- #1

Timoothy

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Why is that information (10.3 miles per hour) rarely mentioned along with other moon facts

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- Thread starter Timoothy
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- #1

Timoothy

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Why is that information (10.3 miles per hour) rarely mentioned along with other moon facts

- #2

mgb_phys

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It's a conspriracy!

ps. Wiki gives it as 4.627 m/s

ps. Wiki gives it as 4.627 m/s

- #3

Timoothy

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It's a conspriracy!

ps. Wiki gives it as 4.627 m/s

"A conspiracy" you say, well that's the best answer I've heard so far!

- #4

russ_watters

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I guess people just don't consider that factoid to be very important.

- #5

Timoothy

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I guess people just don't consider that factoid to be very important.

The link below is the best I've found so far.

http://www.Newton.dep.anl.gov/Newton/askasci/1995/astron/AST047.HTM

- #6

LURCH

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- #7

tony873004

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It's easy to compute: v=d/t = Moon's circumference / period = 2*pi*1737km/(27.32*24) = 16.645 km/hr, same as 10.3 mi/hr or 4.6 m/s.

But at the poles its velocity is 0. To get the speed for any latitude inbetween, multiply the equatorial speed by cos(latitude).

- #8

Timoothy

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Well i suppose so, but I think the 10.3 mph (axial rotation at the equator) figure would be interesting to lots of people such as myself (pushing 60) as well as younger people who are just beginning to learn about the moon.

- #9

Timoothy

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It's easy to compute: v=d/t = Moon's circumference / period = 2*pi*1737km/(27.32*24) = 16.645 km/hr, same as 10.3 mi/hr or 4.6 m/s.

But at the poles its velocity is 0. To get the speed for any latitude inbetween, multiply the equatorial speed by cos(latitude).

Thanks, someone else had already shown me the math so I was able to follow your formula.

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