Question

What maximum height above the surface of Earth does an object attain if it is launched upward at 5.5 km/s from the surface? Express your answer with the appropriate units.

Answer 1

Use potential energy.
Which, per mass of the projectile, is
V/m = -GM/r

Escape speed from Earth's surface is found by
½vₓ² = GM/R
vₓ = √(2GM/R)
where
G = univ. grav. constant = 6.674·10⁻¹¹ m³/(kg·s²)
M = mass of Earth = 5.972·10²⁴ kg
R = radius of Earth = 6371 km = 6.371·10⁶ m

So launching a projectile upward at speed v₁, conservation of energy says that the height h, it will reach obeys:
½v₁² - GM/R = ½0² - GM/(R + h)
R + h = 2GM/(2GM/R - v₁²) = 2GM/(vₓ² - v₁²) = Rvₓ²/(vₓ² - v₁²)
h = Rvₓ²/(vₓ² - v₁²) - R = Rv₁²/(vₓ² - v₁²)

Now, vₓ ≈ 11 km/s, so v₁ ≈ ½vₓ and
h ≈ (¼ / ¾)R = ⅓R

h=2.123*10^6 m