A spacecraft of 140 kg mass is in a circular orbit about the Earth at a height h

Question

A spacecraft of 140 kg mass is in a circular orbit about the Earth at a height h = 2RE. (a) What is the period of the spacecraft's orbit about the Earth? T = h (b) What is the spacecraft's kinetic energy? K = J (c) Express the angular momentum L of the spacecraft about the center of the Earth in terms of its kinetic energy K. (Use the following as necessary: RE for the radius of the Earth, K for the kinetic energy of the satellite, and m for the mass of the satellite.) L = (d) Find the numerical value of the angular momentum. L = J · s

Explanation / Answer

T² = K R^3 where K = 4² /GM
K = 4² / (6.673 x 10^-11 x 5.977 x 10^24) = 9.89818 x 10^-14
Given R = 2Re + Re = 3Re = 3 x 6371 x 10^3 = 1.9113 x 10^+7 m
T² = 9.89818 x 10^14*(1.9113 x 10^+7)^3
T = (9.89818 x 10^14*(1.9113 x 10^+7) ^3
T = 2.63 x 10^+18 s
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K.E =0.5mv² = 0.5*140v² = 0.5*120*(GM/R)
K.E = 0.5 x140 x (6.673 x 10^-11 x 5.977 x 10^24) / (1.9113 x 10^+7)
K.E = 1.46 x 10^9 J
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L = mvr
L² = (mvr) ²
L²/r² = (mv) ²

K = 0.5 mv²
2m K = (mv) ²

L²/r² = 2m K

L = r* (2m K)


L = 1.9113 x 10^+7* (2 x 140 x 1.46 x 10^9)
L = 1.222 x 10^13 Js

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