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: Sam Wormley on 13 Jun 2010 22:43 On 6/13/10 8:10 PM, Sam Wormley 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 I�m 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 In case your interested, the force on our sun from the earth is five orders of magnitude greater than from the Sirius system.
From: Sam Wormley on 13 Jun 2010 22:56 On 6/13/10 9:47 PM, Brad Guth wrote: > 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 I�m 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 Calculate the force on the sun from the earth and the Sirius system and compare.
From: palsing on 13 Jun 2010 23:27 On Jun 13, 8:19 pm, Brad Guth <bradg...(a)gmail.com> wrote: > 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? Since it was so far away at that time, yes, essentially inconsequential. Kinda like your theories... \Paul A
From: Greg Neill on 14 Jun 2010 09:55 Greg Neill wrote: > 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: Oops. Make that an orbital semimajor axis of 1296 AU, (4.114 + 2588)/2. The major axis is the sum of the peri- helion and aphelion distances, and the semimajor axis is half of that. > > 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 = -3.422 x 10^5 J/kg so it is even more "tightly bound" than I'd first calculated. > > Its orbital speed at aphelion is: > > v_ap = sqrt(2*(E + u/r)) with r = 2588 AU > > = 0.586 km/sec = 0.033 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. The conclusion stands.
From: Brad Guth on 15 Jun 2010 01:39
On Jun 13, 8:05 pm, "Greg Neill" <gneil...(a)MOVEsympatico.ca> wrote: > 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. You're wasting time. Just put everything into a public funded supercomputer and let it run. Tell us where that 90% stuff is coming from. ~ BG |