AOS145_HW04_KEY (1) - HW 4 Solution 1 Jupiter For any...

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HW 4. Solution 1. Jupiter For any planet, its solar constant F s is inversely proportional to the square of its distance to the sun, d , i.e., F s 1/ d 2 . The Jupiter-Sun distance is 5.2 times that of Earth-Sun, i.e., 7.8 × 10 8 km/1.5 × 10 8 km = 5.2. Thus the solar constant for Jupiter is F s = (1/5.2) 2 × 1370 W m -2 = 50.7 W m -2 The effective temperature of Jupiter is thus T E = [(1- A ) F s /(4 σ )] ¼ = [(1-0.73) × 50.7 W m -2 /(4 × 5.67 × 10 -8 J m -2 s -1 K -4 )] ¼ = 88 K The energy from solar radiation is σ (88 K) 4 = 3.4 W m -2 (another way to look at it: (1- A ) F s /4 = (1-0.73) × 50.7 W m -2 /4 = 3.4 W m -2 ) Since the observed effective temperature is 134 K, the internal heat source is thus σ (134 K) 4 - σ (88 K) 4 = 14.9 W m -2 , about 4.4 times that from solar radiation (another way to look at it: [ σ (134 K) 4 - σ (88 K) 4 ]/[ σ (88 K) 4 ] = (134/88) 4 – 1 = 4.4) The internal energy source is much larger than that from solar radiation. In comparison, solar radiation accounts for more than 99.99% of the energy supply to the Earth- atmosphere system.
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