# Earths rotational speed vs latitude, and how wrong I was about 1050mph

A project log for 1050mph Virtual Rocket Camera

Warning! No Rocket involved! Collaborative Open Art Project.

Like most people I criticize, I myself started spewing "facts" as an authority about a subject after only a one minute internet search. It seamed pretty straight forward in the first 60 seconds of my ignorance:
Circumference x Time = Speed.
The article below sums up an "apparent solar day," taking into account the Earth's rotation around the sun, but more importantly, how latitude effects your relative speed of the sun to the horizon. Near the equator your speed reaches about 1040mph (1673km/h), but as you move up north (or south away from the equator,) the circumference traveled reduces. (specially "Angular speed" section):
https://en.wikipedia.org/wiki/Earth%27s_rotation#cite_note-38

The simple rough equation to approximate your speed (tangent speed of Earth's rotation according to latitude) is:
Current speed = Speed at the equator x COS(Angel of latitude in degrees)
or
V = Veq Cos(A)

More accurately, if you like math, circumference (C) of the distance traveled at the point of your latitude (A in degrees) is,
C = 2πRcos(A)
Where R is the radius of the Earth at the equator, then,
Velocity = C / T
Where T is time in hours for one "sidereal day," as mentioned here:
http://www.castor2.ca/16_Calc/03_Rotation/index.html

This makes for some interesting estimations by my calculations:

• Equator = 1040mph = 1673km/h
• Florida = 912mph = 1468km/h
• Colorado = 809mph = 1303km/h
• Alaska = 446mph = 718km/h

Which means you only need about half the speed to chase the sun in Nome Alaska as compared to Cape Canaveral Florida.

This has very little to do with the virtual portion of this project, but would make the possibility of a real rocket project much more feasible.

## Discussions 