Question

The acceleration due to gravity at a height $h$ above the Earth's surface is the same as that at a depth $d$ below the surface. If both $h$ and $d$ are much smaller than the radius of Earth $R$, then the relation between $h$ and $d$ is:

• A. $d = 2h$
• B. $h = 2d$
• C. $d = h$
• D. $d = 4h$

Hint:

Near the surface, gravity decreases twice as fast as you go upwards compared to going downwards. Therefore, to experience the same reduction in gravity, the depth must be twice the height ($d = 2h$).

Answer 1

The Correct Option is A

Step 1: Understand the question.
Gravity gets weaker as we go up above the Earth and also as we go down inside the Earth. We are told that at some small height $h$ the gravity equals the gravity at some small depth $d$. We must link $h$ and $d$.

Step 2: Write gravity at a height.
For small height compared to the radius $R$, \[ g_h = g\left(1 - \frac{2h}{R}\right) \] Gravity drops twice as fast with height.

Step 3: Write gravity at a depth.
For a depth $d$ below the surface, \[ g_d = g\left(1 - \frac{d}{R}\right) \] Gravity drops as the depth grows because less mass pulls from below.

Step 4: Set them equal.
The problem says the two are equal. \[ g\left(1 - \frac{2h}{R}\right) = g\left(1 - \frac{d}{R}\right) \]

Step 5: Cancel common parts.
Cancel $g$ and the $1$ on each side. \[ \frac{2h}{R} = \frac{d}{R} \] Now cancel $R$ too. \[ 2h = d \]

Step 6: State the relation.
So the depth is twice the height. \[ \boxed{d = 2h} \]