This is 'Halloween Asteroid' 2015 TB145, as it was approaching Earth, imaged with the Arecibo telescope radar. As it turned out, the space rock measures around 600 meters and rotates once every five hours around its axis. It is thus larger than initially thought.

tiny speck of light even at closest approach, which, when writing these lines, has already occurred. We are still alive. But the sudden appearance of this asteroid (which might be a dormant comet) should be a reminder that we should to to more to detect Near Earth Objects which might pose a potential threat to our planet.

...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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ReplyDelete...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 at mass changing and linear speed continues equal without motor thrust: if mass decreases...radius increases, angular speed decreases, period increases, and viceversa, 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), 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, 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 3th Law...line Sun------→asteroid sweeps equal areas in equal times: Period²/radius medim³) = Constant... Taking into account besides its changes of speed: it brings near accelerates, it brings far decelerates...etc.

ReplyDelete...the orbital radius medium it is proportional to the squared of speed medium: to double speed medium, for once more radius medium, etc...radius medium = (mass*speed medium²)/force (centripetal) medium from Sun gravitational attraction.

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