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Sunday, October 18, 2015

 

Celestia files for Near Earth Asteroid 2015 TB145

Near Earth Asteroid 2015 TB145 at closest approach to Earth as visualised in Celestia. Click to embiggen.Orbits of Earth, Moon and Near Earth Asteroid 2015 TB145 at closest approach to Earth as visualised in Celestia. Click to embiggen.

Asteroid 2015 TB145 will come within 1.3 Earth Moon distances of Earth  on October 31. Despite being somewhere between 280–620 meters in diameter, it will be visible in small amateur instruments at closest approach, although North America will have the best view.  The asteroid is a target for mapping with radar.

For those of us not in North America (or who just want to play around with the asteroid orbit, or who want to argue with people who claim this asteroid will hit us (it won't)) I've made Celestia files for 2015 TB145.  As usual, copy the data here to a plain text file (2015TB145.ssc) and copy the file to the Celestia extras folder.

========= 8< cut =============2015TB145.ssc======= 8< cut =====================
"2015 TB145" "Sol"

#Data from MPC http://cfa-www.harvard.edu/iau/MPEph/MPEph.html
#Epoch 2016 Jan. 13.0 TT = JDT 2457400.5                 MPC
#M  10.48189              (2000.0)            P               Q
#n   0.32423068     Peri.  121.69708     -0.81665165     -0.42486171             T = 2457368.17150 JDT
#a   2.0984565      Node    37.76023     -0.03578994     -0.63822119             q =     0.2938239
#e   0.8599809      Incl.   39.63322     +0.57602011     -0.64200175    Earth MOID = 0.00157 AU
#P   3.04           H   19.8           G   0.15           U   8
#From 62 observations 2015 Oct. 10-15, mean residual 0".36.

{
    Class "asteroid"
    Mesh   "ky26.cmod"
    Texture "asteroid.jpg"
    Radius  0.15 # maximum semi-axis
    MeshCenter [ -0.000718 -0.000099 0.000556 ]

    EllipticalOrbit
    {
    Epoch                         2457400.5      # Epoch 2016 Jan. 13.0 TT = JDT 2457400.5
    Period                             3.04           # P
    SemiMajorAxis          2.0984565      # a
    Eccentricity                0.8599809      # e
    Inclination                  39.63322          # Incl.
    AscendingNode          37.76023       # Node
    ArgOfPericenter       121.69708       # Peri
    PericenterDistance      0.2938239     # q
    MeanAnomaly          10.48189       # M
    }

    RotationPeriod 0.9 #Guess

    Albedo 0.15        #Based on typical Stony asteroids, rotation periods vary from one hour to one day

}
================================8< cut  8<===================================== 

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Comments:
Thank you for the parameters 2015TB145
 
...a possible solution... INCREASING ITS SPEED, gives it a shove, TOWARDS A SIDE IF THERE ARE LITTLE TIME, to what speed gives it thrust towards a side for deflecting it?...speed = space/time...so if want that it passing "close shaving" to 1,000 kms from Earth = 1 million mts, and having approx. 11 days = 1 million seconds, the correction speed will be of: 6,000 kms approx. Earth radius + 1,000 kms = 7,000 kms; 7 million mts/1 million seconds = 7 mts/second (25 kms/hour). If it increases forwards speed, it climbs to a higher orbit, and vice versa: radius = (mass*speed²)/force (centripetal) from Sun gravitational attraction.
 
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...a possible solution (2)...for gives it impulse: a hole in its surface (excavated by astronauts, or by successive nuclear impacts in the same place) and within a nuclear charge: the superficial material of the asteroid, volatilized by nuclear radiations, gamma rays, etc (at Space, on having not air, there is not expansive wave) go out ejected as in a rocket´s nozzle. Also can to impulses a slight asteroid for doing it collides, using it as a missile, against other asteroid or comet very much more massive. War ballistic missiles that there are, they are not useful, are designed for go away from continent to continent... Only can be useful a rocket that can to put the charge at interplanetary orbit, as the multi-phase which put satellites in geostationary orbit (is necessary have them already prepared in a Spatial-Command with all the nuclear weapons that there are in the World controlled and preserved by an International Organism, only for that)...("Doomsday asteroid": W. Cox and H. Chestek).
 
...or...decreasing its mass. Is that perhaps the asteroid orbit will continue equal whether its mass increases at double or decreases at half?. No. (By the Angular momentum, mass*speed*radius, Conservation Law, when mass changes and linear speed continues equal without motor thrust: if mass decreases...radius increases, angular speed decreases, period increases, and vice versa, till it reaches the new stable orbit). If mass increases, bad...it brings near in its radius minimum (perihelion) it "falls" towards the Sun, because with its new major mass also there is a new major Sun gravitational attraction and now it would need major centrifugal force, major speed, for remaining in the same stable orbit (equal angular speed and different linear speed = different centrifugal force, as in Earth poles or equator...equal linear speed and different angular speed = also different centrifugal force, as in "whip" fairground attraction)...and it brings far in its radius maximum (aphelion), it doing a more elliptic new orbit, radius medium decreases... And vice versa: if MASS DECREASES, good...it brings far in its perihelion from Sun, and from Earth, because still conserves speed that it had with the major mass before and at same time decreases the Sun gravitational attraction, it doing a more circular new orbit, radius medium increases. Formulas: for circular orbit...radius = (mass*speed²)/force (centripetal) from Sun Gravitational attraction... For elliptic orbit, besides...the Newton´s Universal Gravitation Law: F = G((m1*m2)/distance²)...the kepler´s Laws...2nd: line Sun------→asteroid sweeps equal areas in equal times...3th: radius medium³(AU)/period²(years) = Constant... Taking into account besides its changes of speed: it brings near accelerates, it brings far decelerates...orbital perturbations from planets, etc...also can colliding among them when go across the asteroids belt, etc. For exact calculation really a complex field...
 
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