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Physics · JEE | GRAVITATION
हिन्दीA particle is fired vertically upward from the Earth's surface with a speed equal to half the escape speed (\( v = \frac{v_e}{2} \)). The maximum height reached by the particle above the Earth's surface in terms of Earth's radius \( R_E \) is:
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Correct answer · Option B
\( \frac{R_E}{3} \)
Explanation
By conservation of energy: \( -\frac{GM_E m}{R_E} + \frac{1}{2}m (\frac{v_e}{2})^2 = -\frac{GM_E m}{R_E + h} \). Since \( v_e^2 = \frac{2GM_E}{R_E} \), solving gives \( h = \frac{R_E}{3} \).
