Question

What would the length of the day be if the earth rotated so rapidly on its axis that objects at the equator were weightless?

Answer 1

In order to solve this question, first we will equate gravitational force and centrifugal force. Then we will replace the value of velocity with the ratio of distance and time and then on putting the required values, we will get the final answer.

Complete Step By Step Answer:
For an object to be weightless at the equator, the gravitational force must be equal to the centrifugal force. This can be written as,
Gravitational force = Centrifugal force
On putting the value of gravitational force and the centrifugal force, we get,
$mg = \dfrac{{m{v^2}}}{R}$
On cancelling $m$ on both the sides, we get,
$g = \dfrac{{{v^2}}}{R}$
Now, on rearranging the terms, we get,
${v^2} = gR$
On taking square root on both the sides, we get,
$v = \sqrt {gR} ......(1)$
We know that $v = \dfrac{d}{t}$
Also, in this question the motion is circular in nature. So, the distance will be the circumference of the circle.
So, $v = \dfrac{{2\pi R}}{T}........(2)$
On equating equation (1) and equation (2), w get,
$\dfrac{{2\pi R}}{T} = \sqrt {gR}$
On taking $T$ on one side and all the other terms on the other side, we get,
$T = \dfrac{{2\pi R}}{{\sqrt {gR} }}$
We know that,
The radius of the earth $R = 6400000m$
On putting this value in the above equation, we get,
$T = \dfrac{{2 \times 3.14 \times 6400000}}{{\sqrt {10 \times 6400000} }}$
On further solving, we get,
$T = \dfrac{{2 \times 3.14 \times 6400000}}{{\sqrt {64000000} }}$
$T = \dfrac{{2 \times 3.14 \times 6400000}}{{8000}}$
On cancelling the common terms in the numerator and the denominator, we get,
$T = 2 \times 3.14 \times 800$
$T = 5024s$
Now, we will convert seconds into minutes,
$T = \dfrac{{5024}}{{60}}\;minutes$
$T = 84\min$
So, 84 minutes should be the length of the day if the earth rotated so rapidly on its axis that objects at the equator were weightless.

Note:
The weight of the objects at the pole is slightly more than the weight of the object at the Equator. This is because the polar radius of the Earth is less as compared to the equatorial radius. The mass of the object remains the same, but the object’s weight changes as its location changes.