Question

A small asteroid that has a mass of 100 kg is moving at 200 m/s when it is 1000 km above the Moon. 1) How fast will the meteorite be traveling when it impacts the lunar surface if it is heading straight toward the center of the Moon

Answer 1

Answer:

v = 200.005 m / s

Explanation:

For this exercise we can use the concept of energy conservation,

Starting point. When the asteroid at h = 1000 km = 1 10⁶ m

               Em₀ = K + U = ½ m v₀² - G m M / (R + h)

Final point. When the asteroid is on the surface of the moon

              Emf = K + U = ½ m v² - G m M / R

As there is no friction the energy is conserved

             Em₀ = Emf

             ½ m v₀² - G m M / (R + h) = ½ m v² - G m M / R

            ½ m (v₀² –v²) = G m M (-1 / R + 1 / R + h)

            ½ (v₀² - v²) = G M (-h / R (R + h))

            v² = v₀² + 2 G M h / R (R + h))

Let's calculate

           v² = 200² + 2 6.67 10⁻¹¹  7.36 10²² / 1.74 (1.74  + 1 ))10¹²

           v2 = 40000 + 2,059

           v = 200.005 m / s