Approximately one billion years ago, the Moon orbited the Earth much closer than it does today. The radius of the orbit was only 24 400 km. The orbital period was only 23 400 s. Today, the average radius is 385 000 km; and the present period is 2.36 × 106 s. Assuming that the orbit of the Moon is circular, calculate the ratio of the speed of the Moon in its ancient orbit to the speed that it has today.

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Answers

2014-05-20T21:33:05+08:00
Speed of the moon (ancient times) : speed of the moon (today) 

calculating first the total distance the moon travels around which is circular, 2 pi r both in ancient times and today. but set first those huge numbers into scientific notations. 

24 400 km = 2.44 * 10^4 km 
23 400 secs = 2.34 * 10^4

385 000km = 3.85 * 10^5 km

*ancient times: 2 pi * 2.44x10^4 
*today: 2 pi * 3.85 * 10^5

then the rate of each which is distance over time it takes to travel the distance, 

*ancient times:  \frac{2 pi * 2.44*10^4 }{2.34*10 x^{4}  }
                    = you can cancel 10^4 here which gives, 
                    =  \frac{2 pi * 2.44 }{2.34 }
                    =  \frac{244 pi}{117}
*today:  \frac{2 pi * 3.85 * 10^5}{2.36 * 10^{6} }
                    =  \frac{2 pi * 3.85 }{2.36 * 10 } }
                    =  \frac{77 pi}{236}

ratio =  \frac{244 pi}{117}  \frac{77 pi}{236}
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you can always substitute pi as 3.14 or 22/7