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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Our Sun shines bright with a luminosity of 3.828 x 1026 Watt. Her energy is responsible for many processes and the habitable temperatures on the Earth that make our life possible.
Calculate the amount of energy arriving on the Earth in a single day.
To how many litres of heating oil (energy density: 37.3 x 106 J/litre) is this equivalent?
The Earth reflects 30% of this energy: Determine the temperature on Earth’s surface.
What other factors should be considered to get an even more precise temperature estimate?
Note: The Earth’s radius is 6370 km; the Sun’s radius is 696 x 103 km; 1 AU is 1.495 x 108 km.
The radiant flux at Earth is 1365 W/m2. Say we move towards the sun to a radius that is 1/3 the radius of the Earth's orbit. By what factor would the radiant flux increase or decrease?
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)
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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- (W/m), in Santa solar radiation into energy- solar radiation (w/m?) 400 300 200 100 time (hrs past midnight) 4 8 12 16 20 ot dengndi loarrow_forwardBy definition, 10,000 K is the upper limit of the Kelvin temperature scale? True O Falsearrow_forwardThe 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_forward
- If 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_forwardEarth's daylight surface disk absorbs about 1047 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? (Compare ratio of disk area to spherical surface area)arrow_forwardThe 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_forward
- Estimate 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_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_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_forward
- In contrast to Venus, the coldest temperature yet measured on the surface of any body in the solar system is -235°C. This temperature was detected by Voyager 2 as it passed by Neptune's largest moon, Triton. What is Triton's surface temperature in degrees Fahrenheit and in kelvins? The warmest temperature in Antarctica was recorded at the Vanda Station on the Scott Coast. On January 5, 1974 early summer in the southern hemisphere-a high temperature of 15°C was reached. Express this temperature in degrees Fahrenheit and kelvins. Answer must be given as calculated and then to the side given in significant figure answer formatarrow_forwardHow is the distance from the sun for planets in our solar system related to the mean temperature of each planet? To find out, a scatterplot that relates the natural log of the distance of each planet (including Pluto) from the sun in millions of miles and the natural log of the mean planetary temperature in Kelvin was created. In(Temperature) vs. In(Distance) 6.8 6.6 6.4 6.2 6 5.8 5.6 5.4 5.2 4.8 4.6 4.4 4.2 4 4 6 7 8. In(Distance) Predictor Coef 7.9009 SE Coef P Conatant 0.4381 18.03 0.000 In Distance -0.4536 0.0706 -6.42 0.004 s = 0.3446 R-Sq = 85.5 R-8q (adj) = 83.2% Based on the scatterplot and computer output, a reasonable estimate of mean temperature in Kelvin for Saturn, which is 886.7 million miles away from the sun is: O 4.822 degrees Kelvin because ý = -0.4536(In 886.7) + 7.9009 = 4.822. O 124.2 degrees Kelvin because in y = -0,4536(in 886,7) + 7.9009 =4,822 and e4.822 = 124.2. O 709.0 degrees Kelvin because In y = - 0.4536(log 886.7) + 7.9009 =6.564 and e6.564 = 709.0. O…arrow_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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