If the radius of the earth shrinks by 3.5% ( mass remains constant) then how would the value of acceleration due to gravity change?

Explanation

To find out how the acceleration due to gravity changes when the radius of the Earth shrinks by 3.5% while keeping the mass constant, we can follow these steps:

### Step-by-Step Solution:

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

2. **Identify the Change in Radius**:
Given that the radius of the Earth shrinks by 3.5%, we can express this mathematically as:
\[ \Delta R = -0.035R \]
This means the new radius \( R' \) can be expressed as:
\[ R' = R - \Delta R = R(1 - 0.035) = 0.965R \]

3. **Calculate the Change in Acceleration due to Gravity**:
Since the mass \( M \) remains constant, we can substitute the new radius into the formula for \( g \):
\[ g' = \frac{GM}{(R')^2} = \frac{GM}{(0.965R)^2} \]
Simplifying this gives:
\[ g' = \frac{GM}{0.931225R^2} = \frac{g}{0.931225} \]

4. **Determine the Percentage Change in \( g \)**:
The percentage change in \( g \) can be calculated as:
\[ \frac{\Delta g}{g} = \frac{g' - g}{g} = \frac{\frac{g}{0.931225} - g}{g} = \frac{1 - 0.931225}{0.931225} \]
This simplifies to:
\[ \frac{\Delta g}{g} = \frac{0.068775}{0.931225} \approx 0.0738 \]
Converting this to a percentage:
\[ \Delta g \approx 7.38\% \]

5. **Conclusion**:
Therefore, the acceleration due to gravity increases by approximately 7.38% when the radius of the Earth shrinks by 3.5%.

### Final Answer:
The acceleration due to gravity increases by approximately 7.38%.

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