The fate of the Earth?
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From: YKhan on 12 Sep 2009 00:43 On Sep 10, 8:31 pm, "N:dlzc D:aol T:com \(dlzc\)" <dl...(a)cox.net> wrote: > Mass loss due to fusion (and then photon escape to infinity) and > solar wind is currently on the order of 1 part in 10^14 per year. > So probably a good guess on your part. Well, it wasn't just a guess, I had asked a similar question in one of these newsgroups a couple of years back, and I remembered the answer was that it's miniscule until the Sun puffs up. It was mentioned that the Sun is losing 4 billion tons of mass per second due to fusion, and that's still nothing! Yousuf Khan
From: oriel36 on 12 Sep 2009 05:43 On Sep 11, 8:09 pm, Sam Wormley <sworml...(a)mchsi.com> wrote: > oriel36 wrote: > > > Depending on the average orbital displacment based on  the Earth year > > and subsequently,Earth's orbital distance for each year Fomalhaut b > > moves,the given ratio of 115 AU would translate into an orbital period > > of 115 Earth years to complete a circuit.That article has Fomalhaut at > > 70 AU so its orbital period would correspond to 70 Earth years or an > > annual orbital displacement of 360 deg/70 or a little over an annual > > motion of 5 degrees. > >   Gerald--You need to: > >   1. Understand Newton's version of Kepler's third law, and the fact >    that it applies all over the cosmos including the solar system. > Kepler never wrote a law,he tried to demonstrate a correlation between the orbital period of a planet and its distance from the Sun and for any reasonably intelligent person it is easy enough to understand how he went about it - "And so if any one take the period, say, of the Earth, which is one year, and the period of Saturn, which is thirty years,and extract the cube roots of this ratio and then square the ensuing ratio by squaring the cube roots, he will have as his numerical products the most just ratio of the distances of the Earth and Saturn from the sun. For the cube root of 1 is 1, and the square of it is 1; and the cube root of 30 is greater than 3, and therefore the square of it is greater than 9. And Saturn, at its mean distance from the sun, is slightly higher than nine times the mean distance of the Earth from the sun." Kepler As it happens,the AU now represents the ration between 1 Earth radius and 1 Earth orbital circumference so that whatever AU is assigned for any planet in our solar system or any other should match the orbital circumference compared to that of the Earth. >   2. Get past the archaic thinking of the 17th century. > The job of genuine investigators,and sadly I have yet to meet one,is to indeed get past the late 17th century obfuscation and use modern imaging to light up the means by which the original astronomers went about things and far from being a dull and tedious affair,their approaches are lively and within the scope of everyone with an interest in how planetary dynamics and terrestrial sciences mesh.If they are scared of mathematicians who have done a great job and screwing up everything then they deserve what they get,as far as I am concerned ,the normal faculty of geometric judgments with an eye for physical considerations is the only thing necessary. So have you get this Sam,at first appearances Kepler looks right but at great distances from the Sun the cracks begin to appear and then they become fully fledged canyons,people see this when a value of 70 AU is assigned for Fomalhaut b in one major website (APOD) and 115 AU in another (Wikipedia). >   3. Learn some basic mathematics, such as algebra, geometry, >    trigonometry and the calculus. > Mathematicians or rather empiricists have had a great innings,3 centuries worth,but now must yield to the power of modern imaging and occupy themselves with other things that do not involve planetary dynamics,structural astronomy or the cause/effects which link astronomy with terrestrial sciences. It is truly amazing to live among people who positively refuse to believe the Earth rotates once in 24 hours and opt for a 'sidereal time' value but there is a catastrophic loss of information in terms of planetary dynamics by reasoning out a calendar/clockwork solar system by mangling the antecedent principles which create a 24 hour value in concordance with planetary geometry and rotational characteristics.How is it possible,with the history of longitude and watches in full view,to believe people who consider themselves intelligent to go along with junk that only people with no self respect would follow,I am not talking about the latest/greatest fads,but basic,basic planetary information. >  -Sam > >  >>   Fomalhaut's mass is about 2 solar masses. >  >>   Fomalhaut-b's orbital RADIUS is 115 astronomical units, Gerald.  a = 115 AU. >  >>   Fomalhaut-b's orbital PERIOD is 872 years, Gerald.  T = 872 years. >  >> >  >>   And using Kepler's third law, we get, >  >> >  >>    T^2 = (2Ï)^2 a^3 / G M >  >>   (872 yr)^2 = (2Ï)^2 (115 AU )^3 / G (3.978 à 10^30 kg) >  >> >  >>   Kepler's third law works beautifully for the Fomalhaut system! The >  >>   observations agree beautifully with Kepler's law of Harmony! >  >
From: oriel36 on 12 Sep 2009 06:19 On Sep 11, 10:26 pm, dow <williamsdavi...(a)gmail.com> wrote: > On Sep 11, 1:36 pm, oriel36 <kelleher.ger...(a)gmail.com> wrote: > > > > > > > On Sep 11, 4:25 pm, dow <williamsdavi...(a)gmail.com> wrote: > > > > I published this stuff in a magazine called The Orrery, which is now > > > defunct. Maybe some people here remember it, and even have back > > > issues. > > > The Orrery is an appropriate name for Newton's non functional > > treatment of planetary motion .You must not have got the memo that > > Kepler's correlation between orbital periods and distance from the > > Sun,by using the cube root of the orbital period and then squaring > > it,does not work ,this is not an opinion but a 100% observational and > > geometric certainty - > > >http://www.phys.ncku.edu.tw/~astrolab/mirrors/apod_e/ap081114.html > > > Depending on the average orbital displacment based on the Earth year > > and subsequently,Earth's orbital distance for each year Fomalhaut b > > moves,the given ratio of 115 AU would translate into an orbital period > > of 115 Earth years to complete a circuit.That article has Fomalhaut at > > 70 AU so its orbital period would correspond to 70 Earth years or an > > annual orbital displacement of 360 deg/70 or a little over an annual > > motion of 5 degrees. > > What are you babbling about? In our solar system, with all the planets > revolving around the sun, the orbital periods of the planets are very > accurately proportional to the 3/2 powers of their distances from the > sun. However, Formalhaut is not the sun. It is more massive than the > sun, so its gravity is stronger, making its planet move faster than a > planet the same distance from our sun would move. Here is what you do,come back to me with an orbital period for Fomalhaut b and a corresponding AU value. http://en.wikipedia.org/wiki/Fomalhaut_b http://www.phys.ncku.edu.tw/~astrolab/mirrors/apod_e/ap081114.html By Kepler's reckoning a 872 year orbital period has a cube root of 9.55 and this value squared is 90.75 AU which is a long way from the APOD value given as 70 AU or the Wikipedia value of 115 AU. You can apply your own silly reasoning to it anyway,as far as I can see the only perverted pleasure any of you get is saying anything without the slightest restraint or objection as I watch as they scramble to cook the books,the latest APOD has it as 23 times the Sun/ Jupiter distance - "Astronomers now identify, the tiny point of light in the small box at the right as a planet about 3 times the mass of Jupiter orbiting 10.7 billion miles from the star (almost 14 times the Sun-Jupiter distance)." http://www.phys.ncku.edu.tw/~astrolab/mirrors/apod_e/ap081114.html "Astronomers now identify, the tiny point of light in the small box at the right as a planet about 3 times the mass of Jupiter orbiting 10.7 billion miles from the star (almost 23 times the Sun-Jupiter distance). " http://apod.nasa.gov/apod/ap081114.html You can't transfer > Kepler's laws from one solar system to another without taking into > account the different masses of their stars. Anyone with even a > rudimentary knowledge of orbital dynamics would know that. > The operative word is rude rather than rudimentary,if I could get supposedly intelligent people to determine basic planetary facts and the Pi proportion which links in with planetary radius and orbital comparisons I would consider it a major miracle but I cannot.I only see scrambling to adjust things which I myself jusrt accept with ease,the floudering around with the AU value for the Fomalhaut system planet is just one instance among many. > The word "orrery" originally meant a clockwork model of the solar > system. The magazine of the same name was devoted to computer models > of astronomical systems. > Ah,you are a newbie and hardly know much about Ra/Dec observing and those guys who think if they magnify an object it makes them an astronomer.You are probably a mathematician lost in a jungle created by Isaac and happy enough with your lot but basically the system you follow is indeed a clockwork one but also a calendar based one.Isaac may have told you it is 365 days 5 hours 49 minutes but the 'predictive' nature of Ra/Dec observing,such as determining what day an eclipse occurs,is based on the equable 365/366 day calendar system.
From: Sam Wormley on 12 Sep 2009 10:52 oriel36 wrote: > On Sep 11, 8:09 pm, Sam Wormley <sworml...(a)mchsi.com> wrote: >> oriel36 wrote: >> >>> Depending on the average orbital displacment based on the Earth year >>> and subsequently,Earth's orbital distance for each year Fomalhaut b >>> moves,the given ratio of 115 AU would translate into an orbital period >>> of 115 Earth years to complete a circuit.That article has Fomalhaut at >>> 70 AU so its orbital period would correspond to 70 Earth years or an >>> annual orbital displacement of 360 deg/70 or a little over an annual >>> motion of 5 degrees. >> Gerald--You need to: >> >> 1. Understand Newton's version of Kepler's third law, and the fact >> that it applies all over the cosmos including the solar system. >> > > Kepler never wrote a law,he tried to demonstrate a correlation between > the orbital period of a planet and its distance from the Sun and for > any reasonably intelligent person it is easy enough to understand how > he went about it - No excuse, Gerald--You need to learn algebra and understand Newton's version of Kepler's third law, and the fact that it applies all over the cosmos including the solar system. Fomalhaut's mass is about 2 solar masses. Fomalhaut-b's orbital RADIUS is 115 astronomical units, Gerald. a = 115 AU. Fomalhaut-b's orbital PERIOD is 872 years, Gerald. T = 872 years. And using Kepler's third law, we get, T^2 = (2Ï)^2 a^3 / G M (872 yr)^2 = (2Ï)^2 (115 AU )^3 / G (3.978 Ã 10^30 kg) Kepler's third law works beautifully for the Fomalhaut system! The observations agree beautifully with Kepler's law of Harmony!
From: Sam Wormley on 12 Sep 2009 10:54
oriel36 wrote: > By Kepler's reckoning a 872 year orbital period has a cube root of > 9.55 and this value squared is 90.75 AU which is a long way from the > APOD value given as 70 AU or the Wikipedia value of 115 AU. > Kepler's reckoning applies ONLY to our solar system, Gerald. Gerald--You need to learn algebra and understand Newton's version of Kepler's third law, and the fact that it applies all over the cosmos including the solar system. Fomalhaut's mass is about 2 solar masses. Fomalhaut-b's orbital RADIUS is 115 astronomical units, Gerald. a = 115 AU. Fomalhaut-b's orbital PERIOD is 872 years, Gerald. T = 872 years. And using Kepler's third law, we get, T^2 = (2Ï)^2 a^3 / G M (872 yr)^2 = (2Ï)^2 (115 AU )^3 / G (3.978 Ã 10^30 kg) Kepler's third law works beautifully for the Fomalhaut system! The observations agree beautifully with Kepler's law of Harmony! |