12. A m = 71.2 kg object is released from rest at a distance h = 0.713515 R abov
ID: 2032830 • Letter: 1
Question
12. A m = 71.2 kg object is released from rest at a distance h = 0.713515 R above the Earth’s surface. The acceleration of gravity is 9.8 m/s 2 . For the Earth, RE = 6.38 × 106 m, M = 5.98 × 1024 kg. The gravitational acceleration at the surface of the earth is g = 9.8 m/s 2 . Find the speed of the object when it strikes the Earth’s surface. Neglect any atmospheric friction. Caution: You must take into account that the gravitational acceleration depends on distance between the object and the center of the earth. Answer in units of m/s.
Explanation / Answer
Using law of conservation of energy
Total energy at h = Total energy at the surface of earth
(-G*m*mE)/(RE+h) = (-G*m*mE)/(RE) + 0.5*m*v^2
(-6.67*10^-11*71.2*5.98*10^24)/((6.38*10^6)+(0.713515*6.38*10^6)) = ((-6.67*10^-11*71.2*5.98*10^24)/(6.38*10^6))+(0.5*71.2*v^2)
v = 7215 m/s = 7.215 km/s
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