Question

The depth at which acceleration due to gravity becomes $\frac{g}{n}$ is [$R = \text{radius of earth}$, $g = \text{acceleration due to gravity}$, $n = \text{integer}$]

• A. $\frac{R(n-1)}{n}$
• B. $\frac{(n-1)}{nR}$
• C. $\frac{Rn}{(n-1)}$
• D. $\frac{n}{R(n-1)}$

Hint:

The gravity-depth profile is linear: it drops to $1/2$ at $R/2$, drops to $1/4$ at $3R/4$, and drops to $0$ at depth $R$ (the center). You can quickly verify the general formula by plugging in $n=2 \implies d=R/2$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find a formula for the specific depth ($d$) below the Earth's surface where the gravitational acceleration decreases to exactly $\frac{1}{n}$ of its surface value ($g$).

Step 2: Key Formula or Approach:
The variation of acceleration due to gravity with depth ($d$) is strictly linear, given by the formula:
$$g_d = g\left(1 - \frac{d}{R}\right)$$
where $R$ is the radius of the Earth.

Step 3: Detailed Explanation:
The problem states that the gravity at the unknown depth $d$ is $g_d = \frac{g}{n}$.
Substitute this condition into the depth variation formula:
$$\frac{g}{n} = g\left(1 - \frac{d}{R}\right)$$
Cancel $g$ from both sides:
$$\frac{1}{n} = 1 - \frac{d}{R}$$
Rearrange the equation to isolate the term containing depth ($d$):
$$\frac{d}{R} = 1 - \frac{1}{n}$$
Find a common denominator on the right side:
$$\frac{d}{R} = \frac{n - 1}{n}$$
Multiply both sides by $R$ to solve for $d$:
$$d = \frac{R(n - 1)}{n}$$

Step 4: Final Answer:
The depth is $\frac{R(n-1)}{n}$, matching option (A).