Step 1: Understanding the Concept:
The acceleration due to gravity (\(g\)) is a measure of the Earth's gravitational pull on an object at its surface.
However, this value changes as we move away from the surface, either upwards into space or downwards into the Earth's interior.
When moving inside the Earth, we must consider the "Shell Theorem."
As we descend to a depth \(d\), we are essentially standing on a smaller sphere of radius \((R - d)\).
The spherical shell of Earth that is "above" us exerts no net gravitational force on us because the pulls from all sides cancel out.
Therefore, only the mass of the inner sphere (of radius \(R-d\)) contributes to gravity.
Because mass decreases with the cube of the radius (\(M \propto r^3\)) while the distance squared in the denominator of the gravity formula (\(g = GM/r^2\)) only decreases with the square, the net result is that gravity decreases linearly with depth.
Key Formula or Approach:
The standard formula for the variation of acceleration due to gravity at a depth \(d\) below the Earth's surface is:
\[ g_d = g \left( 1 - \frac{d}{R} \right) \]
where:
\(g_d\) is the acceleration due to gravity at depth \(d\).
\(g\) is the acceleration due to gravity at the surface (\(\approx 9.8 \text{ m/s}^{2}\)).
\(d\) is the depth from the surface.
\(R\) is the radius of the Earth.
Step 2: Detailed Explanation:
The problem asks for the specific depth \(d\) where the gravity is exactly half of its surface value.
Mathematically, we set the condition:
\[ g_d = \frac{g}{2} \]
Now, we substitute this condition into our depth variation formula:
\[ \frac{g}{2} = g \left( 1 - \frac{d}{R} \right) \]
To solve for \(d\), we first divide both sides by \(g\) (since \(g\) is a non-zero constant):
\[ \frac{1}{2} = 1 - \frac{d}{R} \]
Next, we rearrange the equation to isolate the term containing \(d\).
Move \(\frac{d}{R}\) to the left side and \(\frac{1}{2}\) to the right side:
\[ \frac{d}{R} = 1 - \frac{1}{2} \]
\[ \frac{d}{R} = \frac{1}{2} \]
Finally, multiply both sides by the Earth's radius \(R\):
\[ d = \frac{R}{2} \]
This result confirms that gravity decreases at a constant rate as we go deeper.
At the surface (\(d=0\)), gravity is \(g\).
At half the radius (\(d=R/2\)), gravity is \(g/2\).
At the center (\(d=R\)), gravity becomes zero.
This is fundamentally different from moving above the surface, where gravity follows an inverse-square law and never truly reaches zero.
Step 3: Final Answer:
The depth at which gravity becomes half of its surface value is exactly half of the Earth's radius, which is \(R/2\).
This matches option (A).