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5) Two heavenly bodies both with radii of 6×103 miles and masses of 9×1024 kg ar

ID: 2030239 • Letter: 5

Question

5) Two heavenly bodies both with radii of 6×103 miles and masses of 9×1024 kg are separated by 4×107 miles. They are initially at rest. How fast are they moving just before their surfaces collide. Assume that you can ignore any effects having to do with the existence of atmospheres and that nothing significant exists in the space between the planets. Correct, computer gets: 1.25E+04 mi/hr Hint: Consider the initial "scene" before the planets gain any speed and the final scene just when the surfaces are about to collide. What types of energy are relevant in each scene? Are the centers of the planets still separate in the final scene? Will that separation (if it exists) play a significant role? 6) Consider the same setup as before, only one planet is twice the mass of the other (their radii are still the same size). What is the speed of the more massive planet just before their surfaces collide?

Explanation / Answer

5) given

two heavenly bodies

R = 6*10^3 miles = 9 656 064 m

M = 9*10^24 kg

d = 4*10^7 miles = 64 373 760 000 m

initial KE = 0

initial PE = -GM^2/d

final KE = Mv^2 ( where v is velocoty if the two bodies)

final PE = -GM^2/2R

hence

from conservation of energy

-GM^2/d = Mv^2 - GM^2/2R

hence

GM(1/2R - 1/d) = v^2

v = 5574.4747432834 m/s

v = 12469.74 mph

for the second case

from conservation of energy

-2GM^2/d = 0.5Mv1^2 + Mv2^2 - 2GM^2/2R

also, from conservation of moemntum

2MV2 = MV1

hence

V1 = 2V2

hence

-2GM^2/d = 2Mv2^2 + Mv2^2 - 2GM^2/2R

2GM(1/2R-1/d) = 3v2^2

v2 = 4551.539568 m/s = 10181.504 mph

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