PHYS 101 Lecture Notes - Lecture 11: Spring Scale, Ellipse, Circular Motion

• School: McGill University
• Department: Physics
• Course: PHYS 101
• Professor: Kenneth J Ragan

Document Summary

Distance to moon : 3. 85 x 108 m/6. 4 x 106 m = . We can now understand satellite dynamics, having seen both circular motion and the law of gravitation. Solution: the equator. that is, its orbital period is 24. 0 hours. Firstly, is the satellite motionless? no, of course: it"s orbiting the earth; ie: uniform circular motion ! So we have: a = v 2 /r = fgrav/m = gmearth/r 2. (2 r/t)2/r = gmearth/r 2 which simplifies to r 3 . V = 2 r/t v = 3076 m/s. Now solve for r: r = [(6. 67 x 10-11 x 6. 0 x 1024/4 2) (24 x 60 x 60)2](1/3) r = 4. 23 x 107 m = 42,300 km. A typical low-earth-orbit satellite (shuttle, spy, imaging, etc ) orbits at ~200 km. 2) each planet moves such that a line from the planet to the sun sweeps out equal areas in equal times.