(29)Orbits and Kepler's Laws

(29)Orbits and Kepler's Laws - Lecture 29 Orbits. Kepler's...

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Lecture 29 Orbits. Kepler’s laws.
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A. Closer to the smaller star B. Closer to the larger star C. It depends on the mass of the spaceship. ACT: Gravitational forces Two stars of masses M and 2M are separated by a distance d. A spaceship travels between the two stars. The net gravitational force on the spaceship is zero when the spaceship is: d M 2M
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1 2 1 2 ; I n t er ms of magnit ude: net net F F F F F F = + = - + r r r m F 2 F 1 M 1 M 2 F 2 > F 1 (M 2 > M 1 , and distance to M 2 is shorter) M 1 M 2 m F 2 F 1 F 2 = F 1 (M 2 > M 1 , but distance to M 2 is larger)
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F 2 d M 1 M 2 m F 1 x d - x = - + = 1 2 0 net F F F The actual calculation: = 1 2 Mm F G x ( 29 = - 2 2 2Mm F G d x ( 29 - + = - 2 2 2 0 Mm Mm G G x d x - - + = 2 2 2 0 x dx d ( 29 ( 29 = - = < = - + 2 1 0.41 2 2 1 d x d d x d ( 29 = - 2 2 1 2 x d x
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Circular orbits A communications satellite orbits at 400 km above the surface of the Earth. What is its period?
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The only force acting on the satellite is the gravitational attraction of Earth (“neglecting” the attraction of other bodies, like the Sun) = = 2 S E S 2 g M m v F G m r r F g = E M v G r π = 2 But we also know t hat r v T = E 2 M r G r T = + E wit h r R h r = 6.38 × 10 6 + 400 × 10 3 = 6.78 × 10 6 m M E = 5.98 × 10 24 kg G = 6.67 × 10 -11 N m 2 kg -2 = = = 2 3 E 4 5545 s 1.54 h r T GM Important formulas for circular orbits
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A geosynchronous satellite orbits above the Equator with T = 24 h. How high above the Earth’s surface does such a satellite have to
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This note was uploaded on 03/27/2008 for the course PHYS 221 taught by Professor Herrera-siklody during the Fall '08 term at Iowa State.

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(29)Orbits and Kepler's Laws - Lecture 29 Orbits. Kepler's...

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