The acceleration due to gravity on the earth's surface at the poles is g and angular velocity of the

Q: Why is gravity less at the equator?

Because the Earth is rotating, an object at the equator travels in a large circle and experiences an outward centrifugal force. This force partially counteracts the inward pull of gravity, making the effective gravity slightly less.

Q: Why doesn't rotation affect gravity at the poles?

The centrifugal force depends on the distance from the axis of rotation ($F_c = mr\omega^2$). At the poles, you are exactly on the axis of rotation, so $r = 0$, and the centrifugal force is zero.

Q: When can we use the binomial approximation $(1+x)^n \approx 1+nx$?

This approximation is valid only when the value of $x$ is much smaller than 1 ($x \ll 1$). In this problem, since $h \ll R$, the fraction $h/R$ is very small, making the approximation highly accurate.

Q: What if the problem didn't say $h \ll R$?

If $h$ wasn't small, we couldn't use the binomial approximation. We would have to solve the exact equation $g - R\omega^2 = g / (1 + h/R)^2$, which would lead to a more complex algebraic expression involving square roots.

Q: Does the mass of the object matter?

No. As you can see in the first step of the calculation, the mass $m$ appears on both sides of the equation and cancels out. This means a feather and a bowling ball would experience the same percentage change in weight.

Question

The acceleration due to gravity on the earth's surface at the poles is $g$ and angular velocity of the earth about the axis passing through the pole is $\omega$. An object is weighed at the equator and at a height $h$ above the poles by using a spring balance. If the weights are found to be same, then $h$ is ($h \ll R$, where $R$ is the radius of the earth)

Options:

(A) $\frac{R^2\omega^2}{2g}$

(B) $\frac{R^2\omega^2}{g}$

(C) $\frac{R^2\omega^2}{4g}$

(D) $\frac{R^2\omega^2}{8g}$

Sigma Percentile · Accepted Answer

This problem is a beautiful interplay of two distinct physical phenomena that alter the effective acceleration due to gravity: the Earth's rotation and altitude.

### Analyzing the Setup

Imagine you are standing on the Earth. If you are at the equator, the Earth is spinning beneath you. This rotation creates a centrifugal force that pushes you slightly outward, effectively reducing the pull of gravity. The effective gravity at the equator is given by:

$$g_{eq} = g - R\omega^2$$

where $g$ is the true gravity at the poles, $R$ is the Earth's radius, and $\omega$ is the angular velocity.

Now, imagine you travel to the North Pole and climb a very tall ladder to a height $h$. Because you are on the axis of rotation, you don't experience any centrifugal force. However, you are now further away from the center of the Earth. According to Newton's Law of Universal Gravitation, gravity weakens with distance. The gravity at a height $h$ is:

$$g_h = \frac{g}{\left(1 + \frac{h}{R}\right)^2} = g\left(1 + \frac{h}{R}\right)^{-2}$$

### The Master Equation

The problem states that a spring balance shows the same weight in both scenarios. Since weight is simply mass times the effective gravity ($W = mg_{eff}$), we can equate the two expressions we just found:

$$m(g - R\omega^2) = mg\left(1 + \frac{h}{R}\right)^{-2}$$

The mass $m$ cancels out immediately, reminding us of the equivalence principle—the trajectory (or in this case, the relative weight loss) is independent of the object's mass.

$$g - R\omega^2 = g\left(1 + \frac{h}{R}\right)^{-2}$$

### The Power of Approximation

Here is where we use a crucial piece of information given in the problem: $h \ll R$. The height $h$ is much smaller than the radius of the Earth. This allows us to use the **Binomial Approximation**, which states that for very small $x$, $(1 + x)^n \approx 1 + nx$.

Applying this to our equation, we get:

$$\left(1 + \frac{h}{R}\right)^{-2} \approx 1 - \frac{2h}{R}$$

Substituting this back into our master equation:

$$g - R\omega^2 = g\left(1 - \frac{2h}{R}\right)$$

### Final Calculation

Now, it's just a matter of simple algebra. Let's expand the right side:

$$g - R\omega^2 = g - \frac{2gh}{R}$$

The $g$ terms on both sides cancel each other out perfectly. The negative signs also cancel out, leaving us with:

$$R\omega^2 = \frac{2gh}{R}$$

Finally, we rearrange the terms to solve for $h$:

$$h = \frac{R^2\omega^2}{2g}$$

This elegant result shows exactly how high you need to climb at the pole to experience the same 'weight loss' as you would by simply standing at the spinning equator.

Select Answer:

Visualized Solution

\text{Visualizing the Setup}

\text{Weight at Equator}

\text{Weight at Height } h \text{ above Pole}

\text{Equating the Weights}

\text{Binomial Approximation}

\text{Simplifying the Equation}

\text{Solving for } h

\text{Final Expression for } h

\text{What if } h \text{ is large?}

The Sigma Insight: Acceleration due to Gravity and its Variation

Analyzing the Setup

The Master Equation

The Power of Approximation

Final Calculation

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