Question

6. A meteorite heads straight for the Moon. If it is traveling at 1.85 km/s when very far away, with what speed does it hit the Moon’s surface? (The mass and radius of the Moon are 7.35x1022 kg and 1740 km respectively.)

a) 2.86 km/s b) 2.94 km/s c) 3.01 km/s d) 3.17 km/s e) 3.34 km/s

Answer 1

Using Energy conservation between meteorite and Moon:

KEi + PEi = KEf + PEf

KEi = (1/2)*m1*Vi^2

m1 = mass of meteorite

Vi = Initial speed of meteorite = 1.85 km/sec = 1850 m/sec

PEi = Potential energy = 0, since meteorite is very far away from moon

KEf = (1/2)*m1*Vf^2

Vf = final speed of meteorite when it hits moon's surface = ?

PEf = -G*m1*M/R^2

M = mass of moon = 7.35*10^22 kg

R = Radius of moon = 1740 km = 1.74*10^6 m

Using these values:

(1/2)*m1*Vi^2 + 0 = (1/2)*m1*Vf^2 - G*m1*M/R

Vf = sqrt (Vi^2 + 2*G*M/R)

Vf = sqrt (1850^2 + 2*6.67*10^-11*7.35*10^22/(1.74*10^6))

Vf = 3009.58 m/sec = 3010 m/sec = 3.01*10^3 m/sec

Vf = final speed of meteorite = 3.01 km/sec

Correct option is C.

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