Question

What is the launch speed of a projectile that rises vertically above the Earth to an altitude equal to twice 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.)

Answer 1

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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Note: I used g=9.8 m/s^2 and R=6.37*10^6 m
I converted GM/R^2 into g. Instead of that, you can also use G, M, R directly.