Let's analyze how the acceleration due to gravity changes when the Earth's radius shrinks but its mass stays the same. The formula for the acceleration due to gravity \(g\) on the surface of a spherical body like Earth is given by:

$$ g = \frac{GM}{R^2} $$

Where:

• \(G\) is the universal gravitational constant.
• \(M\) is the mass of the Earth.
• \(R\) is the radius of the Earth.

In this problem, we are told that the mass of the Earth (\(M\)) remains the same, and the gravitational constant (\(G\)) is always constant. The radius of the Earth (\(R\)) is said to shrink by 1%.

Let the original radius be \(R_1\). The original acceleration due to gravity is:

$$ g_1 = \frac{GM}{R_1^2} $$

The new radius \(R_2\) is 1% less than the original radius \(R_1\). So, \(R_2 = R_1 - 1\% \text{ of } R_1\).

$$ R_2 = R_1 - \frac{1}{100} R_1 = R_1 - 0.01 R_1 = (1 - 0.01) R_1 = 0.99 R_1 $$

The new acceleration due to gravity \(g_2\) with the new radius \(R_2\) and the same mass \(M\) is:

$$ g_2 = \frac{GM}{R_2^2} $$

Now, substitute \(R_2 = 0.99 R_1\) into the equation for \(g_2\):

$$ g_2 = \frac{GM}{(0.99 R_1)^2} = \frac{GM}{(0.99)^2 R_1^2} = \frac{1}{(0.99)^2} \frac{GM}{R_1^2} $$

We know that \( g_1 = \frac{GM}{R_1^2} \). So, we can write \(g_2\) in terms of \(g_1\):

$$ g_2 = \frac{1}{(0.99)^2} g_1 $$

Let's calculate \((0.99)^2\):

$$ (0.99)^2 = (1 - 0.01)^2 = 1^2 - 2(1)(0.01) + (0.01)^2 = 1 - 0.02 + 0.0001 = 0.9801 $$

So, the new gravity is:

$$ g_2 = \frac{1}{0.9801} g_1 $$

To find the percentage change in gravity, we calculate \( \frac{g_2 - g_1}{g_1} \times 100\% \).

$$ \frac{g_2 - g_1}{g_1} = \frac{\frac{1}{0.9801} g_1 - g_1}{g_1} = \frac{g_1 \left( \frac{1}{0.9801} - 1 \right)}{g_1} = \frac{1}{0.9801} - 1 $$

Now calculate the value:

$$ \frac{1}{0.9801} - 1 \approx 1.02039 - 1 = 0.02039 $$

The percentage change is approximately \(0.02039 \times 100\% = 2.039\%\).

This means the acceleration due to gravity increases by approximately 2.039% when the radius shrinks by 1%.

However, the question asks by how much \(g\) would "decrease". Our calculation shows an increase. Given the options, the closest value to the magnitude of the change (which is about 2%) is option 2 (2%). This suggests the question might be asking for the approximate percentage change, and the word "decrease" might be an error in the question phrasing or implies a magnitude of change relative to the original value.

Let's consider the approximation for small changes. If \(g \propto R^{-2}\), then for a small percentage change \( \frac{\Delta R}{R} \), the percentage change in \(g\) is approximately \( -2 \times \frac{\Delta R}{R} \). Here, \( \frac{\Delta R}{R} = -1\% = -0.01 \) (since the radius shrinks). So, the percentage change in \(g\) is approximately \( -2 \times (-0.01) = +0.02 \), which means a 2% increase.

Both the exact calculation (approx 2.039% increase) and the approximation for small changes (approx 2% increase) show that gravity increases. However, the magnitude of this change is approximately 2%.

Based on the options provided and the calculated magnitude of the change in gravity, the nearest value is 2%.

Understanding Acceleration Due to Gravity

Acceleration due to gravity, often denoted as \(g\), is the acceleration experienced by an object falling freely near the surface of a massive body like Earth. It depends on the mass of the body and the distance from its center. The formula \(g = \frac{GM}{R^2}\) shows that \(g\) is directly proportional to the mass (\(M\)) and inversely proportional to the square of the radius (\(R\)) (or distance from the center).

• If mass \(M\) increases, \(g\) increases.
• If distance \(R\) increases, \(g\) decreases.
• If distance \(R\) decreases, \(g\) increases.

Calculating Percentage Change in Gravity

When dealing with small percentage changes, we can often use approximations based on calculus or binomial expansion. For a function like \( y = x^n \), a small fractional change \( \frac{\Delta x}{x} \) results in a fractional change \( \frac{\Delta y}{y} \approx n \frac{\Delta x}{x} \). In our case, \( g \propto R^{-2} \), so \(n=-2\). If the radius changes by \( \Delta R \), the change in gravity \( \Delta g \) is approximately:

$$ \frac{\Delta g}{g} \approx -2 \frac{\Delta R}{R} $$

If the radius shrinks by 1%, \( \frac{\Delta R}{R} = -0.01 \). $$ \frac{\Delta g}{g} \approx -2 \times (-0.01) = +0.02 $$

This indicates a fractional increase of 0.02, or a 2% increase. This approximate method gives a result very close to the exact calculation for small changes.

Revision Table: Key Gravitation Concepts

Additional Information: Impact of Radius and Mass on Gravity

The value of acceleration due to gravity \(g\) is a fundamental property determining the weight of objects and the behavior of falling bodies on a planet's surface. It depends directly on the planet's mass and inversely on the square of its radius. This inverse square relationship means that changes in radius have a more significant impact on \(g\) than linear changes in mass (for the same percentage change).

For example, if the Earth's mass doubled (\(M \to 2M\)) with the same radius, \(g\) would double (\(g \to 2g\)). But if the Earth's radius halved (\(R \to R/2\)) with the same mass, \(g\) would become four times stronger (\(g \to \frac{GM}{(R/2)^2} = \frac{GM}{R^2/4} = 4 \frac{GM}{R^2} = 4g\)).

In this problem, the radius shrinks, meaning the surface is closer to the center of the Earth. According to the inverse square law, bringing the surface closer to the mass center increases the gravitational acceleration at the surface, assuming the mass distribution remains effectively constant within the new radius (which is implied by the mass remaining the same). Thus, a shrinking radius leads to an increase in \(g\).

The magnitude of the change calculated, approximately 2%, aligns with one of the given options, even though the direction implied by the word "decrease" in the question seems contrary to the physics.