UNDERSTANDING THE UNIVERSE(LL)-W/CODE
3rd Edition
ISBN: 9780393869903
Author: PALEN
Publisher: NORTON
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Chapter 7, Problem 45QAP
To determine
The temperature of Eris.
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Saturn is even further at 1434X109m from the sun. Calculate the approximate strength of the greenhouse effect on Jupiter. The observed surface temperature is 133K, the albedo is 0.47, the solar constant is 15W/m2.
What is the value of the IR transmission factor (f) for a Venus-like planet, if the
measured average surface temperature is 706 K? Assume the following values for the
planet: E, = 2.60x10³ W m² and a = 0.71.
Would the average temperature increase or decrease if f decreased?
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Chapter 7 Solutions
UNDERSTANDING THE UNIVERSE(LL)-W/CODE
Ch. 7.1 - Prob. 7.1CYUCh. 7.2 - Prob. 7.2CYUCh. 7.3 - Prob. 7.3CYUCh. 7.4 - Prob. 7.4CYUCh. 7.5 - Prob. 7.5CYUCh. 7 - Prob. 1QAPCh. 7 - Prob. 2QAPCh. 7 - Prob. 3QAPCh. 7 - Prob. 4QAPCh. 7 - Prob. 5QAP
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- The fraction of the energy flux received which is reflected into space is the albedo of Venus, av, which is about 0.76. The fraction of the energy flux which is absorbed is then (1-av) = 1. - 0.76 = 0.24. So the amount of energy actually absorbed by Venus in each second is Lv = (1-av)Ev. Lv = (1-av)Ev = ___________________ ergs/s And next calculate the effective temperature of Venus: Tv4 = (Lv/(4pdv2))/s = Lv/(4spdv2) = __________________ K4 and taking the square root of Tv4 twice in succession we get the effective Temperature Tv: Tv = [Lv/(4spdv2)]0.25 = _________________ K Calculate Venus' emittance assuming that the Venus' actual temperature, Tvr, is 472o C = 745 K: ev = Lv/(4pdv2s Tvr4) = __________________ .arrow_forwardEarths daylight surface disk absorbs about 1045 W per m2 from the Sun. Using 6400 km for the Earths radius, how much of this radiative power is emitted by each square meter of the spherical Earth ?arrow_forward(b) Show that the equilibrium temperature of Venus is right for liquid water, given that it lies = 3.85 × 1026 W) and assuming a at a distance of d = 1.09 × 108 km from the Sun (Lo mean albedo of 0.45. Explain why, in reality, Venus is too hot for liquid water.arrow_forward
- You decide to go on an interstellar mission to explore some of the newly discovered extrasolar planets orbiting the star ROTOR. Your spacecraft arrives in the new system, in which there are five planets. ROTOR is identical to the Sun (in terms of its size, mass, age and composition). From your observations of these planets, you collect the following data: Density Average Distance from star (AU] Planet Mass Radius Albedo Temp. [C] Surf. Press. MOI Rotation [Earth = 1] (Earth = 1] [g/cm³] [Atm.] Period (Hours] Factor SIEVER EUGENIA 4.0 0.001 2.0 0.1 5.0 1.0 0.3 20 0.8 N/A 3.0 0.2 N/A 0.3 0.4 0.35 20 10 500 1000 5.0 4.0 0.5 0.8 0.4 0.7 -50 MARLENE CRILE 1.0 1.0 3.0 8.0 1,5 0.0 0.50 0.50 0.25 150 0.4 JANUS 100 12 0.1 10 -80 0.2 200 Figure 1: А Rotor 850 890 900 Wavelength (nm) A Sun В C 860 900 910 Wavelength (nm) 2414 a asarrow_forwardPlanetary scientists are excited about the possibility that some of the moons in the outer solar system might have a global subsurface ocean of liquid water. It's uncertain whether or not Jupiter's moon Callisto has a subsurface ocean, but some scientists think it might. Let's imagine that a short distance below the surface, Callisto has a water layer 20,000 meters thick. Callisto has a radius of 2.410x106 meters. Earth has a radius of 6.378x106 meters, and its oceans are equivalent to a layer of water 3000 meters thick all over the surface of the planet. Which of the following is an accurate comparison of the Earth’s amount of liquid water with this estimate of the amount of water Callisto might have? Group of answer choices If this scenario is correct, Earth and Callisto have roughly the same amount of liquid water. If these scientists are right, Callisto has about one-sixth as much liquid water as the Earth does. If these scientists are right, Callisto has about 11 times as…arrow_forwardCalculate the energy flux density, Fm , at the average distance of Mars from the Sun, rm , (energy flux divided by surface area of sphere). Mars' distance from the Sun = rm = 2.279 x 1013 cmFm = L /(4prm2) = ________________ ergs/s Next Calcuate the Amount of Solar Energy absorbed by Mars is the surface area of Mars which is facing the Sun (1/2 of Mars' surface area = 4pdm2 / 2 = 2pdm2 ) . Where dm = 3.398 x 106 cm is the radius of Mars. So Mars receives :arrow_forward
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