The fate of the Earth?
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From: palsing on 12 Sep 2009 20:30 On Sep 12, 3:19 am, oriel36 <kelleher.ger...(a)gmail.com> wrote: > "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 I give up, how does this translate to only 70 AU? Again you show you ignorance of really simple math. 10.7 billion miles divided by 93 million miles per AU results in about 115 AU as an answer, in good agreement with other sources. Tell us, how did you arrive at only 70 AU? Every time you open your mouth you again demonstrate just how ignorant you are concerning just about anything scientific ... as always, you are totally alone with your theories, and remain completely unteachable.
From: Peter Webb on 12 Sep 2009 21:47 "dow" <williamsdavid65(a)gmail.com> wrote in message news:88b4d30b-b67d-44c1-8230-ded0d6fc2835(a)k26g2000vbp.googlegroups.com... On Sep 11, 10:16 am, dlzc <dl...(a)cox.net> wrote: > Dear dow: > > On Sep 10, 9:09 pm, dow <williamsdavi...(a)gmail.com> wrote: > > > > > > > On Sep 10, 8:44 am, Yousuf Khan <bbb...(a)yahoo.com> wrote: > > > > dow wrote: > > > > It is now thought that the sun probably will not > > > > engulf the earth and moon when it becomes a > > > > red giant. The loss of solar mass to the solar > > > > wind will reduce the maximum radius of the > > > > sun and will also cause the earth's orbit to > > > > spiral outward. Both effects will reduce the > > > > probability of the earth beig engulfed. > > > > The solar mass loss due to the solar winds > > > probably won't be significant *until* the Sun goes > > > red giant. Current solar winds don't even represent > > > a fraction of a percent of total solar mass over the > > > billions of years of Sun's existence. > > > The process of going red giant won't be > > instantaneous. > > We hope. We've not seen it happen that way elsewhere... > > > The sun will swell over a period of millions of years. > > The process will be accompanied by an increase > > in the rate of loss of material to the solar wind. > > With less energy driving the efflux, are you sure? Have we seen other > red giants spewing contents? Hard to see, but theory says it must be so. > > > By the time the sun is getting close to engulfing > > the earth, a lot of mass will have been lost, > > reducing the maximum radius of the sun, and > > also causing the earth's orbit to have spiralled > > outward. > > Er, no. Earth's orbit, neglecting other factors, will become more > elliptical, moving both further from our current average distance, and > *closer*. We have the "wrong" angular momentum for this distance and > a less massive Sun, to "spiral out". Er, no. See my last message. Are you just guessing, or do you have calculations to back up what you say? Lots of people read my Orrery articles, and nobody showed that I was wrong. ___________________________ I know you are wrong. If the Sun suddenly lost 90% of it mass, then the Earth would move into a highly elliptical or hyperbolic orbit, with its minimum distance from the Sun 1 AU and a velocity at that distance the same as it is now. This is conservation of kinetic energy plus angular momentum. Therefore the ellipticity of an orbit is not preserved if the body being orbitted loses mass. If nobody pointed out this simple thought experiment proves your simulations are incorrect, its probably because nobody read your article.
From: oriel36 on 13 Sep 2009 07:53 On Sep 12, 10:23 pm, Sam Wormley <sworml...(a)mchsi.com> wrote: > oriel36 wrote: > > On Sep 12, 3:54 pm, Sam Wormley <sworml...(a)mchsi.com> wrote: > >> 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. > > > The AU is not only a discrete distance,it is also a ratio of one > > planetary  radius to one orbital circumference hence it all operates > > against the background of the Pi proportion. > >   No Gerald -- > No what !,this is corruption on a scale the world has not seen before and would be considered treason in any other sphere of existence.The fact that I am considered a madman for demonstrating how planetary dimensions and rotational characteristics are organised around daily rotation once in 24 hours only highlights just how desperate the situation actually is. The 4 listed forums in this thread must contain at least one sane person who has a grasp of the consequences which arise from distorting basic planetary facts or geometric proportions to suit a mantra spouting empirical cult but I no longer consider it a matter of contention,as there is none,but simply a standard by which all disciplines involved in celestial/terrestrial phenomena can move forward. >   In 1976, the International Astronomical Union revised the definition >   of the AU (a unit of distance) for greater precision, Yeah,just like they tried to 'define' a planet and ended up with a mess and now rely on the usual dull idea that people will forget the stupidity and accept 8 planets,for me it is just a visible fraction of corruption.The same with 'defining' a second out of astronomical context and the mess that created and now this attempt to tamper with the Pi proportion. The ratio of 1;1 represents the proportion of the Earth's orbital radius with its circumference,if you want to move the goalposts on this fundamental gauge then be my guest but all planetary orbits using the AU will be determined from a purely geometric correlation just as Kepler's dictated 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 The error using Kepler's reckoning becomes telescoped with distance from the Sun or any star and that is why the AU values for Fomalhaut b are wildly out of sync with orbital periods,firstly because Kepler's correlation does not work and secondly,while using Earth's orbital dimensions as a gauge is perfectly fine,the orbital circumference must match the orbital radius and the AU value assigned to it. defining it as >   that length for which the Gaussian gravitational constant (k) takes >   the value 0.01720209895 when the units of measurement are the >   astronomical units of length, mass and time. > >   An equivalent definition is the radius of an unperturbed circular >   Newtonian orbit about the Sun of a particle having infinitesimal mass, >   moving with a mean motion of 0.01720209895 radians per day, or >   that length for which the heliocentric gravitational constant (the >   product GMâ) is equal to (0.01720209895)^2 AU^3/d^2. > I am sure this impresses or scares the hell out of some people but then again,it comes from the same crowd who have created a trainwreck out of every single discipline and insight that was exciting and productive. >   It is approximately equal to the mean EarthâSun distance and is a >   convenient distance unit in astronomy for planetary systems. > >   It is NOT a ratio! > Scream all you like,1;1 is a ratio and the geometric proportions of orbital radius;orbital circumference of 1;1 is based on the Earth's orbital radius on one side and its orbital circumference/period represents the other half of the ratio.This is why a 115 AU value for Fomalhaut be will transfer to an orbital circumference that is 115 times that of the Earth with a corresponding orbital period and that value will be much less than 872 years.I am not going to hold you hand and take you through the full explanation ,as far as I am concerned,the opportunity is there to start from scratch and work through the orbital characteristics of the Fomahaut system with ease. > Appealing to the status of > > > Fomalhaut as a different solar system in order to invalidate the 1 to > > 1 ratio of planetary radius/orbital circumference using Earth values > > is like trying to tamper with the oldest and most basic correlation of > > all,the proportion between diameter and circumference otherwise known > > as Pi. > >   No Gerald--Fomalhaut has a different mass from the sun. >   Kepler's Third law in this form > >   T^2 = (2Ï)^2 a^3 / G M > >   takes the mass of the parent body into account and is >   accurate for all planetary system. Your untutored self >   has trouble with the concept because you don't understand >   simple algebra. > >   Do som self education, Gerald! > Tell me how long it takes the Earth to turn once and remind me just how desperate the situation is. > > > > > > > > If people were honest and interested,they would inquire as to what > > Isaac was up to in attempting to rope Kepler's reasoning on orbital > > periods and distance from the Sun into his agenda but few would have > > the skills and the background to go through the elaborate mess which > > really has greed at the bottom of it or more likely,are entirely > > content to continue with this destructive charade based on distorting > > astronomy and the genuine insights based on planetary dynamics. > > >>   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. > > > If Jupiter's orbital radius is 5 times that of the Earth,its orbital > > circumference is also 5 times that of the Earth,the same for Saturn > > which is 9.5 times distant from the Sun as the Earth is and also has > > an orbital circumference 9.5 times that of the Earth. > > > I spotted the error in Kepler's reasoning which becomes telescoped as > > the distances from Earth become greater hence this explicit bungling > > with Fomalhaut b with values which range from 70 AU to 115 AU,a range > > that is so large that it is better just to start from scratch and work > > things out by direct observation.The sampling needed to determine > > where Fomalhaut b is in its orbit would take about 20 years or more > > but nothing is going to get done as long as this empirical mess > > continues. > > >>   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. > > > An AU is a discrete orbital radius distance and also a discrete > > orbital circumference, > >   No Gerald--In 1976, the International Astronomical Union revised the >   definition of the AU (a unit of distance) for greater precision, defining >   it as that length for which the Gaussian gravitational constant (k) takes >   the value 0.01720209895 when the units of measurement are the >   astronomical units of length, mass and time. > >   An equivalent definition is the radius of an unperturbed circular >   Newtonian orbit about the Sun of a particle having infinitesimal mass, >   moving with a mean motion of 0.01720209895 radians per day, or >   that length for which the heliocentric gravitational constant (the >   product GMâ) is equal to (0.01720209895)^2 AU^3/d^2. > >   It is approximately equal to the mean EarthâSun distance and is a >   convenient distance unit in astronomy for planetary systems. > >   It is NOT a ratio! > > I have been through all this before > >   Gerald--You keep getting it wrong! Learn some algebra! > > and will not > > > > > do so again,the orbital period calculated in Earth years/orbital > > distance must correspond to a ratio of the Earth's radius otherwise > > you will have problems with the Universal correlation between diameter > > and circumference otherwise known as  Pi. > > >>   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! > > > Repeat this as many times as you like,what looks good at close > > range,and in Kepler's era,things only went as far as Saturn,but his > > correlation between orbital periods and orbital radii does not work.- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: dow on 13 Sep 2009 11:06 On Sep 12, 9:47 pm, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > "dow" <williamsdavi...(a)gmail.com> wrote in message > > news:88b4d30b-b67d-44c1-8230-ded0d6fc2835(a)k26g2000vbp.googlegroups.com... > On Sep 11, 10:16 am, dlzc <dl...(a)cox.net> wrote: > > > > > > > Dear dow: > > > On Sep 10, 9:09 pm, dow <williamsdavi...(a)gmail.com> wrote: > > > > On Sep 10, 8:44 am, Yousuf Khan <bbb...(a)yahoo.com> wrote: > > > > > dow wrote: > > > > > It is now thought that the sun probably will not > > > > > engulf the earth and moon when it becomes a > > > > > red giant. The loss of solar mass to the solar > > > > > wind will reduce the maximum radius of the > > > > > sun and will also cause the earth's orbit to > > > > > spiral outward. Both effects will reduce the > > > > > probability of the earth beig engulfed. > > > > > The solar mass loss due to the solar winds > > > > probably won't be significant *until* the Sun goes > > > > red giant. Current solar winds don't even represent > > > > a fraction of a percent of total solar mass over the > > > > billions of years of Sun's existence. > > > > The process of going red giant won't be > > > instantaneous. > > > We hope. We've not seen it happen that way elsewhere... > > > > The sun will swell over a period of millions of years. > > > The process will be accompanied by an increase > > > in the rate of loss of material to the solar wind. > > > With less energy driving the efflux, are you sure? Have we seen other > > red giants spewing contents? > > Hard to see, but theory says it must be so. > > > > > > By the time the sun is getting close to engulfing > > > the earth, a lot of mass will have been lost, > > > reducing the maximum radius of the sun, and > > > also causing the earth's orbit to have spiralled > > > outward. > > > Er, no. Earth's orbit, neglecting other factors, will become more > > elliptical, moving both further from our current average distance, and > > *closer*. We have the "wrong" angular momentum for this distance and > > a less massive Sun, to "spiral out". > > Er, no. See my last message. Are you just guessing, or do you have > calculations to back up what you say? Lots of people read my Orrery > articles, and nobody showed that I was wrong. > > ___________________________ > I know you are wrong. If the Sun suddenly lost 90% of it mass, then the > Earth would move into a highly elliptical or hyperbolic orbit, with its > minimum distance from the Sun 1 AU and a velocity at that distance the same > as it is now. This is conservation of kinetic energy plus angular momentum. > > Therefore the ellipticity of an orbit is not preserved if the body being > orbitted loses mass. > > If nobody pointed out this simple thought experiment proves your simulations > are incorrect, its probably because nobody read your article.- Hide quoted text - > > - Show quoted text - I said (but you decided to ignore) that the planet will spiral outwad, keeping the eccentricity of its orbit constant, *if the loss of mass from the star is very slow compared with the orbital period of the planet*, so only a tiny fraction of the mass of the star is lost during each revolution of the planet around it. Obviously, a sudden loss of 90% of he star's mass does not satisfy this condition. It's easy to prove that a sudden loss of 50% or more of a star's mass, which can happen in an explosion, will cause planets that were in circular orbits around it to escape. This was in my Orrery articles, which were very thoroughly read and critiqued by many people. The editor of the magazine, Greg Neill, went so far as to re-do the entire calculation, using a completely different method from mine. I did an iterative simulation of the planet's path as the star's mass changes. Greg did an iterative numerical integration of a function that would show any changes in eccentricity. (I don't believe there is any general analytical method to solve the problem. Numerical iteration is always required.) In the end, Greg's conclusion was the same as mine. *Under the conditions I specified above*, the eccentricity of the orbit (measured at the perihelion of each revolution) remains constant. Like several readers who wrote in to say so, I was surprised by this conclusion. Initially, I had expected the eccentricity to change, using vague reasoning such as yours. But, in science, hunches often turn out to be wrong. Load up some programming language and tackle the problem yourself. dow
From: Sam Wormley on 13 Sep 2009 14:45
oriel36 wrote: > > Tell me how long it takes the Earth to turn once and remind me just > how desperate the situation is. > > The earth rotates exactly 360° in 86,164.09+ seconds. It's a direct observation anybody can make and is referred to a sidereal day. In a solar day (86,400 seconds) the earth rotates about 361° and everybody knows this from the ancients to the present. I have to admit, Gerald, that you are in desperate need of further education. |