What is the speed of the moon's motion?

  • Context: High School 
  • Thread starter Thread starter Timoothy
  • Start date Start date
Timoothy
Messages
33
Reaction score
0
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
 
Astronomy news on Phys.org
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.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 16 ·
Replies
16
Views
6K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 7 ·
Replies
7
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
5K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 27 ·
Replies
27
Views
5K