# Question #a288d

Jul 17, 2015

#### Explanation:

The idea behind this problem is that Earth's roration produces a centripetal acceleration, which is directed towards the center of rotation, that reduces the gravitational acceleration, $g$.

This centripetal acceleration is defined by the equation

$a = \omega \cdot {r}^{2}$, where

$\omega$ - the angular velocity of the Earth;
$r$ - the radius of the Earth.

The force that creates this centripetal acceleration is called the centripetal force and is pointed towards the center of ration as well.

${F}_{\text{centripetal}} = m \cdot a = m \cdot \omega \cdot {r}^{2}$

Now, an object that is in a rotating reference frame, i.e. a reference frame that is rotating with the system, will be acted upon by the centrifugal force.

This "force" (it is not a force per se) points outwards from the center of rotation and actually replaces the centripetal force.

${F}_{\text{centrifugal}} = m \cdot \omega \cdot {r}^{2}$

This means that, for a person that's rotating at the equator, this centifugal force will actually oppose gravity.

This means that the weight of the person at the equator will actually be

$\cancel{m} \cdot {g}_{e q} = \cancel{m} \cdot g - \cancel{m} \cdot \omega \cdot {r}^{2}$

For future reference, ${g}_{e q}$ is actually called the acceleration due to gravity at the equator.

As you can see, the magnitude of ${g}_{e q}$ depends on $\omega$, which is the angular speed of rotation.

This bigger $\omega$ gets, the smaller ${g}_{e q}$ will become, since you're subtrating an increasingly bigger number, $\omega \cdot {r}^{2}$, from $g$.

As a conclusion, the faster the Earth spins, the lower ${g}_{e q}$ will be, which in turn means that a person's weight decreases.