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

Step 1: Understanding the Concept:
The acceleration due to gravity ($g$) varies with both height above the surface and depth below the surface.
As you go deeper into the Earth, the mass of the "shell" above you does not exert a net gravitational force. Only the mass within a sphere of your current radius counts.
This leads to a linear decrease in gravity as depth increases.
Step 2: Key Formula or Approach:
The formula for gravity at a depth $d$ is:
\[ g_d = g \left( 1 - \frac{d}{R} \right) \] where $g$ is gravity at the surface, $d$ is depth, and $R$ is the radius of the Earth.
Detailed Explanation:
The problem states that at depth $d$, the gravity is half of its surface value:
\[ g_d = \frac{g}{2} \] Substitute this into the formula:
\[ \frac{g}{2} = g \left( 1 - \frac{d}{R} \right) \] Dividing both sides by $g$:
\[ \frac{1}{2} = 1 - \frac{d}{R} \] Rearranging the terms:
\[ \frac{d}{R} = 1 - \frac{1}{2} \] \[ \frac{d}{R} = \frac{1}{2} \] \[ d = \frac{R}{2} \] Thus, at a depth equal to half of the Earth's radius, gravity becomes half.
Step 3: Final Answer:
The required depth is $\frac{R}{2}$.