More than 90 percent of objects found in the vast outer-solar system reservoir may have been born around other stars
in [Physics]
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From: Brad Guth on 13 Jun 2010 22:47 On Jun 13, 6:10 pm, Sam Wormley <sworml...(a)gmail.com> wrote: > On 6/13/10 7:49 PM, Brad Guth wrote: > > > > > On Jun 13, 4:12 pm, palsing<pnals...(a)gmail.com> wrote: > >> On Jun 13, 12:37 pm, Brad Guth<bradg...(a)gmail.com> wrote: > > >>> You do realize Im working with the tidal radii grip that Sirius has > >>> on our solar system, and not the other way around... > > >> This statement alone shows that you do not really understand the > >> science. The escape velocity formula cited by Gregg do not imply a > >> 'direction' ... > > >> \Paul A > > > Perhaps that's his problem. > > > You simply can't place something as massive as the original Sirius > > star/solar system nearby anything else, without that situation causing > > Newtonian interactions. > > > Ever heard of a barycenter? > > > Do you think such a barycenter wasn't created? > > > ~ BG > > F = m1 m2 G / r^2 > > Calculate the force between the Sirius system and the solar system > and let us know what the answer is. > > r = 2.64 ± 0.01 pc > m1 = 1.9891 × 10^30 kg > m2 = 5.9628 × 10^30 kg http://www.1728.com/gravity.htm http://www.wsanford.com/~wsanford/calculators/gravity-calculator.html What's next? ~ BG
From: Greg Neill on 13 Jun 2010 23:02 Brad Guth wrote: > Your politically correct and/or faith-based approved obfuscation and > denial of those pesky Newtonian laws of gravity are noted. Every equation I've shown you has been pure Newtonian physics. That you cannot recognize it is a telling advertisement for your lack of education in the field. > > If not due to gravity and those Newtonian laws, why is 2005-VX3 or > most any other elliptical trekking Oort item still with us? Because they are gravitationally bound. 2005 VX3 has an orbital semimajor axis of 2592 AU, with perihelion of 4.114 AU and aphelion of 2588 AU. That makes its specific mechanical energy: a = 2592 u = 39.474 AU^3/yr^2 (Sun's grav. parameter in convenient units) E = -u/(2*a) = -1.711 x 10^5 J/kg Its orbital speed at aphelion is: v_ap = sqrt(2*(E + u/r)) with r = 2588 AU = 0.586 km/sec Escape velocity at that distance is: v_esc = sqrt(2*u/r) = 0.828 km/sec So its speed at aphelion is less than escape velocity at that distance. As expected. > > Didn�t you notice that nifty ratio of 2005-VX3 being 83e6:1 less bound > than our solar system is to Sirius? What does that mean? What is your measure of "boundedness"? The only practical measure is that of total specific mechanical energy: More negative means more tightly bound (the deeper in a 'gravity well' it is). > > Are you suggesting that long-period elliptical trajectories are either > conditional or bogus? No, but apparently you are. *ALL* bound orbits must have gravitational potential energy which is larger in magnitude than the kinetic energy so that the net mechanical energy of the orbit is negative. That's basic Newtonian physics. > > Do long-period or deep elliptical treks wear out or vanish over time? How does a "trek" wear out? What's a trek in phsyics? > > You do realize I�m working with the tidal radii grip that Sirius has There is no physics meaning to the words "tidal radii grip". > on our solar system, and not the other way around. This would have > been most important as of the beginning stellar mass of 12.5, and of > course worse yet if you�d care to deal with that molecular cloud worth > <12.5e6 Ms that managed to produce such massive and vibrant stars. You're heading off into fairy land again. > > Can�t your public funded simulators do any better job, of adding and/ > or subtracting mass in order to see what happens over various > distances, proper motion trajectories and at different velocities? Researchers are not interested in modelling your fantasies just because you wish it were so. They are more conserned with modelling empirically verifiable scenarios.
From: Greg Neill on 13 Jun 2010 23:05 Brad Guth wrote: > On Jun 13, 6:10 pm, Sam Wormley <sworml...(a)gmail.com> wrote: >> F = m1 m2 G / r^2 >> >> Calculate the force between the Sirius system and the solar system >> and let us know what the answer is. >> >> r = 2.64 � 0.01 pc >> m1 = 1.9891 � 10^30 kg >> m2 = 5.9628 � 10^30 kg > > http://www.1728.com/gravity.htm > http://www.wsanford.com/~wsanford/calculators/gravity-calculator.html > > What's next? You haven't answered yet. Give us your numeric result.
From: Brad Guth on 13 Jun 2010 23:19 On Jun 13, 6:36 pm, palsing <pnals...(a)gmail.com> wrote: > On Jun 13, 5:49 pm, Brad Guth <bradg...(a)gmail.com> wrote: > > > On Jun 13, 4:12 pm, palsing <pnals...(a)gmail.com> wrote: > > > > On Jun 13, 12:37 pm, Brad Guth <bradg...(a)gmail.com> wrote: > > > > > You do realize Im working with the tidal radii grip that Sirius has > > > > on our solar system, and not the other way around... > > > > This statement alone shows that you do not really understand the > > > science. The escape velocity formula cited by Gregg do not imply a > > > 'direction' ... > > > > \Paul A > > > Perhaps that's his problem. > > The problem is all yours because you just don't get it. > > > You simply can't place something as massive as the original Sirius > > star/solar system nearby anything else, without that situation causing > > Newtonian interactions. > > > Ever heard of a barycenter? > > > Do you think such a barycenter wasn't created? > > > ~ BG > > Well, to start with, the original 'massive' Sirius system was nowhere > near the solar system when it was born, so that's a non-starter. What > it 'was' then doesn't affect what it 'is' now, about 3 solar masses. > > Strictly speaking, a barycenter can be defined as a simple center of > mass, so sure, there is technically a barycenter between the Sirius > system and the solar system. But the term barycenter is most commonly > used to describe the center of mass of two or more bodies orbiting > around each other, such as the Earth and the Moon. Since we are not in > orbit with Sirius (and there is absolutely no doubt about this, no > matter how much 'intuitive intelligence' you think you have), the > 'barycenter' between systems could more correctly be referred to as > the center of mass... but who would care about this? Between the solar > system and all other objects in the galaxy there are billions and > billions of such 'barycenters', they are just numbers and, for the > most part, meaningless. > > I'm surprised that you aren't saying similar things about the Alpha > Centauri system. It is about 2 solar masses total, compared to Sirius' > 3 solar masses, but it is only half as far away. Clearly it exerts a > greater gravitational tug on us than does Sirius. What, not as > romantic? > > \Paul A Alpha Centauri didn't evolve so quickly, didn't go nova or otherwise rogue or spit anything out, wasn't nearly as massive in its beginning, nor was it as nearby to start off with. The molecular cloud that created those impressive Sirius stars may have been worth <12.5e6 Ms (more if you'd care to include Sirius C). Are you suggesting that kind of molecular mass is also inconsequential? ~ BG
From: Sam Wormley on 13 Jun 2010 22:39
On 6/13/10 8:10 PM, Sam Wormley wrote: > r = 2.64 � 0.01 pc > m1 = 1.9891 � 10^30 kg > m2 = 5.9628 � 10^30 kg In case your interested, the force on our sun from the earth is five orders of magnitude greater than from the Sirius system. |