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# 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: d M 2 M 1 2 1 2 ; In terms of magnitude: net net F F F F F F = + = − + G G G 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) 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 ( ) = 2 2 2 Mm F G d x ( ) + = 2 2 2 0 Mm Mm G G x d x + = 2 2 2 0 x dx d ( ) ( ) = = < = − + 2 1 0.41 2 2 1 d x d d x d ( ) = 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? 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 that r v T π = E 2 M r G r T = + E with 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 π = = = 2 3 E 4 5545 s 1.54 h r T GM Important formulas for circular orbits Example: Geosynchronous orbit A geosynchronous satellite orbits above the Equator with T = 24 h. How high above the Earth’s surface does such a satellite have to orbit?
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