Solution 7: Escape Velocity
1. Effective Gravity:
$W_{eq} = \frac{1}{2} W_{pole}$. \[ mg – m\omega^2 R = \frac{1}{2}mg \implies \omega^2 R = \frac{g}{2} \]
$W_{eq} = \frac{1}{2} W_{pole}$. \[ mg – m\omega^2 R = \frac{1}{2}mg \implies \omega^2 R = \frac{g}{2} \]
2. Orbital Velocity vs Gravity:
Given peripheral velocity $v_0 = \omega R$. \[ \frac{v_0^2}{R} = \omega^2 R = \frac{g}{2} \implies 2v_0^2 = gR \]
Given peripheral velocity $v_0 = \omega R$. \[ \frac{v_0^2}{R} = \omega^2 R = \frac{g}{2} \implies 2v_0^2 = gR \]
3. Escape Velocity:
\[ v_{esc} = \sqrt{2gR} = \sqrt{2(2v_0^2)} = \sqrt{4v_0^2} = 2v_0 \]
\[ v_{esc} = \sqrt{2gR} = \sqrt{2(2v_0^2)} = \sqrt{4v_0^2} = 2v_0 \]
Answer: (b) 2v₀