Question

For a body of mass \(m\), the acceleration due to gravity at a distance \(R\) from the surface of the earth is \(\frac{g}{4}\). Its value at a distance \(\frac{R}{2}\) from the surface of the earth is (\(R =\) radius of the earth, \(g =\) acceleration due to gravity)

• A. \(\frac{g}{8}\)
• B. \(\frac{9g}{4}\)
• C. \(\frac{4g}{9}\)
• D. \(\frac{g}{2}\)

Hint:

Acceleration due to gravity at height \(h\): \(g_h = \frac{g}{(1 + h/R)^2}\) for \(h \ll R\)? Actually this formula is exact for any \(h\) above surface. For \(h = R\), denominator = 4. For \(h = R/2\), denominator = \((1.5)^2 = 2.25 = 9/4\), so \(g_h = 4g/9\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Acceleration due to gravity at a height \(h\) above surface: \(g_h = \frac{g}{\left(1 + \frac{h}{R}\right)^2}\).

Step 2: Key Formula or Approach:
Given: at distance \(R\) from surface means \(h = R\), so \(g_h = \frac{g}{(1+1)^2} = \frac{g}{4}\). This matches given. Need \(g\) at \(h = \frac{R}{2}\).

Step 3: Detailed Explanation:
For \(h = \frac{R}{2}\): \(g_h = \frac{g}{\left(1 + \frac{R/2}{R}\right)^2} = \frac{g}{\left(1 + \frac{1}{2}\right)^2} = \frac{g}{\left(\frac{3}{2}\right)^2} = \frac{g}{\frac{9}{4}} = \frac{4g}{9}\).

Step 4: Final Answer:
Option (C) is correct.