Problem 16 - A Complete Resource Book in Physics for JEE Main

Question

A body of mass \(m\) rises to a height \(h=R / 5\) from the earth's surface where \(R\) is radius of the earth. If \(g\) is acceleration due to gravity at the earth surface, the increase in potential energy is (A) \(m g R\) (B) \((4 / 5) m g R\) (C) \((1 / 6) m g R\) (D) \((6 / 7) m g R\)

Step-by-Step Solution

Answer

The increase in potential energy is (C) \(\frac{1}{6} \cdot g \cdot R \cdot m\).

Step 1: Write the potential energy formula on Earth's surface

Potential energy at Earth's surface: \[PE = - \frac{G \cdot M \cdot m}{R}\]

Step 2: Write the potential energy formula at the height h above Earth's surface

Potential energy at height h: \[PE_{h} = - \frac{G \cdot M \cdot m}{(R + h)}\]

Step 3: Determine the increase in potential energy

Increase in potential energy: \(PE_{h} - PE = \Delta PE\)

Step 4: Substitute given values

We know:\(\\ h = \frac{R}{5} \\\\ \Delta PE = - \frac{G \cdot M \cdot m}{(R + \frac{R}{5})} - (- \frac{G \cdot M \cdot m}{R})\)

Step 5: Simplify the expression for ΔPE

Simplify the expression by multiplying both the numerator and denominator of both fractions by 5 to eliminate the fractions in the expression: \(\\ \Delta PE = - \frac{5 G \cdot M \cdot m}{(6R)} - (- \frac{6 \cdot G \cdot M \cdot m}{6R}) \\\\ \Delta PE = - \frac{5 G \cdot M \cdot m}{6R} + \frac{6 G \cdot M \cdot m}{6R}\) Combine the two terms: \(\\ \Delta PE = \frac{(6 - 5) G \cdot M \cdot m}{6R} = \frac{G \cdot M \cdot m}{6R}\)

Step 6: Express the potential energy increase in terms of g

We know that: \(\\ g = \frac{G \cdot M}{R^{2}} \\ \Rightarrow G \cdot M = g \cdot R^{2} \\\\ \Delta PE = \frac{G \cdot M \cdot m}{6R} = \frac{g \cdot R^{2} \cdot m}{6R}\)

Step 7: Final calculation

Simplify the expression for the potential energy increase: \(\\ \Delta PE = \frac{1}{6} \cdot g \cdot R \cdot m\) The increase in potential energy is (C) \(\frac{1}{6} \cdot g \cdot R \cdot m\).