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) ______.
Show 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!
Step 1: Understanding the Concept:
The acceleration due to gravity at a depth $d$ below the Earth's surface is given by the formula $g_d = g \left( 1 - \frac{d}{R} \right)$. Step 2: Formula Application:
Given $g_d = \frac{g}{n}$.
$\frac{g}{n} = g \left( 1 - \frac{d}{R} \right) \implies \frac{1}{n} = 1 - \frac{d}{R}$. Step 3: Explanation:
Rearranging to solve for $d$:
$\frac{d}{R} = 1 - \frac{1}{n} = \frac{n-1}{n}$.
$d = \frac{R(n-1)}{n}$. Step 4: Final Answer:
The depth is $\frac{R(n-1)}{n}$.