Speeds of the moon's motions

  • Thread starter Timoothy
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  • #1
Timoothy
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I've only found 3 websites that even mention the moon's rotation speed, much less discuss it.

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

Answers and Replies

  • #2
mgb_phys
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It's a conspriracy!

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.
 
  • #6
LURCH
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Generally, I think people find it more useful to measure the moon's rotation in comparison with its orbital period. "Miles" and "meters" are strictly man-made units of measure, so measuring the rotation of the moon in miles per hour or meters per second might seem (to some people) rather arbitrary. Referring to the time it takes the moon to complete one revolution on its axis as "one lunar orbit" is a much more relevant bit of information, for most applications.
 
  • #7
tony873004
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That's the Moon's rotational speed at the equator.
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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Generally, I think people find it more useful to measure the moon's rotation in comparison with its orbital period. "Miles" and "meters" are strictly man-made units of measure, so measuring the rotation of the moon in miles per hour or meters per second might seem (to some people) rather arbitrary. Referring to the time it takes the moon to complete one revolution on its axis as "one lunar orbit" is a much more relevant bit of information, for most applications.


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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That's the Moon's rotational speed at the equator.
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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