Astro 1:HW 5 Solutions (Winter 08)

Astro 1:HW 5 Solutions (Winter 08) - UCSB Winter 2008 Astro...

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UCSB Winter 2008 Astro 1 - Homework #5 Solutions Kevin Moore 2/9/08 1 How can we use the Moon’s orbital motion, plus the distance an apple falls in one second, to determine the dependence of the gravitational force formula on the distance between two bodies? The basic idea here is to realize that the distance an object falls in one second is directly proportional to acceleration, which is in turn proportional to the force of gravity. You didn’t need to do the actual calculation to find this, but I’ll go through it and show you can deduce that the force of gravity goes as 1 /r 2 . You may know from physics that for constant acceleration, distance is given by d = d o + v o t + 1 2 at 2 Thus the distance an object falls in one second is d Earth = 1 2 a (1 s 2 ), indeed directly proportional to acceleration. We also know that F = ma , so d F for an object falling in one second. Since the force of gravity is directly proportional to mass (see prob. 2), it doesn’t matter whether it’s the Moon or an apple up there - they all fall at the same rate. First we’ll get the distance an object falls on Earth’s surface: here a = 9 . 8 m/s 2 so d = 4 . 9 m Getting the distance the Moon ’falls’ in 1 sec is more involved: First we need to find how far the Moon moves in its orbit in one second: It’s angular speed is (in radians/sec) ω = 2 π 27 . 3 days 1 day 24 hr 1 hr 3600 s = 2 . 66 × 10 - 6 rad sec The Earth-Moon distance is D moon = 384 , 400 km so the Moon travels about (384 , 400)(2 . 66 × 10 - 6 ) = 1 . 02 km in 1 sec.
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