Quiz2

Question 5 Star A has twice the temperature as Star B, but is twice as far away. If they have the same radius, what is the ratio of A’s flux at Earth to B’s? Answer: 4 Recall that the brightness of a star would scale with the fourth power of the temperature T , i.e. ∝ T 4 . If the Star A has twice the temperature as Star B (provided they are at the same distance), this implies it would be 2 4 = 16 times brighter. Meanwhile, it is twice as far, meaning it would scale ∝ 1 /r 2 where r is the distance, as such the flux would be reduced by a factor of 1 / 2 2 = 1 / 4 . Combining the two factors, we have a net ratio of 16 × 1 / 4 = 4 . Question 6 The average surface temperature of planet Venus (which experiences a strong greenhouse effect) is 737 K. At what wavelength does its blackbody spectrum peak? Answer: 3.9 μ m We apply Wien’s displacement law, i.e. λ peak = b T = 2 . 9 × 10 - 3 737 = 3 . 9 × 10 - 6 m = 3 . 9 μ m . Question 7 Assume the Earth’s blackbody spectrum as seen from space peaks at a wavelength of 10 um (10 micrometers). What total luminosity does it emit, expressed as a fraction of one solar luminosity? Answer: 7 × 10 - 10 We first calculate the temperature of Earth via Wien’s displacement law: T = b λ peak = 2 . 9 × 10 - 3 10 × 10 - 6 = 290 K . Now that the temperature is known, we can calculate the luminosity via the Stefan-Boltzmann law: L = 4 πR 2 σT 4 = 4 π (6 . 371 × 10 6 ) 2 × (5 . 67 × 10 - 8 ) × 290 4 ≈ 2 × 10 17 W , where we took the Earth’s radius to be 6371 km. Since the solar luminosity is 3 . 85 × 10 26 W , a mere division would yield 5 . 3 × 10 - 10 . The closest answer in the multiple choice is 7 × 10 - 10 . 3