Question

What will be the acceleration due to gravity at a depth \( d \), where \( g \) is the acceleration due to gravity on the surface of the Earth?

• A. \(\frac{g}{\left(1 + \frac{d}{R}\right)^2}\)
• B. \( g \left[ 1 - \frac{2d}{R} \right] \)
• C. \(\frac{g}{\left(1 - \frac{d}{R}\right)^2}\)
• D. \( g \left[ 1 - \frac{d}{R} \right] \)

Hint:

For acceleration due to gravity: - At height \( h \): \( g' = g \left( \frac{R}{R+h} \right)^2 \) - At depth \( d \): \( g' = g \left( 1 - \frac{d}{R} \right) \) - Gravity decreases inside the Earth linearly with depth.

Answer 1

The Correct Option is D

Step 1: Understanding Gravity Variation with Depth The acceleration due to gravity at a depth \( d \) inside the Earth is given by the formula: \[ g' = g \left( 1 - \frac{d}{R} \right) \] where: \( g' \) is the acceleration due to gravity at depth \( d \), \( g \) is the acceleration due to gravity on the surface, \( R \) is the radius of the Earth. Step 2: Explanation of the Formula Gravity decreases linearly as we move inside the Earth because only the mass enclosed within radius \( R-d \) contributes to gravity at depth \( d \), following the equation: \[ g' = g \times \frac{R - d}{R} = g \left( 1 - \frac{d}{R} \right) \] Thus, the correct option is: \[ g' = g \left( 1 - \frac{d}{R} \right) \] which corresponds to option (D).