Question

A projectile is launched vertically from the Earth's surface with an initial speed of 10 km/sec. Neglecting atmospheric drag, how far above the Earth's surface would it go? (Radius of earth = 6380 km)

Answer 1

Mechanical energy of earth projectile system remains conserved.
Mechanical energy = sum of kinetic +potentil energies
                             = mv²/2 + ( -GMm/r)
M is mass of earth , r is distance between center of earth from mass m and v is speed of mass.
Speed of mass at maximum distance = 0. Hence
10000²/2 - GM/R = 0 - Gm/r (r is distance of mass from cenetr of earth)
10⁸/2 - GMR/R² = - GMR²/R²r
10⁸/2 - gR = - gR²/r    (GM/R² = g)
10⁸/2 - 9.8*6.38x10⁷ = - 9.8*6.38²x10¹⁴/r
r = 6.935x10⁷m
distance from earth surface = r - R = 5.55x10⁶m