(a) What is the magnitude of the centripetal acceleration of an object on Earth’s equator due to the rotation of Earth? (b) What would Earth’s rotation period have to be for objects on the equator to have a centripetal acceleration of magnitude$\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$?

Radius of earth at equator$r=6370\times {10}^3\mathrm{m}$

Speed of earth’s rotation at equator v = 460 m/s

This problem deals with the centripetal acceleration. Centripetal acceleration is the acceleration of a body traversing a circular path.Using the formula for the centripetal acceleration in terms of velocity and radius, the centripetal acceleration as well as time period of rotation can be found.

Formula:

For example, if a body has velocity v and radius of a circle is r, thencentripetal acceleration is given as follow:

$a_c=\frac{v^2}{r}$ (i)

By definition, magnitude of centripetal acceleration is given by equation (i), then the centripetal acceleration can be written as,

$$\begin{gathered} a_c=\frac{{460}^2}{6370\times {10}^3} \\ =0.03\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the magnitude of centripetal acceleration is$0.03\mathrm{m}/{\mathrm{s}}^2$

Now, if $a{\text{'}}_c=9.8\frac{\mathrm{m}}{s^2}\mathrm{and}r=6370\times {10}^3\mathrm{m}$,then the speed of rotation can be determined as,

$$\begin{gathered} v{\text{'}}^2=a{\text{'}}_c\times r \\ =9.8\times 6370\times {10}^3\mathrm{m}/\mathrm{s} \\ \mathrm{v}\text{'}=7.9\times {10}^3\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the changed period of rotation will be,

$$\begin{gathered} T\text{'}=\frac{2\pi r}{v\text{'}} \\ =5066.32s \\ \approx 84\min \end{gathered}$$

Hence, the changed period of rotation is 84 min