Lect_29 - ACT Gravitational forces Two stars of masses M...

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Lecture 29 Orbits. Kepler’s laws. 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 2 M 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: 2 12 ; In terms of magnitude: net FF F =+ = −+ GG G m 2 1 1 2 2 > 1 (M 2 > 1 , and distance to 2 is shorter) 1 2 2 1 2 = 1 > 1 , but distance to 2 is larger) 2 1 2 1 x d - x = −+ = 0 The actual calculation: = 1 2 Mm FG () = 2 2 2 dx += 2 2 2 0 = 22 20 xd =−= < =− + 21 0 . 4 1 2 2 1 = 2 2
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Circular orbits A communications satellite orbits at 400 km above the surface of the Earth. What is its period? 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 Mm v FG m rr F g →= E M vG r π = 2 But we also know that T ⇒= E 2 G rT =+ E with rRh = 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 = 23 E 4 5545 s 1.54 h GM Important formulas for circular orbits Example: Geosynchronous orbit A geosynchronous satellite orbits above the Equator with = 24 h. How high above the Earth’s surface does such a satellite have to orbit? A. 6380 km B. 11900 km C. 17900 km D. 35800 km E. 42200 km 2 E 3 2 4 GMT ⇔= We just found the relation between and for any satellite in a circular orbit around the Earth: Watch out!
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This note was uploaded on 04/16/2008 for the course PHYSICS 221 taught by Professor Johnson during the Fall '06 term at Iowa State.

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Lect_29 - ACT Gravitational forces Two stars of masses M...

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