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A uniform spherical planet spins about its axis. A point on its equator moves with speed V due to this rotation. Because of the spin, the effective acceleration due to gravity at the equator is exactly half of its value at the poles. Express the escape velocity of a particle from the pole of this planet in terms of V.
- Vₑ = 2V
- Vₑ = V
- Vₑ = V/2
- Vₑ = 2*sqrt(2) V
Correct answer: Vₑ = 2V
Solution
At the equator the apparent gravity is g_eq = gₚ - omega² R = gₚ - V²/R (using V = omega R). Given g_eq = gₚ/2: gₚ - V²/R = gₚ/2 => gₚ/2 = V²/R => gₚ R = 2 V². Escape velocity from the pole: Vₑ = sqrt(2 gₚ R) = sqrt(2 * 2 V²) = sqrt(4 V²) = 2V.
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