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

The asteroid moves from a high elevation to a lower one therefore potential energy is converted to potential energy.

The speed of the meteorite when it impacts the lunar surface is approximately 1,449.9 m/s

Reasons:

The Kinetic and potential energies are;

 Using the energy conservation equation, we get;

Where;

G = Universal gravitation constant = 6.67408× 10⁻¹¹ m³/(kg·s²)

M = Mass of the Moon = 7.34767309 × 10²² kg

m = Mass of the asteroid = 100 kg

R = Radius of the Moon = 1,737.400 meters

Therefore, we get;

Dividing by m, gives;

Plugging in the values gives;

Which gives;

v₂² = 2× 1051105.6191 = 2102211.2382

v₂ = √(2102211.2382) ≈ 1,449.9

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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