Question

The depth at which the value of acceleration due to gravity becomes ($1/n$) times the value at the surface of the earth is (R = radius of the earth) ______.

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

Hint:

Gravity decreases linearly as you go down a mine shaft ($g_d = g(1 - d/R)$), but it decreases as an inverse square law as you fly up into space ($g_h = g \frac{R^2}{(R+h)^2}$). Always ensure you pick the right formula for altitude vs. depth!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the specific underground depth ($d$) where the local acceleration due to gravity ($g_d$) decreases to a fraction ($1/n$) of its standard surface value ($g$).

Step 2: Key Formula or Approach:
Assuming the Earth is a sphere of uniform density, the variation of acceleration due to gravity strictly inside the Earth at depth $d$ is given by a linear relationship:
$$g_d = g \left( 1 - \frac{d}{R} \right)$$
Where $R$ is the radius of the Earth.

Step 3: Detailed Explanation:
The problem states that $g_d = \frac{1}{n} g$.
Substitute this condition into the depth formula:
$$\frac{1}{n} g = g \left( 1 - \frac{d}{R} \right)$$
Cancel the surface gravity $g$ from both sides:
$$\frac{1}{n} = 1 - \frac{d}{R}$$
Rearrange to isolate the fractional depth $\frac{d}{R}$:
$$\frac{d}{R} = 1 - \frac{1}{n}$$
Find a common denominator on the right side:
$$\frac{d}{R} = \frac{n - 1}{n}$$
Multiply both sides by Earth's radius $R$:
$$d = R \left( \frac{n - 1}{n} \right) = \frac{R(n - 1)}{n}$$

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