The International Year of Astronomy 2009 is a global effort initiated by the International Astronomical Union and UNESCO to help the citizens of the world rediscover their place in the Universe through the day and night-time sky and thereby engage a personal sense of wonder and discovery.

This year coincides with the 400th anniversary of Galileo Galilei’s first real use of the telescope in 1609 to observe and study the heavens. 2009 will also mark the 40th anniversary of mankind’s first landing on the Moon.

Therefore, for families seeking to discover the wonderment of a new activity through interest and wonder can look up to the skies to appreciate, in awe, the wonders of the Universe. This can be done through a small telescope at home or through the access of a computer. Various sites focussing on the wonders of astronomy and the cosmos are out there and are quickly accessible via the net.

Here are a few sites of interest for individual wonderment or to be experienced as a family:

http://portal.unesco.org/en/ev.php-URL_ID=44355&URL_DO=DO_TOPIC&URL_SECTION=201.html

http://www.un.org/apps/news/story.asp?NewsID=29553&Cr=UNESCO&Cr1=

Also check out a site called Sky Charts where an entire Universe can be uploaded on your computer to be enjoyed by the whole family.

Let us know what you think – leave a comment!

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Einstein’s Nemesis: DI Her Eclipsing Binary Stars Solution

The problem that the 100,000 PHD Physicists could not solve

This is the solution to the “Quarter of a century” Smithsonian-NASA Posted motion puzzle that Einstein and the 100,000 space-time physicists including 109 years of Nobel prize winner physics and physicists and 400 years of astronomy and Astrophysicists could not solve and solved here and dedicated to Drs Edward Guinan and Frank Maloney

Of Villanova University Pennsylvania who posted this motion puzzle and started the search collections of stars with motion that can not be explained by any published physics

For 350 years Physicists Astrophysicists and Mathematicians and all others including Newton and Kepler themselves missed the time-dependent Newton’s equation and time dependent Kepler’s equation that accounts for Quantum – relativistic effects and it explains these effects as visual effects. Here it is

Universal- Mechanics

All there is in the Universe is objects of mass m moving in space (x, y, z) at a location

r = r (x, y, z). The state of any object in the Universe can be expressed as the product

S = m r; State = mass x location

P = d S/d t = m (d r/dt) + (dm/dt) r = Total moment

= change of location + change of mass

= m v + m’ r; v = velocity = d r/d t; m’ = mass change rate

F = d P/d t = d¬≤S/dt¬≤ = Force = m (d¬≤r/dt¬≤) +2(dm/d t) (d r/d t) + (d¬≤m/dt¬≤) r

= m Œ≥ + 2m’v +m”r; Œ≥ = acceleration; m” = mass acceleration rate

In polar coordinates system

r = r r(1) ;v = r’ r(1) + r Œ∏’ Œ∏(1) ; Œ≥ = (r” – rŒ∏’¬≤)r(1) + (2r’Œ∏’ + rŒ∏”)Œ∏(1)

F = m[(r”-rŒ∏’¬≤)r(1) + (2r’Œ∏’ + rŒ∏”)Œ∏(1)] + 2m'[r’r(1) + rŒ∏’Œ∏(1)] + (m”r) r(1)

F = [d¬≤(m r)/dt¬≤ – (m r)Œ∏’¬≤]r(1) + (1/mr)[d(m¬≤r¬≤Œ∏’)/d t]Œ∏(1) = [-GmM/r¬≤]r(1)

d¬≤ (m r)/dt¬≤ – (m r) Œ∏’¬≤ = -GmM/r¬≤; d (m¬≤r¬≤Œ∏’)/d t = 0

Let m =constant: M=constant

d¬≤r/dt¬≤ – r Œ∏’¬≤=-GM/r¬≤ —— I

d(r¬≤Œ∏’)/d t = 0 —————–II

r¬≤Œ∏’=h = constant ————– II

r = 1/u; r’ = -u’/u¬≤ = – r¬≤u’ = – r¬≤Œ∏'(d u/d Œ∏) = -h (d u/d Œ∏)

d (r¬≤Œ∏’)/d t = 2rr’Œ∏’ + r¬≤Œ∏” = 0 r” = – h d/d t (du/d Œ∏) = – h Œ∏'(d¬≤u/d Œ∏¬≤) = – (h¬≤/r¬≤)(d¬≤u/dŒ∏¬≤)

[- (h¬≤/r¬≤) (d¬≤u/dŒ∏¬≤)] – r [(h/r¬≤)¬≤] = -GM/r¬≤

2(r’/r) = – (Œ∏”/Œ∏’) = 2[Œª + ·ªâ œâ (t)] – h¬≤u¬≤ (d¬≤u/dŒ∏¬≤) – h¬≤u¬≥ = -GMu¬≤

d¬≤u/dŒ∏¬≤ + u = GM/h¬≤

r(Œ∏, t) = r (Œ∏, 0) Exp [Œª + ·ªâ œâ (t)] u(Œ∏,0) = GM/h¬≤ + AcosŒ∏; r (Œ∏, 0) = 1/(GM/h¬≤ + AcosŒ∏)

r ( Œ∏, 0) = h¬≤/GM/[1 + (Ah¬≤/Gm)cosŒ∏]

r(Œ∏,0) = a(1-Œµ¬≤)/(1+ŒµcosŒ∏) ; h¬≤/GM = a(1-Œµ¬≤); Œµ = Ah¬≤/GM

r(0,t)= Exp[Œª(r) + ·ªâ œâ (r)]t; Exp = Exponential

r = r(Œ∏ , t)=r(Œ∏,0)r(0,t)=[a(1-Œµ¬≤)/(1+ŒµcosŒ∏)]{Exp[Œª(r) + √¨ œâ(r)]t} Nahhas’ Solution

If Œª(r) ‚âà 0; then:

r (Œ∏, t) = [(1-Œµ¬≤)/(1+ŒµcosŒ∏)]{Exp[·ªâ œâ(r)t]

Œ∏'(r, t) = Œ∏'[r(Œ∏,0), 0] Exp{-2·ªâ[œâ(r)t]}

h = 2œÄ a b/T; b=a‚àö (1-Œµ¬≤); a = mean distance value; Œµ = eccentricity

h = 2œÄa¬≤‚àö (1-Œµ¬≤); r (0, 0) = a (1-Œµ)

Œ∏’ (0,0) = h/r¬≤(0,0) = 2œÄ[‚àö(1-Œµ¬≤)]/T(1-Œµ)¬≤

Œ∏’ (0,t) = Œ∏'(0,0)Exp(-2·ªâwt)={2œÄ[‚àö(1-Œµ¬≤)]/T(1-Œµ)¬≤} Exp (-2iwt)

Œ∏'(0,t) = Œ∏'(0,0) [cosine 2(wt) – ·ªâ sine 2(wt)] = Œ∏'(0,0) [1- 2sine¬≤ (wt) – ·ªâ sin 2(wt)]

Œ∏'(0,t) = Œ∏'(0,t)(x) + Œ∏'(0,t)(y); Œ∏'(0,t)(x) = Œ∏'(0,0)[ 1- 2sine¬≤ (wt)]

Œ∏'(0,t)(x) ‚Äì Œ∏'(0,0) = – 2Œ∏'(0,0)sine¬≤(wt) = – 2Œ∏'(0,0)(v/c)¬≤ v/c=sine wt; c=light speed

Œî Œ∏’ = [Œ∏'(0, t) – Œ∏'(0, 0)] = -4œÄ {[‚àö (1-Œµ) ¬≤]/T (1-Œµ) ¬≤} (v/c) ¬≤} radians/second

{(180/œÄ=degrees) x (36526=century)

Œî Œ∏’ = [-720×36526/ T (days)] {[‚àö (1-Œµ) ¬≤]/ (1-Œµ) ¬≤}(v/c) = 1.04¬∞/century

This is the T-Rex equation that is going to demolished Einstein’s space-jail of time

The circumference of an ellipse: 2œÄa (1 – Œµ¬≤/4 + 3/16(Œµ¬≤)¬≤—) ‚âà 2œÄa (1-Œµ¬≤/4); R =a (1-Œµ¬≤/4)

v (m) = ‚àö [GM¬≤/ (m + M) a (1-Œµ¬≤/4)] ‚âà ‚àö [GM/a (1-Œµ¬≤/4)]; m<<M; Solar system

v = v (center of mass); v is the sum of orbital/rotational velocities = v(cm) for DI Her

Let m = mass of primary; M = mass of secondary

v (m) = primary speed; v(M) = secondary speed = ‚àö[Gm¬≤/(m+M)a(1-Œµ¬≤/4)]

v (cm) = [m v(m) + M v(M)]/(m + M) All rights reserved. [email protected]