The fate of the Earth?
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From: dow on 11 Sep 2009 11:25 On Sep 11, 12:13 am, Sam Wormley <sworml...(a)mchsi.com> wrote: > dow wrote: > > > The process of going red giant won't be instantaneous. 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. 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. > > > A simple calculation based on conservation of angular momentum shows > > that the radius of the orbit will be inversely proportional to the > > sun's remaining mass, so if 25 percent of the sun's present mass is > > lost, which is plausible, then the earth's orbital radius will be 4/3 > > of what it is now, which would probably be enough to keep the earth > > from being engulfed. > > > dow > > You need to show your calculation, and the assumed mass loss as > a function of time and radius.- Hide quoted text - > > - Show quoted text - I assumed you could do it. Okay... G = gravitational constant M = mass of star (sun) m = mass of planet (earth) (m << M) R = radius of planet's orbit (assumed circular) A = orbital angjular momentum of planet w = orbital angular velocity of planet A = m.w.(R^2) (standard formula for angular momentum) rearranged: w = A/(m.(R^2)) ... Equation 1 G.M/(R^2) = R.(w^2) (gravitational accel'n = centripetal accel'n ) rearranged: w^2 = G.M/(R^3) ...... Equation 2 Taking w from Equ'n 1 and substituting in Equ'n 2: (A^2)/((m^2),(R^4)) = G.M/(R^3) rearranged: (A^2)/(G.(m^2)) = M.R The left side of this last equation consists only of constants (assuming A to be constant), so we can substitute the whole thing with a single constant, K: K = M.R or: R = K/M So the orbital radius is inversely proportional to the star's mass. This assumes that the orbit is always circular. Back in 2003, I ran some simulations of what would happen if a planet were in an elliptical orbit around a star that is slowly (compared with the orbital period) losing mass. The result was that the eccentricity of the orbit would not change, but its dimensions would be inversely proportional to the mass of the star. This was true no matter whether the loss of mass was at a constant rate or a non- constant one, provided that the rate did not have a periodic component that was synchronized with the motion of the planet. If there is such a periodic component, the eccentricity of the orbit changes. 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. dow
From: dow on 11 Sep 2009 11:37 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. > > > A simple calculation based on conservation of > > angular momentum shows that the radius of > > the orbit will be inversely proportional to the > > sun's remaining mass, > > *average* radius. > > > so if 25 percent of the sun's present mass is > > lost, which is plausible, then the earth's orbital > > radius will be 4/3 of what it is now, which would > > probably be enough to keep the earth from > > being engulfed. > > Likely not. What is more, if "half" the sky is a heat sink at 3000K > or so, we are toast anyway. Absolutely. I did say that the planet's surface would be drastically affected. We are just surface mould. > > David A. Smith- Hide quoted text - > > - Show quoted text -
From: oriel36 on 11 Sep 2009 13:36 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. Newton's agenda was the work of a greedy man for while his contemporaries and those before him were discussing the causes of tides in tandem with planetary dynamics,he fudged the data to make it fit the calendar driven Orrery based on the predictive nature of Ra/ Dec.It is a great injustice even if you know nothing of the elaborates scheme at the expense of astronomy and most of terrestrial sciences so you and the others can engage in useless speculation instead of dealing with real issues as people once did,don't take my word for it,try the work 20 years before Newton conjured up that mess known as the 'principia'. Again,the following excellent letter of Wallis is an example of how men once reasoned and speculated and I pray to God that people will eventually do so again - http://books.google.com/books?id=RyBOsLIi2SMC&pg=PA219&dq=aequation+dayes#v=onepage&q=aequation%20dayes&f=false Telling men to grow up is too insulting,all the same,some should take time out to set aside equations for a while and get to know what the great astronomers thought,how they sometimes made mistakes such as Kepler's correlation whaich became the 'inverse square law' and how modern observations and techniques overcome these obstructive bottlenecks. > dow- Hide quoted text - > > - Show quoted text -
From: Sam Wormley on 11 Sep 2009 15:09 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. 2. Get past the archaic thinking of the 17th century. 3. Learn some basic mathematics, such as algebra, geometry, trigonometry and the calculus. -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 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: dow on 11 Sep 2009 17:26
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. 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 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. (Further drivel snipped) dow |