# What would the period of rotation of Earth have to be for objects on the equator to have a centripetal acceleration with a magnitude of 9.80 ms^-2?

May 23, 2016

Fascinating question! See the calculation below, which shows that the rotational period would be $1.41$ $h$.

#### Explanation:

To answer this question we need to know the diameter of the earth. From memory it's about $6.4 \times {10}^{6}$ $m$. I looked it up and it averages $6371$ $k m$, so if we round it to two significant figures my memory is right.

The centripetal acceleration is given by $a = {v}^{2} / r$ for linear velocity, or $a = {\omega}^{2} r$ for rotational velocity. Let's use the latter for convenience.

Remember that we know the acceleration we want and the radius, and need to know the period of rotation. We can start with the rotational velocity:

$\omega = \sqrt{\frac{a}{r}} = \sqrt{\frac{9.80}{6.4 \times {10}^{6}}} = 0.00124$ $r a {\mathrm{ds}}^{-} 1$

To find the rotational period, we need to invert this to give $\text{seconds"/"radian}$, then multiply by $2 \pi$ to get seconds per full rotation (since there are $2 \pi$ radians in a full rotation).

This yields $5077.6$ $s {\text{rotation}}^{-} 1$.

We can divide this by 3600 to convert to hours, and find $1.41$ hours. This is much faster than the current period of $24$ $h$.