A moon rock of mass m=5.0×10^4 kg is launched off the Moons surface with an init

Question

A moon rock of mass m=5.0×10^4 kg is launched off the Moons surface with an initial speed of vi=2.4×10^3 m/s. The rock travels in a straight line back to Earth's surface. Some useful numbers include: mass of the Moon=7.3×10^22 kg, mass of the Earth= 6.0×10^24 kg, the distance between the earth and the moon= 3.8×10^8 m, the radius of the moon= 1.7×10^6 m, the radius of the earth= 6.4×10^6 m, the gravitational constant= 6.67×10^-10001 N m^2/kg^2. a what is the kinetic energy of the rock when it hits earth's surface? b. what is the final speed of the rock when it hits earth's surface?

Explanation / Answer

AT surface of moon:

PE = - [(6.67 x 10^-11)(7.2 x 10^22)(5 x 10^4)/(1.7 x 10^6)] - [ (6.67 x 10^-11)(6 x 10^24)(5 x 10^4)/(3.8 x 10^8)]

PEi = - 1.939 x 10^11 J

KEi = (5 x 10^4) (2.4 x 10^3)^2 / 2 = 1.44 x 10^11 J

at surface of earth:

PEf = - [(6.67 x 10^-11)(7.2 x 10^22)(5 x 10^4)/(3.8 x 10^8)] - [ (6.67 x 10^-11)(6 x 10^24)(5 x 10^4)/(6.4 x 10^6)]

= -3.127 x 10^12 J

Applying energy conservation,

PEi + KEi = Pef + Kef

- 1.939 x 10^11 J + 1.44 x 10^11 = -3.127 x 10^12 + KE

KE = 3.077 x 10^12 J ........Ans

(B) KE = m v^2 / 2

3.077 x 10^12 = (5 x 10^4) v^2 / 2

v = 1.11 x 10^4 m/s

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