If you’ve been staring at the Sun lately, then you may have noticed it looks a wee bit bigger today than it did a few days ago. That’s not because the UV light from the Sun is frying your retina; it’s actually true. Today is perihelion, the time when the Earth is closest to the Sun in its orbit.
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Sitting here on this whirling blob of mud, we actually go around the Sun in an ellipse, not a circle. It’s almost a circle, though. The deviation of an ellipse from perfect circelness is called the eccentricity, and runs from 0 (a true circle) to 1 (which would actually be a parabola, kind of like a circle stretched out infinitely to one side). The formula for eccentricity is pretty simple:
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The a in the equation is the semimajor axis of the ellipse, or half the long dimension. b is the semiminor axis — half the short width. For a circle, a = b, so the equation works out to 0 as it should. As the ellipse gets more oval, a gets bigger and b gets smaller, and the eccentricity approaches the value of 1.
For the Earth’s orbit, the eccentricity is a miniscule 0.0167, meaning that a and b are pretty close to being the same value. The semimajor axis of the Earth is about 149,598,000 kilometers (93 million miles). Using the value of the eccentricity and plugging it into the equation, the semiminor axis of the Earth is about 149,577,000 km.
So you might think that the Earth gets as far as 149,598,000 km from the Sun, and as close as 149,577,000, a difference of about 21,000 kilometers (13,000 miles). But that’s not correct!
That would be true if the Sun sat at the center of the Earth’s orbital ellipse. It doesn’t. Ellipses are funny when you’re talking gravity and orbits; the more massive object (the Sun) sits at the focus of the less massive object’s (the Earth’s) elliptical orbit. The focus is not really the center, it’s offset from the center by the distance (a2 – b2)1/2. Here’s our diagram again with the focus labeled:
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Plugging and chugging in our values for a and b, we get the Sun being 2.5 million km (1.6 million miles) from the center of the Earth’s orbit. That means the farthest we can get from the Sun is the semimajor axis a plus the distance of the focus from the center: 149,598,000 + 2,500,000 = (roughly) 152,000,000 km (94.4 million miles).
The closest we can get is the semimajor axis minus the focus distance = 149,598,000 – 2,500,000 = (roughly) 147,000,000 km (91.3 million miles). Note I’m not trying to be hugely accurate here; I just want an idea of these numbers.
So over the course of the year, the Earth ranges from about 147 million km from the Sun at its closest to about 152 million km, a difference of about 5 million km (3 million miles), or a difference of a little over 3%.
When the Earth is precisely at the point in its orbit closest to the Sun, that’s perihelion. That happened today, January 4, 2009 at 15:00 UT (10:00 a.m. Eastern time). Aphelion, when we’re farthest away, won’t be until July 4.
What does this mean? Well, that translates directly into a 3% change in the apparent size of the Sun in the sky over the year. Honestly? You’d never notice, staring at the Sun or not. You’d need a telescope and careful measurements to see the difference.
Temperature? In fact, yeah, when we’re at perihelion the Earth gets a little more light and heat from the Sun, and less at aphelion. Yet here in the northern hemisphere we’re in the dead of winter. Obviously, the distance of the Earth from the Sun doesn’t affect the seasons very much. Why not? Well, that’s a whole ‘nuther story.
But for now, enjoy our solar proximity. Over the next six months we’ll pull away from our nearest star, and then in July we’ll reach the apex of our orbit, and the dance starts all over again.
January 4th, 2009 at 11:33 am
The additional sunlight is actually detectable on spacecraft. We have additional power to deal with around perihelion.
Of course, the guys in geosynchronous orbit have their best power time at equinox…
January 4th, 2009 at 11:41 am
The closest we can get is the semimajor axis minus the focus distance = 149,598,000 – 2,500,000 = (roughly) 147,000,000 km (9.13 million miles).
Surely you mean 91.3 million? Those tricksy decimal points…
January 4th, 2009 at 12:12 pm
D’oh! Stupid decimal point. Who needs ‘em!
Thanks, I fixed it.
January 4th, 2009 at 12:25 pm
Oh, sure, but the judge won’t buy it when I call my my deviance an eccentricity.
January 4th, 2009 at 12:33 pm
Aw, man…I forgot to get my wife a Happy Perihelion card!
January 4th, 2009 at 12:47 pm
Is it not true that the center of mass of the earth-moon system is closest to the sun at the time predicted from the motion of that c.m. in the orbit, so the exact moment of the closest approach of the center of the earth to the sun depends on the phase of the moon?
I think Isaac Asimov wrote a column about this for Fantasy and Science Fiction, a few years before his death.
REH
January 4th, 2009 at 12:55 pm
Doesn’t an eccentricity of 1 describe a line?
January 4th, 2009 at 12:59 pm
[...] grootte van de Zon aan de hemel moet je m’n blog van een jaar geleden even lezen. Bron: Bad Astronomy + Sterrengids 2009. Noot:Eén A.E. is de gemiddelde afstand tussen de aarde en de zon. De precieze [...]
January 4th, 2009 at 1:06 pm
[...] the 20-degree temperatures being recorded outside my door right now, but today the Earth is at the closest point to the sun in its orbit. (I don’t link to Bad Astronomy often enough, but it goes without saying that you should be [...]
January 4th, 2009 at 1:08 pm
@Tercel: or a parabola. It just means one axis is infinite (or infinitely small) compared to the other.
An easy calculation is that the distance from the Earth to the Sun varies by a factor of twice the eccentricity. So that’s 2 × 0.0167 = 0.0334 (or 3.34%).
January 4th, 2009 at 3:19 pm
Your second illustration is not quite accurate. The closest point on the ellipse to the focus should be at end of the major axis; in the illustration, the focus should be much farther to the right.
Of course, geometrically, an ellipse has *two* foci, both along the major axis. But in a planetary orbit, only one of them is physically significant.
January 4th, 2009 at 4:08 pm
I do have a question about this. Does “perihelion” only refer to the earth’s orbit around the sun? Does helion=helios=sun? Or do we talk about perihelion with other orbit’s, say the Moon around the Earth?
January 4th, 2009 at 4:39 pm
Does this make the southern hemisphere summer hotter than the north’s?
January 4th, 2009 at 4:43 pm
Keith, the drawing is an illustration, and not meant to be literal.
OtherRob: any object’s closest point to the Sun is perihelion. Closest to Earth is perigee, and the general term is periapsis.
Chris: It should, but the southern hemisphere has lots of water, mitigating the temperature extremes.
January 4th, 2009 at 5:15 pm
I’m curious: IIRC from my math classes, there are TWO focii in a ellipse, not just one. The Sun is one — what’s at the other? And since there’s nothing, how does that affect the orbit?
Or is it because the Sun is moving too? That if the Sun was utterly at rest we would be orbiting in a circle? What would it take for an orbit to be a perfect circle?
January 4th, 2009 at 5:31 pm
@OtherRob,
For something orbiting the earth, we would normaly use perigee and apogee.
Anyone care to give examples for things orbiting the moon (aposelene?) or mars? or proxima-centauri?
January 4th, 2009 at 6:15 pm
Chris: “Does this make the southern hemisphere summer hotter than the north’s?”
Phil Plait: “It should, but the southern hemisphere has lots of water, mitigating the temperature extremes.”
However, in 10,000 years or so, due to precession of the Earth’s axis with respect to inertial space, it will be the opposite way around as this image illustrates:
Effects of axial precession on the seasons.
Because of gravitational disturbances by the other planets, the shape and orientation of Earth’s orbit are not fixed, and the apsides (that is, perihelion and aphelion) slowly move with respect to a fixed frame of reference, i.e., the Earth’s argument of periapsis slowly shifts. Therefore, the anomalistic year is slightly longer than the sidereal year. It takes about 112,000 years for the ellipse to revolve once relative to the fixed stars.
Because the anomalistic year is longer than the sidereal year, while the tropical year (which calendars attempt to track) is shorter, the two forms of precession add. It takes about 21,000 years for the ellipse to revolve once relative to the vernal equinox; that is, for the perihelion to return to the same date — assuming that one has a calendar that tracks the seasons perfectly! The dates of perihelion and of aphelion advance each year on this cycle, an average of 1 day per 58 years.
This interaction between the anomalistic and tropical cycle is important in the long-term climate variations on Earth, called the Milankovitch cycles.
January 4th, 2009 at 6:15 pm
In rounder numbers, for Earth (apparently 2000),
e = 0.016710219, a = 149597887.5 km, b = 149576999.8 km, f = 2499815.05185 km
max = 152,097,701 km
min = 147,098,074 km
GM(sun)= 1.32712440017987 x 10^20 m^3/sec^(-2)
The orbital radii slightly vary from year to year especially given the Earth-moon barycenter (and no Nordtvedt effect). Megayear excursions are much larger.
http://astrobiology.ucla.edu/OTHER/SSO/
January 4th, 2009 at 6:55 pm
In other news, Spirit and Opportunity have now been operating on Mars’ surface for 5 YEARS. Not bad when you consider they were designed to last at most 90 days…
Also, Dr. BA, have you been keeping an eye on the radical increase in seismic activity in the Yellowstone Caldera? I’m sort of afraid I won’t ever get to show my wife the amazing beauty of Yellowstone, if this keeps up. After all, as you are no doubt aware, the Yellowstone Caldera is overdue for a major explosion (it’s had at least 5, spaced about 65,000 years apart – the last one 70,000 years ago (cue dramatic music)). IF, and that’s a huge ‘if’ Yellowstone were to blow, with the force it has in the past, Boulder might not be far enough for safety (think Brazil). Yellowstone is the main reason we have such bumper wheat, corn, and other crops in the Midwest, it’s buried under volcanic ash – several yards deep in most places. Unless the magma bubble has moved (or subsided), or is moving (or subsiding), Yellowstone WILL blow again, tomorrow, or 25,000 years from tomorrow – sort of a geologic version of Death From the Skies, eh???
January 4th, 2009 at 6:58 pm
Thanks for the explanation, Phil and Nick. I learned something today. Which, of course, is the whole point of this site.
January 4th, 2009 at 7:20 pm
Nick s:
Ask and you shall receive:
(I hope this works!)
January 4th, 2009 at 8:21 pm
Periapsis and apapsis are general terms applicable for any type of celestial body.
January 4th, 2009 at 8:28 pm
Ivan:
Excellent additions to the comments! Keep ‘em coming!
I’m wondering if the farthest approach to a good refractor is Apochromatic…
January 4th, 2009 at 8:41 pm
Hey, is there a higher resolution of that picture? It’s totally stellar! It would make a good wallpaper!
January 4th, 2009 at 9:04 pm
IVAN3MAN:
wow – thanks.
January 4th, 2009 at 9:17 pm
@kryth:-
Indeed – a while ago I went on a spree making wallpapers of the Sun, here’s one I did of that same pic: http://fc43.deviantart.com/fs11/i/2006/243/7/9/Tribute_2_by_eeron.jpg
There’s more where that came from
Cheers, Ian
January 4th, 2009 at 10:23 pm
I enjoyed this post – “have you noticed the sun looks bigger” – hahaha!
I think this is a valuable point though – people often hear things like “the sun is closer” (perihelion), “the sun appears largest” (also a consequence of perihelion), “the sun is like a squashed sphere” and even “the earth is like a squashed sphere”. As a chemist who’s spent all his working life building (or redisigning/rebuilding/refurbishing) scientific instruments, I had the opportunity to work for a short period at a solar observatory (or should I say ex-observatory – there have been no publications from that joint in decades).
Anyway “every one knows …” that the earth is closest to the sun at perihelion – but what does that mean? It’s good to have someone point out to the general public that the apparent change in size of the sun is so small that you need to measure it with care. I recall that a solar telescope I worked with had a set of occulting cones for coronography. There were three cones of different sizes and the operator selected the best sized cone for obscuring the solar disc at that particular time of the year. Most people wouldn’t even be able to see that the cones were different sizes unless they put a caliper to each one.
Now for the sun not being quite a sphere – I’ve seen many hundreds of photos of the sun and they all look round to me. Once again, very careful measurements will reveal that the (projection of the) sun is elliptical. A haphazard measurment would indicate that the image of the sun is circular. It helps a little to have a large photographic plate and a reflecting telescope which produces an image approximately 8 inches across.
Of course even then, tilting the plate could distort the image by a larger amount than the sun’s actual distortion! Just some of the more obvious technical challenges when making the sort of measurements we’re talking about.
The same goes for the earth – in fact in the past decade I’ve often heard “the earth is somewhat pear-shaped” in place of “the earth is somewhat like a squashed ball”. And yet, photos of an entire hemisphere taken from space all reveal what seems to be a perfect circle. After all, what’s a few tens of kilometers difference on a ball with a radius of over 6,300 kilometers? Even measuring a roughly 1/300 difference is quite a challenge. (http://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html) Oh, if only the earth were a simple shape – that would make GPS calculations so much easier!
January 4th, 2009 at 10:44 pm
“Blob of Mud? My Home, This Is!”
(Spoken with a Fozzie Bear Voice by a Little Wrinkled Green Guy…)
January 4th, 2009 at 11:27 pm
Richard Drumm, thanks for the complement! I just wanted to chip in my two pennies’ worth!
Nick s, you’re welcome!
kryth, you can find and download many more images of the Sun, including the one above, in high resolution from the Best Of SOHO Gallery — click on my name for the link — and there just click on the thumbnail picture of your choice and then select the resolution required.
January 4th, 2009 at 11:34 pm
Ivan:
Very cool, where did you get this list? I’ve been working on a list like that for years, and I’ve never seen some of those before beyond my own guesswork. Also, I have seen different forms for Venus and Mars (didn’t save my source, though). What I have is:
Venus – Apaphroditia/Periaphroditia
Mars – Apoares (Apares?)/Periares
I think in general your list is easier to pronounce, though.
January 5th, 2009 at 12:06 am
Ivan, you forgot the term for something orbiting Phil Plait:
Periphilion / Apophilion
January 5th, 2009 at 12:24 am
Dr. Phil Plait:
Due to inverse-square law, the changing Earth-Sun distance results in an increase of about 6.9%* in solar energy reaching the Earth at perihelion relative to aphelion.
*Source — Wikipedia: Earth; Axial tilt and seasons (click on my name for the link).
January 5th, 2009 at 12:58 am
@ Wayne,
At my post above featuring Earth’s orbit, if you click on the blue word “apsides”, the link will direct you to the source of my information; there, scroll down to the “Terminology” section.
@ Adrian Lopez,
Heh! Or how about: Periplaithion/Periplaitee/Periplaitron; Apoplaithion/Apoplaitee/Apoplaitron?
January 5th, 2009 at 3:25 am
I always think perigalacticon sounds like something you might hear while watching Transformers…
January 5th, 2009 at 3:38 am
Hi Robert…YES, you’re right in your thinking.
You said: “Is it not true that the center of mass of the earth-moon system is closest to the sun at the time predicted from the motion of that c.m. in the orbit, so the exact moment of the closest approach of the center of the earth to the sun depends on the phase of the moon?”
The central Centre of Gravity (CoG) foci that the Earth and Moon both orbit around suggests that the Earth would be closer to the Sun when during a Full Moon phase (that is, when the Earth would be between the Sun and the Moon). The foci is actually some hundreds of kilometres below the Earth’s surface, and during a Full Moon phase, this wobble-dance effect slightly puts the Earth into a position closer to the Sun.
If this CoG/Full Moon configuration had happened to coincide with the 4th January 2009 perihelion date Phil gives, then, yes, we would have been even closer to the Sun than stated. However, as First Quarter in 2009 (that is, several days before Full Moon) was on the 4th Jan, then, I think, the CoG position of the Moon was between its Min distance and Max distance from the Sun.
Sorry, if it sounds confusing
John
http://www.moonposter.ie (A poster on important aspects about the Moon + FREE Poster)
January 5th, 2009 at 7:03 am
[...] The sun at perihelion [...]
January 5th, 2009 at 10:32 am
[...] Sometimes I learn such awesome stuff on the internets. Today I learned about Perihelions. Did you know (anyone besides Tadd and Wally) that the Earth’s orbit around the Sun is actually elliptical? I didn’t. Yesterday the Earth was at is closest orbital point, to the Sun. I read about it on Gothamist, but the Gothamist article linked to a pretty sweet column called “Bad Astronomy,” hosted on the Discovery site. The article contains diagrams, and formulas, plus extensive nerd dialogue in the comments section. Check it. [...]
January 5th, 2009 at 11:42 am
Dang, the BA gave us an actual equation. Kind of brings tears to my eyes. We’re so proud of you, Phil.
January 5th, 2009 at 11:56 am
@Richard:
Apochromatic is the farthest you can get from (false) COLOR.
January 5th, 2009 at 12:07 pm
Of course, the drawback is that we in the Southern hemisphere have a shorter summer than you northerners.
January 5th, 2009 at 12:38 pm
Jake’s New Blog: “The article contains diagrams, and formulas, plus extensive nerd dialogue in the comments section.”
To whom are you referring to, sir?
January 5th, 2009 at 1:15 pm
Lab Lemming: “Of course, the drawback is that we in the Southern hemisphere have a shorter summer than you northerners.”
Yes, but it’s a long winter for the penguins in Antarctica!
January 5th, 2009 at 7:01 pm
Me: “Hey, that sun photo looks familiar.”
*scrolls down the page*
“Hey, look at the cover of Phil’s book.”
This must be the exact image used for the cover of DFS, except mirrored top to bottom (the solar flare at bottom left moves to top left of cover).
January 5th, 2009 at 7:22 pm
I also agree that Phil’s ellipse diagram is misleading since the location of the focus is WAY off, so it appears that there could be closer approaches than periapsis (I was going to say perihelion, but clearly this is not the Earth-Sun system, unless the exaggerated eccentricity is due to projection to a viewing plane other than top down, in which case . . .).
I also would have appreciated a mention of the Earth moving fastest (relative to the sun, not some galactic or exogalactic inertial frame, which would be an entirely different discussion) at perihelion, which of course does affect length of seasons.
Because I want to make sure that this post qualifies as ‘nerd dialog’, I would like to point out that while the term ‘extragalactic’ might be considered the correct term, I used ‘exogalactic’ because ‘extra-‘ comes from Latin while both ‘exo-‘ and ‘galactic’ come from Greek. Also, it just sounds better.
January 5th, 2009 at 10:52 pm
Yes, I can confirm this.
I see the sun from my office window, where at noon it shines just above the building across the street. So basically I just see the upper part of the sun. And this parts of the sun have been growing bigger since the third of Januarry.
January 6th, 2009 at 6:50 am
Invader Xan Says:
“January 5th, 2009 at 3:25 am” (Does he what a weird thing to say! Oh well, Jan5th back atchya!)
“I always think perigalacticon sounds like something you might hear while watching Transformers…”
Peri Galacticon?
Isn’t she the new Dr Who companion?
Thinks back to Peri Perpugilliam Brown of the .. pleasing proportions .. with the last Peter Davidson and first Colin Baker Who eps.
Mmmmm … Peri! (Think Homer Simpson drool!)
January 6th, 2009 at 6:55 am
Quoteth the BA :
“If you’ve been staring at the Sun lately, then you may have noticed it looks a wee bit bigger today than it did a few days ago.”
Or you may notice you can’t see anything too well anymore!
NOT recomended Phil, NOT recommended at all.
It sent Galileo blind remember!
Unless you have a good solar filter anyway – saw the Sun in H-alpha filter through a friend’s telescope earlier this year – amazing how red it looks!
January 6th, 2009 at 7:00 am
Bein’Silly, apparently Galileo went blind in old age from cataracts and glaucoma not related to his sun observations in earlier years.
January 6th, 2009 at 11:25 am
Guys, during the Apollo missions, I seem to remember the NASA talking heads referring to “perilune”.
January 6th, 2009 at 2:42 pm
[...] I totally forgot that yesterday was the Perihelion, which is the day of the year when the Earth is closest to the Sun (peri meaning [...]
January 25th, 2009 at 2:32 pm
[...] away, so it doesn’t cover the Sun completely. In this case, the eclipse is happening near perihelion, when the Earth is closest to the Sun, so we’re maximizing the effect: the Moon looks small [...]
January 26th, 2009 at 6:50 am
Kepler (demolish) Vs Einstein’s
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’ = (-720×36526/T) {[√ (1-ε²)]/(1-ε)²} (v/c)² degrees/100 years
Δ w” = (-720×3600/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<
G=6.673×10^-11; M=2×10^30kg; m=.32×10^24kg
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?
January 28th, 2009 at 5:51 am
Einstein’s Nemesis: DI Her Eclipsing Binary Stars Solution
The problem that the 100,000 PHD Einsteinists 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 γ + 2m’v +m”r; γ = acceleration; m” = mass acceleration rate
In polar coordinates system
r = r r(1) ;v = r’ r(1) + r θ’ θ(1) ; γ = (r” – rθ’²)r(1) + (2r’θ’ + rθ”)θ(1)
F = m[(r"-rθ'²)r(1) + (2r'θ' + rθ")θ(1)] + 2m’[r'r(1) + rθ'θ(1)] + (m”r) r(1)
F = [d²(m r)/dt² - (m r)θ'²]r(1) + (1/mr)[d(m²r²θ')/d t]θ(1) = [-GmM/r²]r(1)
d² (m r)/dt² – (m r) θ’² = -GmM/r²; d (m²r²θ’)/d t = 0
Let m =constant: M=constant
d²r/dt² – r θ’²=-GM/r² —— I
d(r²θ’)/d t = 0 —————–II
r²θ’=h = constant ————– II
r = 1/u; r’ = -u’/u² = – r²u’ = – r²θ’(d u/d θ) = -h (d u/d θ)
d (r²θ’)/d t = 2rr’θ’ + r²θ” = 0 r” = – h d/d t (du/d θ) = – h θ’(d²u/d θ²) = – (h²/r²)(d²u/dθ²)
[- (h²/r²) (d²u/dθ²)] – r [(h/r²)²] = -GM/r²
2(r’/r) = – (θ”/θ’) = 2[λ + ỉ ω (t)] – h²u² (d²u/dθ²) – h²u³ = -GMu²
d²u/dθ² + u = GM/h²
r(θ, t) = r (θ, 0) Exp [λ + ỉ ω (t)] u(θ,0) = GM/h² + Acosθ; r (θ, 0) = 1/(GM/h² + Acosθ)
r ( θ, 0) = h²/GM/[1 + (Ah²/Gm)cosθ]
r(θ,0) = a(1-ε²)/(1+εcosθ) ; h²/GM = a(1-ε²); ε = Ah²/GM
r(0,t)= Exp[λ(r) + ỉ ω (r)]t; Exp = Exponential
r = r(θ , t)=r(θ,0)r(0,t)=[a(1-ε²)/(1+εcosθ)]{Exp[λ(r) + ì ω(r)]t} Nahhas’ Solution
If λ(r) ≈ 0; then:
r (θ, t) = [(1-ε²)/(1+εcosθ)]{Exp[ỉ ω(r)t]
θ’(r, t) = θ’[r(θ,0), 0] Exp{-2ỉ[ω(r)t]}
h = 2π a b/T; b=a√ (1-ε²); a = mean distance value; ε = eccentricity
h = 2πa²√ (1-ε²); r (0, 0) = a (1-ε)
θ’ (0,0) = h/r²(0,0) = 2π[√(1-ε²)]/T(1-ε)²
θ’ (0,t) = θ’(0,0)Exp(-2ỉwt)={2π[√(1-ε²)]/T(1-ε)²} Exp (-2iwt)
θ’(0,t) = θ’(0,0) [cosine 2(wt) - ỉ sine 2(wt)] = θ’(0,0) [1- 2sine² (wt) - ỉ sin 2(wt)]
θ’(0,t) = θ’(0,t)(x) + θ’(0,t)(y); θ’(0,t)(x) = θ’(0,0)[ 1- 2sine² (wt)]
θ’(0,t)(x) – θ’(0,0) = – 2θ’(0,0)sine²(wt) = – 2θ’(0,0)(v/c)² v/c=sine wt; c=light speed
Δ θ’ = [θ'(0, t) - θ'(0, 0)] = -4π {[√ (1-ε) ²]/T (1-ε) ²} (v/c) ²} radians/second
{(180/π=degrees) x (36526=century)
Δ θ’ = [-720x36526/ T (days)] {[√ (1-ε) ²]/ (1-ε) 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: 2πa (1 – ε²/4 + 3/16(ε²)²—) ≈ 2πa (1-ε²/4); R =a (1-ε²/4)
v (m) = √ [GM²/ (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; 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-ε²/4)]
v (cm) = [m v(m) + M v(M)]/(m + M) All rights reserved. joenahhas1958@yahoo.com
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