How long would a day be if the Earth were rotating so fast that objects at the equator were apparently weightless?

An object faces the condition of weightlessness when its centripetal acceleration equals the acceleration due to gravity.

The velocity of rotation can be calculated as,

\(v = \frac{{2\pi {R_{{\rm{Earth}}}}}}{T}\)

Here, \({R_{Earth}}\) is the radius of the Earth and its value is \(6.38 \times {10^6}\;{\rm{m}}\) and \(T\) is the time period.

The acceleration due to gravity is given by,

\(\begin{aligned}g &= \frac{{{v^2}}}{{{R_{{\rm{Earth}}}}}}\\g &= \frac{{4{\pi ^2}{R_{{\rm{Earth}}}}}}{{{T^2}}}\\T &= 2\pi \sqrt {\frac{{{R_{{\rm{Earth}}}}}}{g}} \end{aligned}\)

Substitute the values in the above equation,

\(\begin{aligned}T &= 2\pi \sqrt {\frac{{6.38 \times {{10}^6}\;{\rm{m}}}}{{9.8\;{\rm{m/}}{{\rm{s}}^2}}}} \\ &= 5.07 \times {10^3}\;{\rm{s}}\left( {\frac{{1\;{\rm{min}}}}{{60\;{\rm{s}}}}} \right)\\ &= 84.5\;{\rm{min}}\end{aligned}\)

Thus, the length of the day is \(84.5\;min\).