A body weights 49 N on a spring balance at the North pole. What will be its weight recorded on the s

Q: Why is there no centrifugal force at the poles?

At the poles, your distance from the axis of rotation is zero. Since centrifugal force is $mr\omega^2$ (where $r$ is the distance from the axis), a radius of zero means zero centrifugal force.

Q: Does the mass of the body change when moved to the equator?

Absolutely not! Mass is an intrinsic property of the body, representing the amount of matter. It remains $5 \text{ kg}$ everywhere. Only the *weight* (the net force exerted on the weighing scale) changes.

Q: Why do we use $\omega = 2\pi / T$ and what is $T$?

$\omega$ is the angular velocity, which is the angle covered per unit time. The Earth covers one full circle ($2\pi$ radians) in one day ($T = 24 \text{ hours}$). So, $\omega = 2\pi / (24 \times 3600)$ radians per second.

Q: Why did we convert the radius $R$ from kilometers to meters?

In physics, we must always use standard SI units to ensure our final answer is in standard units (Newtons). Since $g$ is in $\text{m/s}^2$, $R$ must be in meters ($6.4 \times 10^6 \text{ m}$).

Q: Could the weight at the equator ever be greater than at the poles?

No, not due to rotation. Rotation always creates an outward centrifugal force that opposes gravity. The only way weight could be greater at the equator is if the Earth were shaped very strangely, but for a standard oblate spheroid, gravity is always weakest at the equator.

Question

A body weights $49	ext{ N}$ on a spring balance at the North pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator ?
(Use, $g = \frac{GM}{R^2} = 9.8	ext{ ms}^{-2}$ and radius of earth, $R = 6400	ext{ km}$)

Options:

(A) $49	ext{ N}$

(B) $48.83	ext{ N}$

(C) $49.83	ext{ N}$

(D) $49.17	ext{ N}$

Sigma Percentile · Accepted Answer

### The Rotating Earth: A Giant Merry-Go-Round

Imagine standing at the North Pole. You are perfectly aligned with the Earth's axis of rotation. You are spinning, yes, but you are not moving in a circle. You are just pirouetting on the spot. Here, the only force you feel is the pure, unadulterated pull of Earth's gravity.

Now, teleport yourself to the equator. Suddenly, you are on the outer edge of a giant merry-go-round, moving at over 1600 kilometers per hour! Because you are moving in a massive circle, your body wants to fly off in a straight line due to inertia. This tendency is what we feel as an outward **centrifugal force**.

### The Master Equation of Effective Gravity

Because of this outward centrifugal force, the net force pulling you towards the center of the Earth is slightly reduced. We call this reduced pull the **effective gravity**, denoted by $g'$.

The formula that governs this is:
$$g' = g - R\omega^2 \cos^2\lambda$$

Here, $\lambda$ is your latitude. At the North Pole, $\lambda = 90^\circ$, so the cosine term vanishes. You feel the full force of gravity. But at the equator, $\lambda = 0^\circ$, and the cosine term is at its maximum. The effective gravity is at its weakest!

### Setting Up the Problem

The problem states that a body weighs $49 	ext{ N}$ at the North Pole. This is its true weight, $w_p = mg$.

At the equator, the weight $w_e$ will be:
$$w_e = m(g - R\omega^2) = mg - mR\omega^2$$

We already know $mg = 49 	ext{ N}$. To find the weight at the equator, we just need to calculate that tiny outward centrifugal term, $mR\omega^2$, and subtract it.

### The Atomic Computation

First, let's find the mass of the body. Since $mg = 49 	ext{ N}$ and $g = 9.8 	ext{ m/s}^2$, we can easily find $m$:
$$m = \frac{49}{9.8} = 5 	ext{ kg}$$

Next, we need the angular velocity of the Earth, $\omega$. The Earth completes one full rotation ($2\pi$ radians) in 24 hours.
$$\omega = \frac{2\pi}{24 	imes 3600 	ext{ s}} \approx 7.27 	imes 10^{-5} 	ext{ rad/s}$$

Now, let's calculate the centrifugal force. The radius of the Earth is $R = 6400 	ext{ km} = 6.4 	imes 10^6 	ext{ m}$.
$$F_c = mR\omega^2 = 5 	imes (6.4 	imes 10^6) 	imes (7.27 	imes 10^{-5})^2$$

If you crunch these numbers, you get:
$$F_c \approx 0.17 	ext{ N}$$

### The Final Answer and a Smart Trick

Subtracting this outward force from the true weight gives us the weight at the equator:
$$w_e = 49 - 0.17 = 48.83 	ext{ N}$$

This perfectly matches option (b).

But wait! Did we really need to do all that math? Look at the physics. We established that the weight at the equator **must be strictly less** than the weight at the poles because of the centrifugal force.

The weight at the pole is $49 	ext{ N}$. Let's look at the options:
(a) $49 	ext{ N}$
(b) $48.83 	ext{ N}$
(c) $49.83 	ext{ N}$
(d) $49.17 	ext{ N}$

Options (a), (c), and (d) are all equal to or greater than $49 	ext{ N}$. Only option (b) is less than $49 	ext{ N}$! In a competitive exam like JEE, spotting these logical constraints can save you precious minutes. Always let the physics guide your math!

### The Way Forward

What if the Earth started spinning faster? The $\omega$ term would increase, making the centrifugal force stronger. Your weight at the equator would continue to drop. If the Earth spun fast enough, the centrifugal force would perfectly cancel out gravity, and you would be completely weightless! This is exactly what happens to astronauts in orbit. Physics is beautifully consistent!

Animated Solution for Physics - Gravitation: A body weights $49 N$ on a spring balance at the North pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator ? (Use, $g = \frac{GM}{R ^{2}} = 9.8 ms^{- 2}$ and radius of earth, $R = 6400 km$ )

Select Answer:

Visualized Solution

$Earth’s Rotation$

$g^{'} = g - R ω^{2} cos^{2} λ$

$w_{e} = m g - m R ω^{2}$

$m = 5 kg$

$F_{c} \approx 0.17 N$

$w_{e} = 48.83 N$

$g = R ω^{2}$

The Sigma Insight: Acceleration due to Gravity and its Variation

The Rotating Earth

The Master Equation of Effective Gravity

Setting Up the Problem

The Atomic Computation

The Final Answer and a Smart Trick

The Way Forward

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