solutions-ps02

solutions-ps02 - Physics 341: Problem Set #2 Solutions 1....

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Physics 341: Problem Set #2 Solutions 1. We showed that for circular orbits, Kepler’s Third Law can be written as P 2 = 4 π 2 GM r 3 where M is the mass of the central object. (a) Use this expression and a Taylor expansion to derive the following approximation for the orbital period of a satellite in “low Earth orbit”, with a constant height h above the surface of the Earth, assuming that h ± R : P P 0 ± 1 + 3 h 2 R ² For an orbit with altitude h , the orbital radius is just r = R + h . Plugging this into Kepler’s Law we get P = ± 4 π 2 GM ² 1 / 2 ( R + h ) 3 / 2 = 2 π ( GM ) 1 / 2 R 3 / 2 ± 1 + h R ² 3 / 2 Now using our Taylor expansion approximation (see notes from Lectures 1 and 2) that (1 + x ) α 1 + αx when x ± 1, we have ± 1 + h R ² 3 / 2 1+ 3 2 h R = P 2 πR 3 / 2 ( GM ) 1 / 2 ± 1 + 3 h 2 R ² = P 0 ± 1 + 3 h 2 R ² (b) What is the constant P 0 (in symbols), and what is its value (in minutes)? From above, we see that P 0 = 2 πR 3 / 2 ( GM ) 1 / 2 This corresponds to the period of something “orbiting” right at the Earth’s surface (that is, h = 0 and r = R ). Plugging in the values, we find P 0 = 2 π (6 . 38 × 10 8 cm) 3 / 2 [(6 . 67 × 10 - 8 cm 3 g - 1 s - 2 )(5 . 97 × 10 27 g)] 1 / 2 = 5074 s = 84 . 6 min 1
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(c) The Hubble Space Telescope orbits the Earth at an altitude of h = 570 km. Com- pare its exact orbital period based on Kepler’s Law with your approximation. For the exact value we have
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solutions-ps02 - Physics 341: Problem Set #2 Solutions 1....

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