If the radius of the earth were to shrink by 1% its mass remaining the same, the acceleration due to gravity on the earth's surface would

Options

A. Increase by 2%
B. Decrease by 2%
C. Increase by 1%
D. Decrease by 1%

Answer

To solve the problem, we need to determine how the acceleration due to gravity (g) changes when the radius of the Earth shrinks by 1% while its mass remains constant.

### Step-by-Step Solution:

1. **Understanding the Formula for Acceleration due to Gravity**:
The formula for acceleration due to gravity at the surface of a planet is given by:
\[ g = \frac{GM}{r^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( r \) is the radius of the Earth.

2. **Identifying the Change in Radius**:
According to the problem, the radius of the Earth shrinks by 1%. This can be expressed mathematically as:
\[ \Delta r = -0.01r \]
Here, \( \Delta r \) is the change in radius (negative because it is a decrease).

3. **Calculating the Change in Acceleration due to Gravity**:
We need to find the percentage change in \( g \) due to the change in radius. The relationship between the change in \( g \) and the change in \( r \) can be derived from the formula:
\[ \frac{\Delta g}{g} = -2 \frac{\Delta r}{r} \]
This indicates that the change in \( g \) is inversely proportional to the square of the radius.

4. **Substituting the Change in Radius**:
Since we know that the radius decreases by 1%, we can substitute:
\[ \frac{\Delta r}{r} = -0.01 \]
Substituting this into the equation gives:
\[ \frac{\Delta g}{g} = -2(-0.01) = 0.02 \]
This means that the percentage change in \( g \) is:
\[ \Delta g = 0.02g \]

5. **Interpreting the Result**:
The positive value indicates that \( g \) increases. Therefore, the acceleration due to gravity increases by 2%.

### Final Answer:
The acceleration due to gravity on the Earth's surface would **increase by 2%**.

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