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Message: <blockquote data-quote="tomBitonti" data-source="post: 7488286" data-attributes="member: 13107">Starting with:That doesn't have enough detail.  Further basic results are also lacking in deail.Finally:Assessments of the energy, mass and size of the Chicxulub Impactor Hector Javier Durand-Manterola and Guadalupe Cordero-TerceroDepartamento de Ciencias Espaciales, Instituto de Geofísica, Universidad Nacional Autonoma de MéxicoArxiv March 19, 2014Since our hero was having difficulty with a much smaller rock having adiameter of 1km, the following uses the lowest values from the rangesof the estimate:Diameter          1.06 10^4 m    Mass:1.0 10^15 kgKinetic Energy (KE):1.3 10^24 JThen:Volume:        6.2 10^11 m^3  Density:1.6 10^3 kg/m^3Speed:5.1 10^4 m/s (51 km/s)In support of these values, Assessments writes:Motion of 51 km/s is 1.84x10^5 km/hr, or 4.4x10^6 km/day.(In comparison, Earths average orbital speed is about 30 km/s.)Working from the equations of motion under constant acceleration,and that relate force, mass, and acceleration:Under constant acceleration:    D = 1/2 A T^2Or:    A = 2 D / T^2Definition of force:    F = M AThen:    A = F / MFrom which:    F / M = 2 D / T^2Or:    F = 2 D M / T^2Putting in:The radius of the Earth:    D = 6.4x10^6 mThe mass of the asteroid:    M = 1.0 10^15 kgThe available time (one hour):    T = 3.6 10^3Results in:    F = 1.0 10^15 kg m / s^2Total impulse is:    I = F T    I = 3.6 10^18 kg m / sFrom: <a href="http://www.b14643.de/Spacerockets_2/United_States_1/Saturn-5/Design/SaturnV.htm" target="_blank">http://www.b14643.de/Spacerockets_2/United_States_1/Saturn-5/Design/SaturnV.htm</a>The total impulse of the Saturn V for Apollo 17 (launched 12-Jul-1972) was:    I(SV) = 8.7x10^9 kg m / sThat is, pushing the asteroid out of the way would require the totalimpulse of 2.4x10^8 (240,000,000, or 240 million) Saturn V's.Notes:The impulse requirements do not vary with the speed of the asteroid!What matters is how long one has to push the asteroid out of the way.That this should be the case can be seen by adopting an asteroid-centricframe of reference, in which case the asteroid is still andthe earth which is in motion.  When looked at from this frame ofreference, all that matters is how long until the Earth and theasteroid collide, not how fast the earth is moving.From the force and total impulse equations:    F = 2 D M / T^2    I = F T = 2 D M / TThe force requirement increases linearly with the distance and mass,and decreases in portion to the square of the available time.  But,the total impulse requirement decreases linearly with the availabletime.  (That is, as the available time increases, while the necessaryforce is reduced, the duration of application of that force increases,moderating a square factor to a linear one.)Requirements for an asteroid having a similar composition but having adifferent radius vary according to the change of radius of theasteroid.  Changing from 10km to 1km reduces the force requirement bya factor of 1000 (10^3).  Note that a "small" 1 km asteroid, stillneeds the total impulse of 240,000 Saturn V's.Thx!TomB</blockquote>

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