A 3.14-kg rock is released from rest a distance 2.26 × 106 m from above the surf

Question

A 3.14-kg rock is released from rest a distance 2.26 × 106 m from above the surface of the Moon. Due to the mutual attraction between the Moon and the rock, the rock falls to the Moon's surface.
(a) What is the change in the gravitational potential energy as the rock falls to the Moon?

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  J
(b) How fast is the rock moving just before it lands on the Moon?

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  m/s

USE THIS HINT: (a) The gravitational potential energy of two masses is given by   Ug=Gm1m2/r,Ug=Gm1m2/r, where   rr is their separation, which here is the distance between the rock and the center of the Moon. You can find the mass and the radius of the Moon in Appendix C.
(b) Energy is conserved, so the loss in potential energy is equal to the gain in kinetic energy.

Explanation / Answer

(a) PE = - G M m / r

change in PE = PEf - PEi = - G M m / R - G M m / r

= -(6.67 x 10^-11 x 7.348 x 10^22 x 3.14)[ 1/(1737 x 10^3) - 1/(1737 x 10^3 + 2.26 x 10^)]

= - 5 x 10^6 J

(B) change in PE + change in Ke = 0

- 5 x 10^6 + 3.14 v^2 /2 = 0

v = 1786 m/s

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