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Unformatted text preview: homework 13 – FIERRO, JEFFREY – Due: Mar 3 2008, 11:00 pm 1 Question 1, chap 9, sect 4. part 1 of 2 10 points The period of the earth around the sun is 1 year and it is at a distance 150 million kilometers from the sun. An asteroid in a circular orbit around the sun is at a distance, 181 million kilometers from the sun. What is the period, T a , of the asteroid orbit? Correct answer: 1 . 3255 year (tolerance ± 1 %). Explanation: From Kepler’s laws, T 2 e r 3 e = T 2 a r 3 a T a = parenleftbigg r a r e parenrightbigg 3 2 T e = parenleftbigg 181 million 150 million parenrightbigg 3 2 1 year = 1 . 3255 year Question 2, chap 9, sect 4. part 2 of 2 10 points What is the orbital velocity of the asteroid? Assume there are 365 days in one year. Correct answer: 27206 . 4 m / s (tolerance ± 1 %). Explanation: v a = 2 π r a T a = 2 π (1 . 81 × 10 11 m) (1 . 3255 year) = 2 π (1 . 81 × 10 11 m) (4 . 18011 × 10 7 s) =27206 . 4 m / s . Question 3, chap 9, sect 5. part 1 of 1 10 points Given: G = 6 . 67259 × 10 11 N m 2 / kg 2 Calculate the work required to move a planet’s satellite of mass 1490 kg from a cir cular orbit of radius 2 R to one of radius 3 R , where 6 . 54 × 10 6 m is the radius of the planet. The mass of the planet is 4 . 52 × 10 24 kg. Correct answer: 5 . 72612 × 10 9 J (tolerance ± 1 %). Explanation: Applying the equation for the total energy, E , of a satellite in a circular orbit, E = GM m 2 r , where r is the radius of the orbit, we obtain for the total initial and final energies E i = GM m 4 R , E f = GM m 6 R . Therefore, the work required to increase the energy of the system is W = E f E i = GM m 12 R = 6 . 67259 × 10 11 N m 2 / kg 2 × 4 . 52 × 10 24 kg × 1490 kg 12 × 6 . 54 × 10 6 m = 5 . 72612 × 10 9 J . If we wish to determine how the energy is dis tributed after the work is done on the system, we find from the equation 1 2 mv 2 = GM m 2 r , that the change in kinetic energy is Δ K = GM m 12 R (it decreases), while the corresponding change in potential energy is Δ U = GM m 6 R (it increases). Thus, the work done on the system is W = Δ K + Δ U =...
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 Spring '08
 Turner
 Mass, Potential Energy, Work, Celestial mechanics

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