Question

What is the launch speed of a projectile that rises vertically above the Earth to an altitude equal to 14 REarth before coming to rest momentarily

Answer 1

Answer:

 v = 1.078 10⁴ m / s

Explanation:

To solve this exercise we can use conservation of mechanical energy

Starting point. Just clearing

          Em₀ = K + U = ½ m v² - G m M / $R_{e}$

Final point. At r = 14R_{e}

        Em_{f} = U = -G m M / 14R_{e}

how energy is conserved

         Em₀ = $Em_{f}$

         ½ m v² - G m M /R_{e} = -G m M / 14R_{e}

        ½ v² = G M / R_{e} (-1/14 + 1)

        v² = 2 G M / R_{e} 13/14

we calculate

       v² = 2 6.67 10⁻¹¹ 5.98 10²⁴ / 6.37 10⁶ 13/14

       v = √ (1,16 10⁸)

       v = 1.078 10⁴ m / s