ε)²} (v/c)² degrees/100 years
Δ w” = (-720x3600/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² seconds of arc by 3600

Δ w" = (-720x36526x3600/T) {[√ (1-ε²]/(1-ε)²} (v/c)² seconds of arc per century
This Kepler's Equation solves all the problems Einstein and all physicists could not solve

The circumference of an ellipse: 2πa (1 - ε²/4 + 3/16(ε²)²- --.) ≈ 2πa (1-ε²/4); R =a (1-ε²/4) v=√ [G m M / (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m<<M; Solar system
Advance of Perihelion of mercury.

G=6.673x10^-11; M=2x10^30kg; m=.32x10^24kg
ε = 0.206; T=88days; c = 299792.458 km/sec; a = 58.2km/sec
Calculations yields:
v =48.14km/sec; [√ (1- ε²)] (1-ε) ² = 1.552
Δ w”= (-720x36526x3600/88) x (1.552) (48.14/299792)²=43.0”/century

Conclusions: The 43" seconds of arc of advance of perihelion of Planet Mercury (General relativity) is given by Kepler's equation better than all of Published papers of Einstein. Kepler's Equation can solve Einstein's nemesis DI Her Binary stars motion and all the other dozens of stars motions posted for past 40 years on NASA website SAO/NASA as unsolved by any physics

Anyone dare to prove me wrong?