Question

The radius of earth is \(R\) and acceleration due to gravity on its surface is \(g\). The height at which the acceleration due to gravity becomes \(\frac{g}{8}\) is:

• A. \(2R\)
• B. \((2\sqrt{2}-1)R\)
• C. \(\sqrt{2}R\)
• D. \(2\sqrt{2}R\)

Hint:

At height, use \( g' = g\left(\frac{R}{R+h}\right)^2 \). Always take square root carefully.

Answer 1

The Correct Option is B

Step 1: Formula for gravity at height.
\[ g' = g \left(\frac{R}{R+h}\right)^2 \]

Step 2: Substitute given condition.
\[ \frac{g}{8} = g \left(\frac{R}{R+h}\right)^2 \]

Step 3: Simplify.
\[ \frac{1}{8} = \left(\frac{R}{R+h}\right)^2 \]

Step 4: Take square root.
\[ \frac{1}{\sqrt{8}} = \frac{R}{R+h} \]

Step 5: Solve for \(h\).
\[ R+h = R\sqrt{8} \]
\[ h = R(\sqrt{8} - 1) \]

Step 6: Simplify.
\[ h = (2\sqrt{2} - 1)R \]

Step 7: Final conclusion.
\[ \boxed{(2\sqrt{2} - 1)R} \] Hence, correct answer is option (B).