What is the Iinear speed of a point on Earth's equator? What if the point was at latitude -15.0° ?

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources


Write a one sentence answer...



Explain in detail...


I want someone to double check my answer

Describe your changes (optional) 200

Apr 16, 2018


At the equator, linear speed with respect to the polar (day/night spin ) axis is 463.82 meter/s = 1521.8 feet/s, At latitude #15^o N and 15^o S#, this speed is 447.77 meter/s = 1469.1 feet/s.


In the day-night spin, any point on the Earth moves around Earth's

axis in its latitude circle.

Speed = (distance traveled in a second)/second..

The circumference of the equator = #2 pi X 6378# km.

The number of seconds in a day = 24 X 3600

So, at the equator, speed = #( 2 pi X 6378 ) / ( 24 X 3600 )# km/s =

0.46382 km/s = 463.82 meter/s = 1521.8 feet/s.

At #+-15^o# latitude Earth's ( a little less than equatorial )

radius is

6374.5 km, nearly.

The latitude-circle circumference

= #2 pi X 6374.5 X cos 15^o# km.

So, the speed is (#2 pi X 6374.5 X cos 15^o#)/(24 X 3600)=

0.44777 km/s = 447.77 meter/s = 1469.1 feet/s
Thanks to Johannes C for pointing to the error in the unit km,

used for meter, in my earlier edition...

Was this helpful? Let the contributor know!
Impact of this question
2220 views around the world
You can reuse this answer
Creative Commons License