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# If earth were to rotate faster than its present speed, then weight of an object:A) Increase at the equator but remain unchanged at polesB) Decreases at the equator but remain unchanged at the polesC) Remain unchanged at the equator but decreases at the polesD) Remain unchanged at the equator but increases at the poles

Last updated date: 14th Sep 2024
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Hint: We know that the weight of a body depends on the gravitational force and other forces acting on the body. For this question, as the rotational speed of Earth increases, the value of acceleration due to gravity will change. This change is different at different points/locations on the surface of the Earth.

Complete step by step solution:
When the Earth rotates or for any rotating body, centrifugal force acts towards the center of the rotation. Since, the body is intact, that is there must be an equal force acting in the opposite direction. This force is called the centrifugal force. The direction of centrifugal force is opposite to the direction of centripetal force. Centrifugal force is pseudo force, it does not exist as there is no gravity, electrical force or magnetic force to bring about this force.

As the weight of a body $= m \times g$ .
Where $m$ is the mass of the body
$g$ is the acceleration due to gravity.
Then under the effect of increased rotational speed,
Weight of body will be $= mg' = mg - (centrifugal\,force)$
Where $g'$ is the decreased acceleration due to gravity.

For rotating Earth, the centrifugal acceleration $= R{\omega ^2}{\cos ^2}\theta$ where $R$ is the radius of the Earth, $\omega$ is the angular velocity and $\theta$ is the angle with the equator.

Remember that this angle is taken as follows:
A line joining the centre of the Earth and any point on the equator is taken as a reference line. Now the other line is drawn by joining the point where the object lies and the centre of the Earth.
$\theta$ is this angle between these two lines.
As you can see, for an object on the equator:
$\theta = 0$
$\Rightarrow \cos \theta = \cos {0^ \circ } = 1$
$\Rightarrow R{\omega ^2}{\cos ^2}\theta = R{\omega ^2}(1)$
$\Rightarrow g' = g - R{\omega ^2}$
Clearly we can see that for an object on the equator the acceleration due gravity $g'$ will decrease with increase in angular velocity $\omega$ . As a result, the weight of the object will also decrease if Earth was to rotate faster.
You can see that only option B satisfies our result. Still let’s check for the weight for an object at poles, North as well as South pole.
Let’s go back to the formula:
$g' = g - R{\omega ^2}{\cos ^2}\theta$
Now the angle with the line joining the equator and the line joining the pole to the centre is zero.
$\Rightarrow \cos \theta = \cos {90^ \circ } = 0$
$\Rightarrow R{\omega ^2}{\cos ^2}\theta = R{\omega ^2}(0) = 0$
This means that there is no acceleration due to centrifugal force for an object at the poles.
As a result, $g' = g$
Therefore, the weight of the body will remain unchanged.

Therefore, option (B) is the correct option.

Note: Remember that the angle of a body on the surface is taken with respect to the equator. Also be careful when interpreting the result, as angular speed increases the weight will decrease for a body on the equator. Also remember that centrifugal force is pseudo force (imaginary force) while centripetal force is a real force. The weight at both the poles will be the same.