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Exercise Solution

Pages 196-197, Problems 2, 4

Problem 2

Mass is equal to density times the volume.    M = 3000 ´ 4/3 ´ p ´ (6.378´106)3        M = 3.3´1024 kg

Surface gravity is the same as the acceleration due to gravity, a = GM/r2       

G is the gravitational constant = 6.67x10-11 Nm2   

M is the mass (of Earth in this case) = 3.3´1024 kg

r (radius of the Earth) = approximately  3.3´1024 kg        approximately, since the earth is not completely round

a = (6.67x10-11 Nm2/kg2)(3.3x1024 kg)/6.378x106 km)2 =  5.4 m/s2         but not used, just for comparison with 9.8 m/s2 (actual)

and, from chapter 2, page 56 (yes!, you need to remember or be able to look up material covered previously)

Vescape = (2GM/r)1/2 = (2(6.67x10-11 )(3.3´1024 )/6.378x106 ))1/2 = 8.3 km/s   

Problem 4

Since the luminosity in Stefan’s law is proportional to T4, compare the Earth at 250 K to Earth at 290 K.  (250 / 290)4 = 0.55.  Subtracting this from 1 gives 0.45 or 45%.