Question

What is the speed of a 1850 kg meteorite when it hits the moon's surface? This meteorite had a velocity of 1.4 x 10^3 m/s heading directly toward the moon. when it was 15000 m above the moon's surface

Answer 1

Answer on Question #84399, Physics / Astronomy | Astrophysics

What is the speed of a 1850 kg meteorite when it hits the moon's surface? This meteorite had a velocity of $1.4 \times 10^{3} \, \text{m/s}$ heading directly toward the moon. when it was 15000 m above the moon's surface.

Answer:

Total mechanical energy is conserved in the system, so

$$KE_i + PE_i = KE_f + PE_f$$

Where, $KE_{i,f}$ and $PE_{i,f}$ are initial and final kinetic energy and initial and final potential energy, respectively.

- m is the masses of the asteroid (1850 kg)

- M is the Moon (=7.35×10²²kg), respectively.

- the radius of the moon and the initial height of the asteroid, respectively, so

$$R = 1.737 \times 10^6 \, \text{m}$$ $$h = 15000 \, \text{m}$$ $$KE_i = \frac{1}{2}mv_i^2 = 1.81 \times 10^9 \, \text{J}$$ $$PE_i = -\frac{GMm}{R + h} = -\frac{6.67 \times 10^{-11} \, \text{N} \cdot \frac{m^2}{kg^2} \times 1850 \, \text{kg} \times 7.35 \times 10^{22} \, \text{kg}}{1.737 \times 10^6 \, \text{m} + 15000 \, \text{m}} = -5.18 \times 10^9 \, \text{J}$$ $$KE_f = \frac{1}{2}mv_f^2 = 925v_f^2$$ $$PE_f = -\frac{GMm}{R} = -\frac{6.67 \times 10^{-11} \, \text{N} \cdot \frac{m^2}{kg^2} \times 1850 \, \text{kg} \times 7.35 \times 10^{22} \, \text{kg}}{1.737 \times 10^6 \, \text{m}} = -5.22 \times 10^9 \, \text{J}$$ $$v_f = \sqrt{\frac{5.22 \times 10^9 \, \text{J} - 3.36 \times 10^9 \, \text{J}}{925}} = 3.5 \times 10^3 \, \text{m/s}$$

Answer: $3.5 \times 10^3 \, \text{m/s}$

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