Question

Question #a288d

Answer 1

The answer is (c) decreases.

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 * 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_"centripetal" = m * a = m * omega * 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_"centrifugal" = m * omega * r^2`

![http://www.calctool.org/CALC/phys/newtonian/centrifugal]()

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) * g_(eq) = cancel(m) * g - cancel(m) * omega * r^2`

For future reference, `g_(eq)` is actually called the acceleration due to gravity at the equator.

As you can see, the magnitude of `g_(eq)` depends on `omega`, which is the angular speed of rotation.

This bigger `omega` gets, the smaller `g_(eq)` will become, since you're subtrating an increasingly bigger number, `omega * r^2`, from `g`.

As a conclusion, the faster the Earth spins, the lower `g_(eq)` will be, which in turn means that a person's weight decreases.