Question

The depth 'd' below the surface of the earth where the value of acceleration due to gravity becomes $\left(\frac{1}{n}\right)$ times the value at the surface of the earth is ($R$ = radius of the earth)

• A. $R \left( \frac{n - 1}{n} \right)$
• B. $R \left( \frac{n}{n + 1} \right)$
• C. $\frac{R}{n}$
• D. $\frac{R}{n^2}$

Hint:

Gravity decreases linearly as you go down into a hole. If you want gravity to drop to a fraction like $\frac{1}{5}$, you have to go down $\frac{4}{5}$ of the way to the center. Structurally, dropping to $\frac{1}{n}$ means you must descend a depth of $\frac{n-1}{n}$ of the radius!

Answer 1

The Correct Option is A

The acceleration due to gravity at a depth $d$ below the Earth's surface is given by the formula: $$g' = g \left(1 - \frac{d}{R}\right)$$ Given that the gravity at depth $d$ becomes $\frac{1}{n}$ times the surface value ($g' = \frac{g}{n}$): $$\frac{g}{n} = g \left(1 - \frac{d}{R}\right)$$ $$\frac{1}{n} = 1 - \frac{d}{R}$$ Rearranging the terms to isolate the depth fraction $\frac{d}{R}$: $$\frac{d}{R} = 1 - \frac{1}{n} = \frac{n - 1}{n}$$ $$d = R \left(\frac{n - 1}{n}\right)$$
Final Answer:
The depth below the surface is $R \left( \frac{n - 1}{n} \right)$, which corresponds to option (A).