91Ó°ÊÓ

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

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,

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

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

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

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

Hence, the changed period of rotation will be,

$$\begin{gathered} T\text{'}=\frac{2\mathrm{π}r}{v\text{'}} \\ =5066.32s \\ ≈84\min \end{gathered}$$

Hence, the changed period of rotation is 84 min