The radius of earth is increased by 2%. By what precentage the acceleration due to gravity will change assuming mass remaining unchanged ?

Explanation

To solve the problem of how the acceleration due to gravity changes when the radius of the Earth is increased by 2%, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Acceleration Due to Gravity**:
The acceleration due to gravity \( g \) at the surface of the Earth is given by the formula:
\[ g = \frac{G M}{R^2} \]
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**:
We are given that the radius of the Earth is increased by 2%. This can be expressed mathematically as:
\[ \Delta R = 0.02 R \]
Hence, the new radius \( R' \) becomes:
\[ R' = R + \Delta R = R + 0.02 R = 1.02 R \]

3. **Substitute the New Radius into the Gravity Formula**:
The new acceleration due to gravity \( g' \) can be calculated as:
\[ g' = \frac{G M}{(R')^2} = \frac{G M}{(1.02 R)^2} \]
Simplifying this gives:
\[ g' = \frac{G M}{1.0404 R^2} = \frac{g}{1.0404} \]

4. **Calculate the Percentage Change in Acceleration Due to Gravity**:
The percentage change in \( g \) can be calculated using the formula:
\[ \text{Percentage Change} = \left(\frac{g' - g}{g}\right) \times 100\% \]
Substituting \( g' \):
\[ \text{Percentage Change} = \left(\frac{\frac{g}{1.0404} - g}{g}\right) \times 100\% \]
This simplifies to:
\[ \text{Percentage Change} = \left(\frac{1 - 1.0404}{1.0404}\right) \times 100\% \]
\[ = \left(-0.0404\right) \times 100\% \approx -4.04\% \]

5. **Conclusion**:
Therefore, the acceleration due to gravity decreases by approximately 4.04% when the radius of the Earth is increased by 2%.

### Final Answer:
The acceleration due to gravity decreases by approximately **4%**.