Question

The acceleration due to gravity becomes \( \frac{g}{2} \) at a height equal to:

• A. \( \frac{R}{4} \)
• B. \( \frac{R}{2} \)
• C. \( \frac{R}{3} \)
• D. \( \frac{R}{5} \)

Hint:

For small heights, use approximation \(g_h \approx g(1 - 2h/R)\) instead of full formula.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Acceleration due to gravity (\(g\)) varies with altitude. For heights close to the Earth's surface, we can use an approximation, but for larger heights, the exact inverse square law must be considered.
Step 2: Key Formula or Approach:
The exact formula for gravity at height \(h\) is:
\[ g' = \frac{g}{\left(1 + \frac{h}{R}\right)^{2}} \]
The linear approximation for small heights (\(h \ll R\)) is:
\[ g' \approx g \left(1 - \frac{2h}{R}\right) \]
Step 3: Detailed Explanation:
The question provides options that suggest a linear approximation might be intended for simplicity in this specific test.
Given \(g' = \frac{g}{2}\).
Using the approximation formula:
\[ \frac{g}{2} = g \left(1 - \frac{2h}{R}\right) \]
\[ \frac{1}{2} = 1 - \frac{2h}{R} \]
\[ \frac{2h}{R} = 1 - \frac{1}{2} = \frac{1}{2} \]
\[ h = \frac{R}{4} \]
(Note: Using the exact formula \(g/2 = g/(1+h/R)^{2}\) would yield \(h = (\sqrt{2}-1)R \approx 0.414R\), which is not in the options. This confirms the use of the approximation).
Step 4: Final Answer:
At a height of \(R/4\), the acceleration due to gravity becomes \(g/2\).