At what depth below the surface of the earth acceleration due to gravity will be half its value at 1600 km above the surface of the earth ?

Options

A. `4.3 xx 10^(6)` m
B. `2.4 xx 10^(6)` m
C. `3.2 xx 10^(6)` m
D. `1.6 xx 10^(6)` m

Explanation

To solve the problem of finding the depth below the Earth's surface where the acceleration due to gravity is half of its value at a height of 1600 km above the surface, we can follow these steps:

### Step 1: Understand the formula for gravitational acceleration
The acceleration due to gravity at a distance \( r \) from the center of the Earth is given by:
\[ g' = g \left(1 - \frac{d}{R}\right) \]
where:
- \( g' \) is the acceleration due to gravity at depth \( d \),
- \( g \) is the acceleration due to gravity at the surface,
- \( R \) is the radius of the Earth,
- \( d \) is the depth below the surface.

### Step 2: Calculate the value of gravity at 1600 km above the surface
The acceleration due to gravity at a height \( h \) above the surface is given by:
\[ g_h = g \left( \frac{R}{R + h} \right)^2 \]
For \( h = 1600 \) km (or \( 1.6 \times 10^6 \) m) and \( R \approx 6400 \) km (or \( 6.4 \times 10^6 \) m):
\[ g_h = g \left( \frac{6400}{6400 + 1600} \right)^2 = g \left( \frac{6400}{8000} \right)^2 = g \left( \frac{4}{5} \right)^2 = g \left( \frac{16}{25} \right) \]

### Step 3: Set up the equation for when gravity is half of \( g_h \)
We want to find the depth \( d \) such that:
\[ g' = \frac{1}{2} g_h \]
Substituting the expression for \( g_h \):
\[ g \left(1 - \frac{d}{R}\right) = \frac{1}{2} \left( g \frac{16}{25} \right) \]
Cancelling \( g \) from both sides:
\[ 1 - \frac{d}{R} = \frac{8}{25} \]

### Step 4: Solve for \( d \)
Rearranging the equation:
\[ \frac{d}{R} = 1 - \frac{8}{25} = \frac{17}{25} \]
Thus,
\[ d = R \cdot \frac{17}{25} \]
Substituting \( R = 6.4 \times 10^6 \) m:
\[ d = 6.4 \times 10^6 \cdot \frac{17}{25} = 4.352 \times 10^6 \text{ m} \]

### Final Answer
The depth below the surface of the Earth where the acceleration due to gravity is half its value at 1600 km above the surface is approximately:
\[ d \approx 4.352 \times 10^6 \text{ m} \text{ or } 4352 \text{ km} \]