Universe: Stars And Galaxies
6th Edition
ISBN: 9781319115098
Author: Roger Freedman, Robert Geller, William J. Kaufmann
Publisher: W. H. Freeman
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Chapter 5, Problem 48Q
To determine
The possibility of a satellite to convert the orbital energy into thermal energy.
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Chapter 5 Solutions
Universe: Stars And Galaxies
Ch. 5 - Prob. 1QCh. 5 - Prob. 2QCh. 5 - Prob. 3QCh. 5 - Prob. 4QCh. 5 - Prob. 5QCh. 5 - Prob. 6QCh. 5 - Prob. 7QCh. 5 - Prob. 8QCh. 5 - Prob. 9QCh. 5 - Prob. 10Q
Ch. 5 - Prob. 11QCh. 5 - Prob. 12QCh. 5 - Prob. 13QCh. 5 - Prob. 14QCh. 5 - Prob. 15QCh. 5 - Prob. 16QCh. 5 - Prob. 17QCh. 5 - Prob. 18QCh. 5 - Prob. 19QCh. 5 - Prob. 20QCh. 5 - Prob. 21QCh. 5 - Prob. 22QCh. 5 - Prob. 23QCh. 5 - Prob. 24QCh. 5 - Prob. 25QCh. 5 - Prob. 26QCh. 5 - Prob. 27QCh. 5 - Prob. 28QCh. 5 - Prob. 29QCh. 5 - Prob. 30QCh. 5 - Prob. 31QCh. 5 - Prob. 32QCh. 5 - Prob. 33QCh. 5 - Prob. 34QCh. 5 - Prob. 35QCh. 5 - Prob. 36QCh. 5 - Prob. 37QCh. 5 - Prob. 38QCh. 5 - Prob. 39QCh. 5 - Prob. 40QCh. 5 - Prob. 41QCh. 5 - Prob. 42QCh. 5 - Prob. 43QCh. 5 - Prob. 44QCh. 5 - Prob. 45QCh. 5 - Prob. 46QCh. 5 - Prob. 47QCh. 5 - Prob. 48QCh. 5 - Prob. 49QCh. 5 - Prob. 50QCh. 5 - Prob. 51Q
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- The average temperature of the atmosphere has increased by 0.4°C over the last thirty years. Estimate how much energy has gone into warming up the planet in this way. Keep in mind that the atmosphere has a mass of 5 × 1018kg, and the specific heat capacity of air is about 1 Jg−1K−1. How do we get to this answer (2×1021J)arrow_forwardWe’re lucky that the earth isn’t in thermal equilibrium with the sun (which has a surface temperature of 5800 K). But why aren’t the two objects in thermal equilibrium?arrow_forwardBy definition, 10,000 K is the upper limit of the Kelvin temperature scale? True O Falsearrow_forward
- The following heat transfer formula quantifies the radiation emitted from the Sun: P = eoA(T4 – T?) Equation 5 where: P= radiated power (Watts) e = emissivity (=1 for ideal radiator; unitless) o = Stefan-Boltzmann constant = 5.67x10-8 W/m2-K+ A = radiating area (m²) T= temperature of radiator (Kelvin) Tc = temperature of surroundings (Kelvin) Q3 Using the following values, together with equation 5, calculate the power emitted by the Sun. sun's surface temperature = 5780 K temperature of the environment that the Sun is located in = 4 K emissivity of the Sun = 1 radius of the Sun = 695,700,000 m Stephen-Boltzmann constant o = 5.67 x 10-8 W/m2-K4 Show your work below-you may use the equation editor or insert a picture of your handwritten work.arrow_forwardIf the temperature of an object goes from 500 K to 9500 K,(c) By what factor is its energy emitted per second increased?If the temperature of an object goes from 900 K to 9000 K,(d) By what factor is its energy emitted per second increased?arrow_forwardEstimate the energy per second (power) radiated from the Sun assuming its surface temperature is 5,800 K and emissivity is 1. Estimate the fraction of this energy that reaches the Earth?arrow_forward
- The planet Venus is different from the earth in several respects. First, it is only 70% as far from the sun. Second, its thick clouds reflect 77% of all incident sunlight. Finally, its atmosphere is much more opaque to infrared light. Calculate the solar constant at the location of Venus, and estimate what the average surface temperature of Venus would be if it had no atmosphere and did not reflect any sunlight.arrow_forwardThe temperature T of the disk x2 + y2 ≤ 1 is given by T = 2x2 − 3y2 − 2x. Findthe hottest and coldest points of the disk.arrow_forwardEarth's daylight surface disk absorbs about 1036 W per m2 from the Sun. Using 6400 km for the Earth's radius, how much of this radiative power is emitted by each square meter of the spherical Earth? Hint: Compare the ratio of the disk area to the spherical surface area.arrow_forward
- Give proper explanation The surface temperature of the Sun is about 5750 K. What is this temperature on the Fahrenheit scale?arrow_forwardfrom d to f please thank youarrow_forwardAt "low" temperatures, the heat capacity of some materials varies with temperature as 3 12π¹ NkB T C (T) ² (£) ². 5 Here N is the number of atoms, kB is Boltzmann's constant, and Tp is the "Debye temperature," which is different for different materials. For example, 2300 K for diamond, which is particularly high. TD = How much energy would it take to raise the temperature of one mole of diamond from 100 K to 300K? O 150 Joules. 60 Joules. O 319 joules. O 630 Joules.arrow_forward
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