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

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Explanation

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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 ${15}^{o} N \mathmr{and} {15}^{o} S$, this speed is 447.77 meter/s = 1469.1 feet/s.

Explanation:

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 = $\frac{2 \pi X 6378}{24 X 3600}$ km/s =

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

At $\pm {15}^{o}$ latitude Earth's ( a little less than equatorial )

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
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Thanks to Johannes C for pointing to the error in the unit km,

used for meter, in my earlier edition...

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