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 = sum of kinetic +potentil energies

= mv^{2}/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

10000^{2}/2 - GM/R = 0 - Gm/r (r is distance of mass from
cenetr of earth)

10^{8}/2 - GMR/R^{2} = -
GMR^{2}/R^{2}r

10^{8}/2 - gR = - gR^{2}/r
(GM/R^{2} = g)

10^{8}/2 - 9.8*6.38x10^{7} = -
9.8*6.38^{2}x10^{14}/r

r = 6.935x10^{7}m

distance from earth surface = r - R = 5.55x10^{6}m

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