Question

In Example, we found the centripetal acceleration of the Earth as it revolves around the Sun. From information on the endpapers of this book, compute the centripetal acceleration of a point on the surface of the Earth at the equator caused by the rotation of the Earth about its axis.

Short answer

The centripetal acceleration is, $0.03369 {\text{m/s}}^2$.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

The radius of the earth is,$R_e=6371 \text{km}$ .

The period of revolution of the earth is, $T=24 \text{h}$.

The formula to calculate velocity of point on the equator

The formula to calculate velocity of point on the equator is given by:

$\mathit{v}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }{\mathbf{R}}_{\mathbf{e}}}{\mathbf{T}}$

Here $v$is the velocity of a point at the equator, $R_e$is the radius of earth and T is the time period of the rotation.

Calculate the centripetal acceleration

Substitute 6371 km for R_e and 24 h for T in above equation to find v.

$\begin{array}{rcl}v& =& \frac{2\pi \left(6371 \text{km}\times \left(\frac{{10}^3 \text{m}}{1 \text{km}}\right)\right)}{24 \text{h}\times \left(\frac{3600 \text{s}}{1 \text{h}}\right)}\\ & =& 463.3 \text{m/s}\end{array}$

Write the formula to calculate centripetal acceleration

$a_{\text{c}}=\frac{v^2}{R_{\text{e}}}$

Here,$a_{\text{c}}$ is the centripetal acceleration.

Substitute all the value in the above equation.

$$\begin{gathered} a_{\text{c}}=\frac{{\left(463.3 \text{m/s}\right)}^2}{6371 \text{km}\times \left(\frac{{10}^3 \text{m}}{1 \text{km}}\right)} \\ {a}_{\text{c}}=0.03369 {\text{m/s}}^2 \end{gathered}$$

Hence the centripetal acceleration is,$0.03369 {\text{m/s}}^2$ .