spacecraft of 150 kg mass is in a circular orbit about the Earth at a height h =
ID: 2302089 • Letter: S
Question
spacecraft of 150 kg mass is in a circular orbit about the Earth at a height h = 5RE.
(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
Explanation / Answer
a)Let r = distance of spacecraft from Earth's center,
ME = Earth's mass
v = speed of satellite
r = RE + h = RE + 4RE = 5 RE------------(a.1)
Gravitational force = G*ME*m/r^2
Gravitational force provides centripetal force and centripetal force = mv^2/r
Therefore G*ME*m/r^2 = mv^2/r
Dividing by m,
G*ME/r^2 = v^2/r
Multiplying by r,
v^2 = G*ME/r--------------------------------(a...
Or v = sqrt(G*ME/r)
T = 2*pi*r/v = 2*pi*r/sqrt(G*ME/r)
= 2*pi*r^(3/2)/sqrt(G*ME)
T = 2*pi*(5RE)^(3/2)/sqrt(G*ME)
You may plug in the values now. As you can see, this is the same as in your text book. But I thought of explaining how this formula comes so that in exam, if you forget the formula, you should not worry.
b) K = 1/2 mv^2 = 1/2 * m * G*ME/r [using equation (a.2)]
Or K = 1/2 * m * G*ME/(5RE)------[using equation (a.1)]
Or K = 1/10 G*m*ME/RE
c)K = 1/2 mv^2
Or v^2 = 2K/m
Or v = sqrt(2K/m)-----------------------(c.1)
L = mvr
Or v = L/(mr)----------------------------(c.2)
From (c.1) and (c.2),
sqrt(2K/m) = L/(mr)
Or L = mr*sqrt(2K/m)
Or L = r*sqrt(2mK)
Using equation (a.1) in the above,
L = 5RE*sqrt(2mK)
Your understanding that your should add Earth's radius to h is correct.
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