Weightless

We have to first realize that by weight-less, we are referring to the apparent weight , which is what is obtained on a scale. The real weight force $mg$ will always be there. The spinning of the Earth causes this discrepancy as it is a non-inertial reference frame.

Since the Earth is spinning there must be a net force inwards(centripetal force). This is caused by the imbalance between the real weight $w$ and the normal force of the Earth on the object $N$ . This normal force $N$ is what a scale measures, thus $W_{apparent} = N$

diag . We start as always with $\sum F = ma$ This net force is centripetal $\sum F =\frac{mv^2}{r}$ From the diagram above $w - N = \frac{mv^2}{R_{E}}$ $w - W_{apparent} = \frac{mv^2}{R_{E}}$ $W_{apparent} = w - \frac{mv^2}{R_{E}}$ When setting $W_{apparent}$ to $0$ and solving for $v^2$ we get. $v^2 = gR_{E}$ Substituting $v = \frac{2 \pi R_{E}}{T}$ we get $T = 2 \pi \sqrt{\frac{R_{E}}{g}}$

To the equator the centrifugal acceleration is:

$a_{cf}={\omega}^2R,\,$ with $\;\omega ={\frac {2\pi} T},\,T\,$ period of rotation;

the hypotesis that any object is weight-less on the equator implies: $|a_{cf}|=|g|$

thus: $\,{\omega}^2R=g,\,\omega = \sqrt {\frac g R},\,T=\frac {2\pi} \omega =2\pi\sqrt {\frac R g}\equiv\frac {2\pi}{60}\sqrt {\frac R g},\,$ in minutes, hence substituting values:

$T=\frac {2\pi}{60}\sqrt \frac {6 \times {10^6}}{9.8}=\boxed {81.94}\,$ minutes