# Question fe905

Jan 23, 2018

The centripetal acceleration over the equator is

${A}_{\text{ctp"=v_"equator"^2/r_"earth}} = 3.4 \setminus \times {10}^{- 2} \frac{m}{s} ^ 2$,

And it fades to zero at the poles.

#### Explanation:

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

${R}_{\text{equator}} = 6.378 \setminus \times {10}^{6}$ m,

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

$v = \frac{2 \cdot \setminus \pi \cdot {R}_{\text{equator}}}{T} = 465.21$ m/s.

Using this value in ${A}_{\text{ctp}} = {v}^{2} / r$, comes

${A}_{\text{ctp"= (465.21 " m/s")^2/{6.378\times10^6 "m"}=3.4\times 10^{-2} " m/s}}^{2.}$

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 ($\setminus \sin \left(\text{Latitude}\right) = \setminus \sin \left({90}^{\setminus} \circ\right) = 0$) 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#.