Question

When the value of acceleration due to gravity \( g' \) becomes \( \frac{g}{3} \) above surface of height \( h \) then relation between \( h \) and \( R \) is \( (R = \text{radius of earth}) \)

• A. \( h = \frac{R}{\sqrt{3}-1} \)
• B. \( h = \frac{\sqrt{3}}{R} \)
• C. \( h = (\sqrt{2}-1)R \)
• D. \( h = (\sqrt{3}-1)R \)

Hint:

For heights small compared to \(R\), use approximation \(g' \approx g(1 - 2h/R)\). But here exact formula is needed. Always square root carefully.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need height \(h\) above Earth’s surface where \(g' = g/3\).

Step 2: Key Formula or Approach:
Acceleration due to gravity at height \(h\): \(g' = g \left( \frac{R}{R+h} \right)^2\).

Step 3: Detailed Explanation:
Set \(g/3 = g \left( \frac{R}{R+h} \right)^2\). Cancel \(g\) (positive): \[ \frac{1}{3} = \left( \frac{R}{R+h} \right)^2 \quad \Rightarrow \quad \frac{1}{\sqrt{3}} = \frac{R}{R+h}. \] Invert: \(\sqrt{3} = \frac{R+h}{R} = 1 + \frac{h}{R}\) ⇒ \(\frac{h}{R} = \sqrt{3} - 1\) ⇒ \(h = (\sqrt{3} - 1)R\).

Step 4: Final Answer:
Thus \(h = (\sqrt{3}-1)R\), option (D).