A spacecraft of 120 kg mass is in a circular orbit about the Earth at a height h
ID: 1553976 • Letter: A
Question
A spacecraft of 120 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? (h)
B. What is the spacecraft's kinetic energy? (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.)
D. Find the numerical value of the angular momentum. (J x s)
Explanation / Answer
m = 120 kg , h = 4RE
RE = 6371 km , mass of earth M = 5.98*10^24 kg
A) from keplers third law
T^2 = 4pi^2*(RE+h)^3/GM
T^2 = 4*3.14^2*(5*6371000)^3/(6.67*10^-11*5.98*10^24)
T = 56534.4 s
T = 56534.4/3600
T =15.7 hr
B) v^2 = (GM/(R+h))
K = 0.5mv^2
K = 0.5*120*(6.67*10^-11*5.98*10^24/(5*6371000))
K = 7.5*10^8 J
(c) L = mvr
L^2 = (mvr)^2
L^2 = 2mKr^2
L = (2mKr^2)^0.5
L = (5RE)(2mK)^0.5
(d)
L = (2*120*7.5*10^8*(5*6371000)^2)^0.5
L = 1.3510^13 J.s
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