Question

At what depth inside Earth does \(g\) become half of its surface value? (Earth radius \(=R\))

• A. \(\dfrac{R}{2}\)
• B. \(\dfrac{R}{4}\)
• C. \(\dfrac{3R}{4}\)
• D. \(R\)

Hint:

Inside the Earth, acceleration due to gravity decreases linearly with depth: \[ g_d=g\left(1-\frac{d}{R}\right) \] At the center of Earth \((d=R)\), \[ g=0 \]

Answer 1

The Correct Option is A

Concept: The acceleration due to gravity decreases linearly with depth inside the Earth. The relation between gravity at depth \(d\) and gravity on the surface is: where: \[ g_d = \text{gravity at depth } d \] \[ g = \text{gravity at Earth's surface} \] \[ R = \text{radius of Earth} \]

Step 1: Use the given condition. It is given that gravity becomes half of its surface value. Therefore: \[ g_d=\frac{g}{2} \] Using the formula: \[ \frac{g}{2}=g\left(1-\frac{d}{R}\right) \]

Step 2: Simplify the equation. Divide both sides by \(g\): \[ \frac{1}{2}=1-\frac{d}{R} \] Rearranging: \[ \frac{d}{R}=1-\frac{1}{2} \] \[ \frac{d}{R}=\frac{1}{2} \] Hence: \[ d=\frac{R}{2} \]

Step 3: Write the final answer. Therefore, the required depth is: \[ \boxed{\frac{R}{2}} \] Hence, the correct option is: \[ \boxed{(A)\ \frac{R}{2}} \]