What is the launch speed of a projectile that rises vertically above the Earth t

Question

What is the launch speed of a projectile that rises vertically above the Earth to an altitude equal to five times the Earth's radius before coming to rest momentarily? (Astronomical data needed for this problem can be found on the inside back cover of the text. Ignore air resistance.)


Gravitational constant: 6.67x10^-11

Radius of Earth: 6.37x10^3 km

mass of earth: 5.97x10^24 kg

radius of moon: 1740 km

mass of moon: 7.35x10^22 kg

dist of moon from earth: 3.84x10^5 km

radius of sun: 6.95x10^5 km

mass of sun: 2.00x10^30 kg

dist. of earth from sun: 1.50x10^8 km

Explanation / Answer

Let M = Earth's mass, R = Earth's radius, m = projectile's mass, v = launch speed, Ki = KE of projectile at surface of Earth Kf = KE of projectile at given altitude Ui = PE of projectile at surface of Earth Uf = PE of projectile at given altitude Ki = 1/2 mv^2 Kf = 0 (because projectile is momentarily at rest) Ui = -GMm/R The given altitude = 2R. So at that altitude, distance of projectile from center of Earth = R + 2R = 3R So Uf = -GMm(3R) By conservation of energy, Ki + Ui = Kf + Uf 1/2 mv^2 - GMm/R = 0 - GMm/(3R) Divide by m 1/2 v^2 - GM/R = - GM/(3R) Or 1/2 v^2 = GM/R - GM(3R) Or 1/2 v^2 = (2/3)*(GM/R) Or v^2 = (4/3)*(GM/R) Or v^2 = (4R/3)*(GM/R^2) But GM/R^2 = g v^2 = 4Rg/3 v = sqrt(4Rg/3) = sqrt(4 * 6.37 * 10^6 * 9.8/3) = 9123 m/s = 9.123 km/s Ans: 9.123 km/s

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