The International Year of Astronomy Begins
I first heard about the proposal for the International Year of Astronomy when it was being discussed at the IAU General Assembly in Prague. Now, almost two and a half years later, the year has finally begun! There seems to be a buzz of excitement and I'm sharing in it.
The year celebrates the 400th anniversary of the first use of the astronomical telescope and also happens to be the 40th anniversary of the Apollo Moon landings so there could hardly be a better year for these cosmic celebrations.
Throughout the year there will be plenty of activities all over the globe so you should check with your National Node for events near you. If there are none, you could consider organising some along with local astronomers or astronomical societies. One big event to look out for is the 100 Hours of Astronomy which aims to be a 100 hour long global star party. Given that it stretches over 100 hours, it might give those of us in cloudy places a chance to participate! On the net you can also participate in the year through podcasts, blogs and twitter (I'm involved in two of those).
If you own a telescope that doesn't usually see the outside of a cupboard, then this is the time to take it out and share the wonders of the night sky (or Sun if you have a solar filter) to friends, family, neighbours, and co-workers. The International Year of Astronomy hopes to let millions of people see the night sky this year and that will only happen if we all do our bit. Go on, it'll be fun.
It's a big universe and it's ours to discover.
Comments: The International Year of Astronomy Begins
What a wonderful idea to have a year, when we look at the skies, not just peer.When the moon and the sun are there for our fun, while the planets and stars seem so near.
Posted by Joseph Lavelle on Thursday 15th Jan 2009 (09:03 UTC)
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 = d²S/dt² = Force = m (d²r/dt²) +2(dm/d t) (d r/d t) + (d²m/dt²) r
= m ó + 2m'v +m"r; ó = acceleration; m'' = mass acceleration rate
In polar coordinates system
r = r r(1) ;v = r' r(1) + r ø' ø(1) ; ó = (r" - rø'²)r(1) + (2r'ø' + rø")ø(1)
F = m[(r"-rø'²)r(1) + (2r'ø' + rø")ø(1)] + 2m'[r'r(1) + rø'ø(1)] + (m"r) r(1)
F = [d²(m r)/dt² - (m r)ø'²]r(1) + (1/mr)[d(m²r²Ã¸')/d t]ø(1) = [-GmM/r²]r(1)
d² (m r)/dt² - (m r) ø'² = -GmM/r²; d (m²r²Ã¸')/d t = 0
Let m =constant: M=constant
d²r/dt² - r ø'²=-GM/r² ------ I
d(r²Ã¸')/d t = 0 -----------------II
r²Ã¸'=h = constant -------------- II
r = 1/u; r' = -u'/u² = - r²u' = - r²Ã¸'(d u/d ø) = -h (d u/d ø)
d (r²Ã¸')/d t = 2rr'ø' + r²Ã¸" = 0 r" = - h d/d t (du/d ø) = - h ø'(d²u/d ø²) = - (h²/r²)(d²u/dø²)
[- (h²/r²) (d²u/dø²)] - r [(h/r²)²] = -GM/r²
2(r'/r) = - (ø"/ø') = 2[û + á»ⰠÃâ° (t)] - h²u² (d²u/dø²) - h²u³ = -GMu²
d²u/dø² + u = GM/h²
r(ø, t) = r (ø, 0) Exp [û + á»ⰠÃâ° (t)] u(ø,0) = GM/h² + Acosø; r (ø, 0) = 1/(GM/h² + Acosø)
r ( ø, 0) = h²/GM/[1 + (Ah²/Gm)cosø]
r(ø,0) = a(1-õ²)/(1+õcosø) ; h²/GM = a(1-õ²); õ = Ah²/GM
r(0,t)= Exp[û(r) + á»ⰠÃâ° (r)]t; Exp = Exponential
r = r(ø , t)=r(ø,0)r(0,t)=[a(1-õ²)/(1+õcosø)]{Exp[û(r) + ì Ãâ°(r)]t} Nahhas' Solution
If û(r) ââ°Ë† 0; then:
r (ø, t) = [(1-õ²)/(1+õcosø)]{Exp[á»ⰠÃâ°(r)t]
ø'(r, t) = ø'[r(ø,0), 0] Exp{-2á»â°[Ãâ°(r)t]}
h = 2à a b/T; b=a√ (1-õ²); a = mean distance value; õ = eccentricity
h = 2Ãa²√ (1-õ²); r (0, 0) = a (1-õ)
ø' (0,0) = h/r²(0,0) = 2Ã[√(1-õ²)]/T(1-õ)²
ø' (0,t) = ø'(0,0)Exp(-2á»â°wt)={2Ã[√(1-õ²)]/T(1-õ)²} Exp (-2iwt)
ø'(0,t) = ø'(0,0) [cosine 2(wt) - á»Ⱐsine 2(wt)] = ø'(0,0) [1- 2sine² (wt) - á»Ⱐsin 2(wt)]
ø'(0,t) = ø'(0,t)(x) + ø'(0,t)(y); ø'(0,t)(x) = ø'(0,0)[ 1- 2sine² (wt)]
ø'(0,t)(x) â✠ø'(0,0) = - 2ø'(0,0)sine²(wt) = - 2ø'(0,0)(v/c)² v/c=sine wt; c=light speed
Ãâ ø' = [ø'(0, t) - ø'(0, 0)] = -4à {[√ (1-õ) ²]/T (1-õ) ²} (v/c) ²} radians/second
{(180/Ã=degrees) x (36526=century)
Ãâ ø' = [-720x36526/ T (days)] {[√ (1-õ) ²]/ (1-õ) ²}(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: 2Ãa (1 - õ²/4 + 3/16(õ²)²---) ââ°Ë† 2Ãa (1-õ²/4); R =a (1-õ²/4)
v (m) = √ [GM²/ (m + M) a (1-õ²/4)] ââ°Ë† √ [GM/a (1-õ²/4)]; m<
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 = √[Gm²/(m+M)a(1-õ²/4)]
v (cm) = [m v(m) + M v(M)]/(m + M) All rights reserved. joenahhas1958@yahoo.com
Posted by joe nahhas on Wednesday 28th Jan 2009 (14:28 UTC)
I'm sure you are convinced by your theories. If you want to convince others, please publish them somewhere more appropriate such as a peer reviewed journal. If you have read around the subject, addressed counter-arguments, agree with the existing data and predict some future observation, people will take it seriously.
Posted by Stuart on Wednesday 28th Jan 2009 (17:52 UTC)
I've never heard of the International Year of Astronomy. Glad I've discovered it. Thanks for the exposure.
Posted by Levinson Axelrod on Thursday 18th Mar 2010 (15:14 UTC)