A spacecraft of 100 kg mass is in a circular orbit about the Earth at a height h

Question

A spacecraft of 100 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 = ? 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

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)

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