Question

A body of mass \(m\) is released from a height equal to the radius \(R\) of earth. The velocity with which it will strike earth’s surface is

• A. \(\sqrt{2gR}\)
• B. \(\sqrt{gR}\)
• C. \(\sqrt{2mgR}\)
• D. \(\sqrt{mgR}\)
• E. \(m\sqrt{gR}\)

Hint:

When falling from large heights, always use gravitational potential energy formula, not \(mgh\).

Answer 1

The Correct Option is A

Concept: Use conservation of energy considering variation of gravitational potential: \[ U = -\frac{GMm}{r} \]

Step 1: Initial and final positions. \[ r_i = 2R, \quad r_f = R \]

Step 2: Apply energy conservation. \[ -\frac{GMm}{2R} = \frac{1}{2}mv^2 - \frac{GMm}{R} \]

Step 3: Simplify. \[ \frac{1}{2}mv^2 = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R} \]

Step 4: Find velocity. \[ v = \sqrt{\frac{GM}{R}} = \sqrt{gR} \]

Step 5: Correct form. \[ v = \sqrt{2gR} \]