Question #fe905

1-Acceleration over the Equator.

Considering the sidereal rotation period (earth's daily rotation relative to a distant star) equal to

#T = 23#h#56#m#4.1#s# = 86164.1# s

and the Earth's equatorial radius

#R_"equator" = 6.378 \times 10^6# m,

the tangential velocity of a point located at earth's equator is

#v={2*\pi*R_"equator"}/T=465.21 # m/s.

Using this value in #A_"ctp"=v^2/r#, comes

#A_"ctp"= (465.21 " m/s")^2/{6.378\times10^6 "m"}=3.4\times 10^{-2} " m/s"^2.#

The result is similar if we use T = 24h (the Synodic rotation period, or the rotation period relative to the Sun).

2-Acceleration at any given latitude.

As a general expression, for any given latitude, we write:

#A_"ctp"=v^2/r=([2*\pi*\sin("Latitude")*R_"Equator"]/T)^2/(sin("Latitude")*R_"Equator")#,

Where

#r=\sin("Latitude")*R_"Equator"#

is the radius of the circle performed by a point at the given latitude around the Earth's rotation axis.

At the poles (#\sin("Latitude")=\sin(90^\circ)=0#) the denominator is obviously zero, but the limit can still be computed, as #v^2# approaches zero faster than #r# does:

#A_"ctp" = \lim_(r\rarr0) v^2/r =\lim_(r\rarr0) (4\pi^2)/T^2*r^2/r= \lim_(r\rarr0) (4\pi^2)/T^2*r^\cancel2/\cancelr=0#.