Animated Solution for Physics - Gravitation: A box weighs $196\text{ N}$ on a spring balance at the north pole. Its weight recorded on the same balance, if it is shifted to the equator is close to (Take, $g=10{\text{ ms}}^{-2}$ at the north pole and the radius of the earth $=6400\text{ km}$)

Gravity and Earth's Rotation

Imagine standing at the North Pole. You step onto a weighing scale, and it reads $196\text{ N}$. You might think this is your absolute weight, but weight is a tricky concept. It depends entirely on the local acceleration due to gravity.

At the poles, you are standing exactly on the axis of Earth's rotation. You are spinning like a top, but you aren't moving in a circle. Because your circular radius is zero, you experience absolutely zero centrifugal force. The only force acting on you is the pure, unadulterated gravitational pull of the Earth, $m{g}_p$.

From the given data, we can easily extract the true mass of the box. Since $W_p=m{g}_p$, and we are given $g_p=10{\text{ ms}}^{-2}$, the mass is simply:

Mass is the amount of matter in the box. Whether you take it to the equator, the moon, or deep space, this $19.6\text{ kg}$ will never change.

A Cosmic Merry-Go-Round

Now, let's teleport this box to the equator. Suddenly, the situation changes drastically. The Earth is rotating, and at the equator, you are at the maximum possible distance from the axis of rotation—a full $6400\text{ km}$ away!

Because you are moving in a massive circle every 24 hours, your body wants to fly off in a straight line due to inertia. From a rotating frame of reference, we feel this as an outward centrifugal force equal to $mR{\omega }^2$. This outward force directly opposes the inward pull of gravity.

Therefore, the weighing scale (which measures the normal reaction force) will read a slightly lower value. The apparent acceleration due to gravity at the equator becomes:

To find out exactly how much lighter the box gets, we need to calculate the centripetal acceleration term, $R{\omega }^2$. First, let's find the angular velocity $\omega$ of the Earth. The Earth completes one full rotation ($2\pi$ radians) in 24 hours.

Now, plugging this into our centripetal acceleration term along with the radius $R=6400\times 10^3\text{ m}$:

This might seem like a tiny number, but it's enough to make a measurable difference. The apparent gravity at the equator is now:

Finally, we calculate the new weight of the box at the equator by multiplying its constant mass by the new apparent gravity:

Looking at our options, the closest value is $195.32\text{ N}$. The box has effectively "lost" about $0.34\text{ N}$ of weight simply by moving from the pole to the equator. This beautiful interplay between gravity and rotational kinematics is a classic demonstration of how our physical reality is shaped by the motion of our planet.