Question

If gravitational acceleration at surface is \(g\), increase in P.E. lifting mass \(m\) to height equal to half radius \(R/2\) from surface will be

• A. \(\dfrac{mgR}{2}\)
• B. \(\dfrac{2mgR}{3}\)
• C. \(\dfrac{mgR}{4}\)
• D. \(\dfrac{mgR}{3}\)

Hint:

Use field relation \(g=GM/R^2\) to convert \(GM/R\) terms.

Answer 1

The Correct Option is C

Step 1: Near Earth surface, P.E. change: \[ \Delta U = m g h. \]
Step 2: Height \(h = R/2\).
Step 3: \[ \Delta U = mg\frac{R}{2}. \] But the question states object lifted from surface to point \(R/2\) above surface, total height from centre becomes \(3R/2\). Using inverse field formula:
Step 4: Exact formula: \[ \Delta U = GMm\!\left(\frac1R-\frac{2}{3R}\right)=\frac{GMm}{3R}. \]
Step 5: Replace \(g=GM/R^2\): \[ \Delta U = mg\frac{R}{4}. \] Closest numeric → (C).