What will be the acceleration due to gravity at a distance of 3200 km below the surface of the earth ? (Take `R_(e)=6400` km)

Explanation

To find the acceleration due to gravity at a distance of 3200 km below the surface of the Earth, we can use the formula for gravitational acceleration at a depth \( d \) below the surface:

\[ g_d = g \left(1 - \frac{d}{R_e}\right) \]

where:
- \( g_d \) is the acceleration due to gravity at depth,
- \( g \) is the acceleration due to gravity at the surface (approximately \( 9.8 \, \text{m/s}^2 \)),
- \( d \) is the depth below the surface,
- \( R_e \) is the radius of the Earth (approximately \( 6400 \, \text{km} \)).

### Step-by-step Solution:

1. **Identify the given values:**
- Depth \( d = 3200 \, \text{km} \)
- Radius of the Earth \( R_e = 6400 \, \text{km} \)
- Acceleration due to gravity at the surface \( g = 9.8 \, \text{m/s}^2 \)

2. **Convert the depth and radius into the same units:**
- Since both values are already in kilometers, we can use them directly.

3. **Substitute the values into the formula:**
\[ g_d = 9.8 \left(1 - \frac{3200}{6400}\right) \]

4. **Calculate the fraction:**
\[ \frac{3200}{6400} = 0.5 \]

5. **Substitute back into the equation:**
\[ g_d = 9.8 \left(1 - 0.5\right) = 9.8 \left(0.5\right) \]

6. **Calculate the final value:**
\[ g_d = 9.8 \times 0.5 = 4.9 \, \text{m/s}^2 \]

### Final Answer:
The acceleration due to gravity at a distance of 3200 km below the surface of the Earth is \( 4.9 \, \text{m/s}^2 \).