A spacecraft of 110 kg mass is in a circular orbit about the Earth at a height h
ID: 1464381 • Letter: A
Question
A spacecraft of 110 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
Given
Mass of the space craft m = 110 Kg
Height h = 2R
Mass of the earth M = 5.98 x 1024 kg
Radius of the earth R = 6.38 x 106 m
Gravitational constant G = 6.67 x 10-11 Nm2/kg2
Acceleration due to gravity g = 9.8 m/s2
Solution
a)
Time period T = 2{(R + h)3/GM}
= 2 x 3.14 x (9R3/GM)
= 5067.30 s
= 1.41 hours
b)
kinetic energy K = GMm/2(R+h)
= 4387.526 x 1013 / 6R
= 114.62 x 107 J
c)
L = mv(R+h)
= 3mvR
= 3R * 2K /v
= 6KR/v
V = (GM/3R)
So
L = 6KR *(3R/GM)
d)
L = 6 x 114.62 x 107 x 6.38 x 106 (3x6.38 x 106 / 6.67 x 10-11 x 5.98 x 1024)
= 9611.47 x 109
= 9.6 x 1012 Js
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