At the equator, a person stands on a spring balance. If the speed...

The apparent weight of a person at the equator is given by $W' = mg - m\omega^2 R$, where $m$ is the mass, $g$ is the acceleration due to gravity, $\omega$ is the angular velocity of the earth, and $R$ is the radius of the earth.

Given that the new apparent weight is half of the true weight, $W' = \dfrac{mg}{2}$.

Substituting this into the equation:

$\dfrac{mg}{2} = mg - m\omega^2 R$

$\Rightarrow m\omega^2 R = \dfrac{mg}{2}$

$\Rightarrow \omega = \sqrt{\dfrac{g}{2R}}$

The new length of the day $T$ is given by $T = \dfrac{2\pi}{\omega}$.

T = 2\pi \sqrt{\dfrac{2R}{g}} = \sqrt{2} \times \left( 2\pi\sqrt{\dfrac{R}{g}} \right)

The quantity $2\pi$\sqrt{\dfrac{R}{g}} represents the time period of a satellite orbiting close to the earth's surface, which is approximately $84.6 \text{ minutes}$.

$T \approx 1.414 \times 84.6 \text{ minutes} \approx 119.6 \text{ minutes}$

Converting minutes to hours:

$T \approx$\dfrac{120}{60} \text{ hours} =$2.0 \text{ hours}$.