Question

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

• A. R/2
• B. R/4
• C. 3R/4
• D. R

Hint:

Gravity decreases linearly with depth inside the Earth ($g_d \propto (R-d)$), unlike the inverse-square reduction seen with altitude. This makes mental calculations very simple for any fractional depth!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks us to find the depth ($d$) below the Earth's surface where the acceleration due to gravity ($g_d$) is exactly half of its value on the surface ($g$).

Step 2: Key Formula or Approach:
The variation of acceleration due to gravity with depth ($d$) inside the Earth is given by the formula: \[ g_d = g \left(1 - \frac{d}{R}\right) \] where: $g$ = acceleration due to gravity on the Earth's surface
$R$ = radius of the Earth
$d$ = depth from the surface

Step 3: Detailed Explanation:

• We are given that the gravity at depth $d$ is half of its surface value:
\[ g_d = \frac{g}{2} \]

• Substitute this value into our depth variation equation:
\[ \frac{g}{2} = g \left(1 - \frac{d}{R}\right) \]

• Divide both sides of the equation by $g$:
\[ \frac{1}{2} = 1 - \frac{d}{R} \]

• Rearrange the terms to solve for $\frac{d}{R}$:
\[ \frac{d}{R} = 1 - \frac{1}{2} \] \[ \frac{d}{R} = \frac{1}{2} \]

• Solve for $d$:
\[ d = \frac{R}{2} \]

• This linear relation shows that gravity decreases at a constant rate as we go deeper into the Earth. It becomes half at a depth of $R/2$ and drops to zero at the center of the Earth ($d=R$).

Step 4: Final Answer:
The depth at which the acceleration due to gravity becomes half of its surface value is $R/2$.