If the radius of the earth were to shrink by one per cent, its mass remaining the same, the accelera

Q: Why does gravity increase when the Earth shrinks? Shouldn't a smaller planet have less gravity?

A smaller planet has less gravity only if its mass also decreases. In this problem, the mass of the Earth remains exactly the same. Because the mass is packed into a smaller volume, any object on the surface is now closer to the Earth's center of mass. Since gravitational force is inversely proportional to the square of the distance, being closer to the center results in a stronger gravitational pull.

Q: Can we use the approximation formula $\\frac{dg}{g} = -2 \\frac{dR}{R}$ for large percentage changes, like if the radius shrinks by $50\%$?

No, the differential approximation is only valid for very small percentage changes (typically less than $5\%$ or $10\%$). For large changes, you must calculate the new gravity directly using $g' = \\frac{GM}{(R')^2}$ and compare it to the original value.

Q: What is the difference between mass being constant and density being constant in this context?

If mass is constant, shrinking the radius brings you closer to the center of the same mass, increasing gravity ($g \\propto 1/R^2$). If density is constant, shrinking the radius means the planet loses a massive amount of volume and therefore mass. The loss of mass dominates, causing gravity to decrease ($g \\propto R$).

Q: Why is the fractional change formula negative: $\\frac{dg}{g} = -2 \\frac{dR}{R}$?

The negative sign represents the inverse relationship between gravity and radius. It mathematically indicates that if the radius decreases (negative change), the gravity must increase (positive change), and vice versa.

Q: Does the mass of the test object on the surface affect the acceleration due to gravity?

No, the acceleration due to gravity $g = \\frac{GM}{R^2}$ depends only on the mass of the planet $M$ and its radius $R$. The mass of the test object $m$ cancels out, which is why all objects fall with the same acceleration in a vacuum.

Question

If the radius of the earth were to shrink by one per cent, its mass remaining the same, the acceleration due to gravity on the earth's surface would

Options:

(A) decrease

(B) remain unchanged

(C) increase

(D) be zero

Sigma Percentile · Accepted Answer

### Introduction to Gravitational Fields

Imagine standing on the surface of a massive planet, feeling the invisible pull that anchors you to the ground. This force, which we call gravity, is one of the fundamental interactions shaping our universe.

In this problem, we explore a fascinating thought experiment: what happens to the acceleration due to gravity on the Earth's surface if the planet were to suddenly shrink by $1\%$ of its radius, while somehow keeping its entire mass perfectly intact?

This question, originally asked in the JEE in 1981, is a beautiful test of physical intuition and basic calculus. Let's dive deep into the physics and mathematics behind it.

### The Governing Law of Gravity

To understand how gravity changes, we must look at the mathematical description of this force. Newton's Law of Universal Gravitation states that the force of attraction $F$ between two spherical masses, $M$ (the Earth) and $m$ (a test mass on the surface), is given by:

$$F = \frac{G M m}{R^2}$$

where $G$ is the universal gravitational constant, and $R$ is the distance between their centers (which is simply the radius of the Earth when the test mass is on the surface).

According to Newton's second law, the acceleration experienced by this test mass is the acceleration due to gravity, $g$. We find this by dividing the force by the mass $m$:

$$g = \frac{F}{m} = \frac{G M}{R^2}$$

This simple yet powerful equation tells us that the local strength of gravity depends on two main factors: the mass of the planet and how close you can get to its center of mass.

### Analyzing the Proportionality

In our problem, we are given a crucial constraint: **the mass of the Earth remains the same**. Since $G$ is a universal constant and $M$ is kept constant, the numerator of our equation is completely fixed.

This allows us to write a direct proportionality relation:

$$g \propto \frac{1}{R^2}$$

This is the famous **inverse-square law**. It reveals that gravity is highly sensitive to the radius of the planet.

If the radius decreases, the denominator gets smaller, which means the value of $g$ must increase. Conversely, if the planet expands, gravity weakens. Since the Earth is shrinking in our scenario, we can immediately rule out any options suggesting that gravity decreases or remains unchanged. The answer must be that gravity **increases**.

### Calculating the Exact Change

To find out exactly how much gravity increases, we can use a powerful mathematical tool from calculus: **logarithmic differentiation** (often used in physics for error and percentage change analysis).

Let's write our gravity equation in exponent form:

$$g = G M R^{-2}$$

Taking the natural logarithm ($\ln$) on both sides, we get:

$$\ln(g) = \ln(G M) - 2 \ln(R)$$

Now, we differentiate both sides. Since $G$ and $M$ are constant, the term $\ln(G M)$ is a constant, and its derivative is zero:

$$\frac{dg}{g} = 0 - 2 \frac{dR}{R}$$

$$\frac{dg}{g} = -2 \frac{dR}{R}$$

This elegant equation relates the fractional change in gravity directly to the fractional change in radius. The factor of $-2$ is the key here—it tells us that any percentage change in the radius will result in a percentage change in gravity that is **twice as large** and in the **opposite direction**.

### Substituting the Values

We are told that the radius of the Earth shrinks by $1\%$. A shrink represents a decrease, so we write:

$$\frac{dR}{R} = -1\% = -0.01$$

Substituting this value into our fractional change equation:

$$\frac{dg}{g} = -2 (-1\%) = +2\%$$

Because the result is positive, it indicates an **increase**. Therefore, if the Earth's radius shrinks by $1\%$ while its mass remains constant, the acceleration due to gravity on its surface will **increase by approximately $2\%$**.

This perfectly matches **Option (c)**.

### The Classic Trap: Constant Density vs. Constant Mass

In competitive exams like JEE, examiners love to twist this question to catch students off guard. What if, instead of keeping the mass constant, the question stated that the **density of the Earth remains constant** as it shrinks?

Let's analyze this scenario. Density $\rho$ is mass divided by volume:

$$\rho = \frac{M}{V} = \frac{M}{\frac{4}{3} \pi R^3}$$

This means the mass of the Earth can be written as:

$$M = \rho \cdot \frac{4}{3} \pi R^3$$

If we substitute this expression for mass back into our formula for $g$, we get:

$$g = \frac{G M}{R^2} = \frac{G \left( \rho \cdot \frac{4}{3} \pi R^3 \right)}{R^2}$$

$$g = \frac{4}{3} \pi G \rho R$$

Under the condition of constant density, we find that:

$$g \propto R$$

In this case, gravity is **directly proportional** to the radius! If the Earth were to shrink under constant density, its gravity would actually **decrease** because the loss of mass outweighs the effect of getting closer to the center.

Always read the question carefully to identify what is being held constant!

Animated Solution for Physics - Gravitation: If the radius of the earth were to shrink by one per cent, its mass remaining the same, the acceleration due to gravity on the earth's surface would

Select Answer:

Visualized Solution

Visualizing the Shrinking Earth

Newton's Law of Gravitation

Formula for Acceleration due to Gravity

Establishing Proportionality

Using Logarithmic Differentiation

Substituting the Percentage Change

Calculating the Final Change

The Way Forward

The Sigma Insight: Acceleration due to Gravity and its Variation

Introduction to Gravitational Fields

The Governing Law of Gravity

Analyzing the Proportionality

Calculating the Exact Change

Substituting the Values

The Classic Trap

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