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The above question may sound naive. Still I would be thankful to someone who helps me in understanding the concept.

In the following link it is mentioned that earth's rotation speed with respect to latitude varies as cosine of the latitude.

http://image.gsfc.nasa.gov/poetry/ask/a10840.html

My question is how the rotation speed varies with cosine of latitude?

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At any latitude, the Earth completes one rotation per day.

At the equator, the circumference of the Earth is about 40000 km, so the speed of rotation is 40000 km/day or 463 m/s.

If you pick a line of higher latitude and look at it on a globe, you will see that the line of latitude is smaller than the equator.

One rotation completed in a day is therefore a shorter distance; at 60 degrees latitude the line is about (cos 60 x 40000) = 34600 km long, so the speed of rotation is 34600 km/s or 400 m/s.

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  • $\begingroup$ thanks for the quick response. The circumference of the latitude becomes smaller as we move away from the equator. But why does it obey the cosine function ? $\endgroup$ – Soumajit Sep 4 '16 at 16:11
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    $\begingroup$ Because the Earth is approximately spherical, so it's cross section is approximately circular. Draw a circle, and a radius line from the center to the edge. This is a side view cross section of the Earth. The angle made between the horizontal and the radius line is the angle of latitude. The distance from the vertical to the point where the radius meets the circle is how the cosine is defined. $\endgroup$ – Russell Borogove Sep 4 '16 at 16:17
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    $\begingroup$ If you need precision, you need to factor in the shape of the earth. More info: en.wikipedia.org/wiki/Earth_ellipsoid $\endgroup$ – Cem Kalyoncu Apr 13 '17 at 8:33
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Linear speed in circular motion depends on angular velocity (here: 1 rotation per day, for all latitudes) and radius of the circle - distance from the axis of rotation.

$ v = \omega r$

enter image description here

If $\alpha$ is your latitude, the radius of the circle you make with each revolution of Earth, is $r$. As you can see above, basic trigonometry, ${r \over R} = cos \alpha$. The closer you are to the pole, the more each degree of latitude decreases the radius of the circle you make.

Indirect consequence is that moderate distance of the space center from the equator diminishes delta-V much less than larger distance.

Take:

French Guiana Space Centre is 5 degrees North; Cape Canaveral is 28 degrees North; Baikonur is 45 degrees North, Plesetsk is 63 degrees North.

It would superficially seem that the velocity losses between Guiana and Canaveral (22° apart) are higher than between Canaveral and Baikonur (17°) or Baikonur and Plesetsk (18°). But this is not the case -

$\cos 5° - \cos 28° = 0.11;$

$ \cos 28° - \cos 45° = 0.17;$

$ \cos 45° - \cos 63° = 0.25 $

so the 17 degrees Baikonur loses to Canaveral is way worse than 22 degrees Canaveral loses to Guiana.

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