A spacecraft of 130 kg mass is in a circular orbit about the Earth at a height h
ID: 1412836 • Letter: A
Question
A spacecraft of 130 kg mass is in a circular orbit about the Earth at a height h-4RE. (a) What is the period of the spacecraft's orbit about the Earth? T= 6.4 (b) What is the spacecraft's kinetic energy? (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.) (d) Find the numerical value of the angular momentum. Use its definition to express the period of the spacecraft's motion and apply Newton's second law to the spacecraft to determine its orbital speed. Then use this orbital speed to calculate the kinetic energy of the spacecraft. You can relate the spacecraft's angular momentum to its kinetic energy and moment of inertia. eBookExplanation / Answer
T² = K R^3 where K = 4² /GM
K = 4² / (6.673e-11*5.977e24) = 9.89818e-14
Given R = 4Re + Re = 5Re = 5*6371e3 = 3.1855e+7 m
T² = 9.89818e14*(3.1855e+7) ^3
T = (9.89818e14*(3.1855e+7) ^3
T = 5.65e+18 s
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K.E =0.5mv² = 0.5*130v² = 0.5*130*(GM/R)
K.E = 0.5*130*6.673e-11*5.977e24/1.9113e+7
K.E = 1.35e9 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)
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L = 3.1855e+7* (2*130* 1.25e9)
L = 3.1855e+7* (2*130* 1.25e9)
L = 1.81601e13 Js
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