# How Far is Mercury from the Sun

by Fraser Cain
on May 8, 2008

The orbit of Mercury is elliptical. This means that its orbit travels the path of an ellipse, and isn’t always the same distance from the Sun.

When Mercury makes its closest approach to the Sun, astronomers call this perihelion. So, the perihelion of Mercury is 46 million kilometers (29 million miles), or 0.307 astronomical units (1 AU is the distance from the Sun to the Earth).

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The furthest point of a planet’s orbit is called the aphelion. So in the case of Mercury, its furthest point stretches out to 70 million km (44 million miles), or 0.466 astronomical units.

Qué tan lejos está Mercurio del Sol

*Related*

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Kepler (demolish) Vs Einstein’s

Areal velocity is constant: r² ?’ =h Kepler’s Law

h = 2? a b/T; b=a? (1-?²); a = mean distance value; ? = eccentricity

r² ?’= h = S² w’

Replace r with S = r exp (? wt); h = [r² Exp (2iwt)] w’

w’ = (h/r²) exp [-2(i wt)]

w’= (h/r²) [cosine 2(wt) – ? sine 2(wt)] = (h/r²) [1- 2sine² (wt) – ? sin 2(wt)]

w’ = w'(x) + ? 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)= [2?a.a? (1-?²)]/Ta² (1-?) ²= [2?? (1-?²)]/T (1-?) ²

? w’ = (d w/d t – h/r²] = -4? {[? (1-?²)]/T (1-?) ²} (v/c) ² radian per second

? w’ = (- 4? /T) {[? (1-?²)]/ (1-?) ²} (v/c) ² radians

? w’ = (-720/T) {[? (1-?²)]/ (1-?) ²} (v/c) ² degrees; Multiplication by 180/?

? w’ = (-720×36526/T) {[? (1-?²)]/(1-?)²} (v/c)² degrees/100 years

? w” = (-720×3600/T) {[? (1-?²)]/ (1-?) ²} (v/c) ² seconds of arc by 3600

? w” = (-720x36526x3600/T) {[? (1-?²]/(1-?)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: 2?a (1 – ?²/4 + 3/16(?²)²- –.) ? 2?a (1-?²/4); R =a (1-?²/4) v=? [G m M / (m + M) a (1-?²/4)] ? ? [GM/a (1-?²/4)]; m<<M; Solar system

Advance of Perihelion of mercury.

G=6.673×10^-11; M=2×10^30kg; m=.32×10^24kg

? = 0.206; T=88days; c = 299792.458 km/sec; a = 58.2km/sec

Calculations yields:

v =48.14km/sec; [? (1- ?²)] (1-?) ² = 1.552

? 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?