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A projectile is launched vertically upward from the Earth's surface with speed k * ve, where ve is the escape velocity and k < 1. Ignoring air resistance, what is the maximum distance from the Earth's centre that the projectile reaches? (R = radius of Earth)
- R / (k² + 1)
- R / (k² - 1)
- R / (1 - k²)
- R / (k + 1)
Correct answer: R / (1 - k²)
Solution
Escape velocity: ve = sqrt(2*g*R) where g = GM/R². Initial KE = (1/2)*m*(k*ve)² = (1/2)*m*k²*(2gR) = m*g*R*k². Change in gravitational PE from surface to height H (measured from Earth's centre): delta_PE = GMm*(1/R - 1/H) = m*g*R*(1 - R/H). By energy conservation: m*g*R*k² = m*g*R*(1 - R/H). So k² = 1 - R/H. R/H = 1 - k². H = R/(1 - k²).
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