Find the acceleration due to gravity at a point 64km below the surface of the earth. R = 6400 k, g on the surface of the earth `= 9.8 ms^(-2)`.

Explanation

To find the acceleration due to gravity at a point 64 km below the surface of the Earth, we can use the formula for the variation of gravity with depth:

\[ g' = g \left(1 - \frac{d}{R}\right) \]

Where:
- \( g' \) = acceleration due to gravity at depth \( d \)
- \( g \) = acceleration due to gravity on the surface of the Earth
- \( d \) = depth below the surface
- \( R \) = radius of the Earth

### Step-by-Step Solution:

1. **Identify the given values**:
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity on the surface)
- \( d = 64 \, \text{km} = 64 \times 10^3 \, \text{m} \) (depth below the surface)
- \( R = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \) (radius of the Earth)

2. **Substitute the values into the formula**:
\[ g' = 9.8 \left(1 - \frac{64 \times 10^3}{6400 \times 10^3}\right) \]

3. **Calculate the fraction**:
\[ \frac{64 \times 10^3}{6400 \times 10^3} = \frac{64}{6400} = \frac{1}{100} = 0.01 \]

4. **Substitute back into the equation**:
\[ g' = 9.8 \left(1 - 0.01\right) = 9.8 \times 0.99 \]

5. **Perform the multiplication**:
\[ g' = 9.8 \times 0.99 = 9.702 \, \text{m/s}^2 \]

6. **Round the result**:
\[ g' \approx 9.7 \, \text{m/s}^2 \]

### Final Answer:
The acceleration due to gravity at a point 64 km below the surface of the Earth is approximately \( 9.7 \, \text{m/s}^2 \).

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