A satellite in a circular orbit around the earth with a radius 1.015 times the m

Question

A satellite in a circular orbit around the earth with a radius 1.015 times the mean radius of the earth is hit by an incoming meteorite. A large fragment (m = 69.0 kg) is ejected in the backwards direction so that it is stationary with respect to the earth and falls directly to the ground. Its speed just before it hits the ground is 369.0 m/s. Find the total work done by gravity on the satellite fragment. RE 6.37·103 km; Mass of the earth= ME 5.98·1024 kg.

Calculate the amount of that work converted to heat.

Explanation / Answer

M = 5.98x10^24 kg, R =6.37x10^3 km , m =69 kg

Ri =1.015R

Rf =R, vi =0, vf = 369 m/s

G =6.67x10^-11 N.m^2/kg^2

initial potential energy Ui= -GMm/R1

Ui = - (6.67x10-11x5.98x1024x69)/(1.015x6.37x106)

Ui = - 4.2567x109 J

final potential energy Uf= -GMm/R2

Uf = - (6.67x10-11x5.98x1024x69)/(6.37x106)

Uf = - 4.32053x109 J

Ki =0

Kf = (1/2)mv2^2 =(1/2)(69)(369)(369)

Kf = 0.469755x107 J

Amount of heat converted into heat is equal to change in mechanical energy

W = Ui+Ki -Uf-Kf

W = - 4.2567x109 +0 + 4.32053x109 -0.469755x107

Amount of heat converted into heat W =5.913x107 J

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