7) A meteorite of mass 1.6 x 10^3 kg is in a circular orbit around the Earth at

Question

7) A meteorite of mass 1.6 x 10^3 kg is in a circular orbit around the Earth at an altitude of 4.2 x 10^6 m above its surface. The poor meteorite suddenly suffers a head-on collision with a much smaller meteorite and loses 2.0% of its kinetic energy without changing its direction of motion or its total mass. (I.e., assume the smaller meteorite is vaporized in the collision.)
a) What basic physics principles apply to the motion of the more massive meteorite? Describe how they do so. [6]
b) Describe the shape of the meteorite’s orbit after the collision. [6]
c) Find the distance of closest approach of the meteorite to the Earth after the collision. [8]

Explanation / Answer

conservation of energy and momentum. so that we always have. M=mrxv = constant. and E=-GmM/r+mv^2/2 = constant. b) it is eclipse. c) it is eclipse we that. E=-GMm/2a where a of the eclipse. so that E=-GmM/r+GMm*0.98/2r=-0.51GmM/r. so that 1/2a=0.51/r so a=r/(2*0.51)=0.98r. so that c=r-0.98r=0.02r. so nearest distance. a-c=0.96r.

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