What is the speed of the moon's motion?

  • Context: High School 
  • Thread starter Thread starter Timoothy
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Discussion Overview

The discussion revolves around the speed of the moon's rotation and its relevance in various contexts. Participants explore the numerical values associated with the moon's rotational speed and question why this information is not more commonly referenced.

Discussion Character

  • Exploratory
  • Debate/contested
  • Technical explanation

Main Points Raised

  • Some participants note that the moon's rotational speed is rarely mentioned, with one suggesting it is a conspiracy.
  • Several participants provide the value of the moon's rotational speed as 10.3 miles per hour or 4.627 m/s, citing different sources.
  • One participant argues that the moon's rotation is more relevant when compared to its orbital period rather than expressed in man-made units like miles per hour or meters per second.
  • Another participant explains the calculation of the moon's rotational speed at the equator and notes that it is zero at the poles, suggesting a formula to find speeds at different latitudes.
  • Some express that the rotational speed figure could be interesting to a wider audience, including younger learners.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the importance of the moon's rotational speed or its presentation. Multiple viewpoints exist regarding its relevance and the best way to express this information.

Contextual Notes

There are varying assumptions about the significance of the moon's rotational speed, and the discussion includes different methods of measurement and calculation without resolving the implications of these differences.

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
 
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It's a conspriracy!

ps. Wiki gives it as 4.627 m/s
 
mgb_phys said:
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!
 
I guess people just don't consider that factoid to be very important.
 
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.
 
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).
 
LURCH said:
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.
 
tony873004 said:
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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