The Cosmic Perspective (9th Edition)
9th Edition
ISBN: 9780134874364
Author: Jeffrey O. Bennett, Megan O. Donahue, Nicholas Schneider, Mark Voit
Publisher: PEARSON
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Question
Chapter 15, Problem 52EAP
a
To determine
The apparent brightness measured from the Sun if we are located at half earth’s distance from the Sun.
b
To determine
The apparent brightness measured from the Sun if we are located at two times earth’s distance from the Sun.
c
To determine
The apparent brightness measured from the Sun if we are located at five times earth’s distance from the Sun.
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Earth is about 150 million kilometers from the Sun (1 Astronomical Unit, or AU), and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Using these two facts and the inverse square law for light, determine the apparent brightness that we would measure for the Sun if we were located at the following positions.
a) At the orbit of Jupiter (780 million km from the Sun).
Earth is about 150 million kilometers from the Sun (1 Astronomical Unit, or AU), and the apparent brightness of the Sun in our sky is about 1300 watts/m^2.
Using these two facts and the inverse square law for light, determine the apparent brightness that we would measure for the Sun if we were located at the following positions.
b) At the orbit of Jupiter (780 million km from the Sun).
Earth is about 150 million kilometers from the Sun (1 Astronomical Unit, or AU), and the apparent brightness of the Sun in our sky is about 1300 watts/m^2.
Using these two facts and the inverse square law for light, determine the apparent brightness that we would measure for the Sun if we were located at the following positions.
a) At the orbit of Venus (67 million km from the Sun).
b) At the orbit of Jupiter (780 million km from the Sun).
c) At the mean distance of Pluto (40 Astronomical Units).
Chapter 15 Solutions
The Cosmic Perspective (9th Edition)
Ch. 15 - Prob. 1VSCCh. 15 - Prob. 2VSCCh. 15 - Prob. 3VSCCh. 15 - Prob. 4VSCCh. 15 - Prob. 5VSCCh. 15 - Prob. 6VSCCh. 15 - Prob. 1EAPCh. 15 - Prob. 2EAPCh. 15 - Prob. 3EAPCh. 15 - Prob. 4EAP
Ch. 15 - Prob. 5EAPCh. 15 - Prob. 6EAPCh. 15 - Prob. 7EAPCh. 15 - Prob. 8EAPCh. 15 - Prob. 9EAPCh. 15 - Prob. 10EAPCh. 15 - Prob. 11EAPCh. 15 - Prob. 12EAPCh. 15 - Prob. 13EAPCh. 15 - Prob. 14EAPCh. 15 - Prob. 15EAPCh. 15 - Prob. 16EAPCh. 15 - Prob. 17EAPCh. 15 - Prob. 18EAPCh. 15 - Prob. 19EAPCh. 15 - Prob. 20EAPCh. 15 - Prob. 21EAPCh. 15 - Prob. 22EAPCh. 15 - Prob. 23EAPCh. 15 - Prob. 24EAPCh. 15 - Prob. 25EAPCh. 15 - Prob. 26EAPCh. 15 - Prob. 27EAPCh. 15 - Prob. 28EAPCh. 15 - Prob. 29EAPCh. 15 - Prob. 30EAPCh. 15 - Prob. 31EAPCh. 15 - Prob. 32EAPCh. 15 - Prob. 33EAPCh. 15 - Prob. 34EAPCh. 15 - Prob. 35EAPCh. 15 - Prob. 36EAPCh. 15 - Prob. 37EAPCh. 15 - Prob. 40EAPCh. 15 - Prob. 42EAPCh. 15 - Prob. 44EAPCh. 15 - Prob. 45EAPCh. 15 - Prob. 46EAPCh. 15 - Prob. 47EAPCh. 15 - Prob. 48EAPCh. 15 - Prob. 49EAPCh. 15 - Prob. 50EAPCh. 15 - Prob. 52EAPCh. 15 - Prob. 53EAPCh. 15 - Prob. 54EAPCh. 15 - Prob. 55EAPCh. 15 - Prob. 56EAPCh. 15 - Prob. 57EAPCh. 15 - Prob. 58EAPCh. 15 - Prob. 59EAP
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- Earth is about 150 million kilometers from the Sun (1 Astronomical Unit, or AU), and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Using these two facts and the inverse square law for light, determine the apparent brightness that we would measure for the Sun if we were located at the following positions. a) At the mean distance of Pluto (40 Astronomical Units).arrow_forwardHow Do We Know? Why is it important that a theory make testable predictions?arrow_forwardThe light a planet receives from the Sun (per square meter of planet surface) decreases with the square of the distance from the Sun. So a planet that is twice as far from the Sun as Earth receives (1/2)2=0.25 times (25%) as much light and a planet that is three times as far from the Sun receives (1/3)2=0.11 times (11%) as much light. How much light is received by the moons of Jupiter and Saturn (compared to Earth), worlds which orbit 5.2 and 9.5 times farther from the Sun than Earth?arrow_forward
- Do the previous problem again, this time using the information that the Sun is 150,000,000 km away. You will get a very large number of km as your answer. To get a better feeling for how the distances compare, try calculating the time it takes light at a speed of 299,338 km/s to travel from the Sun to Earth and from Alpha Centauri to Earth. For Alpha Centauri, figure out how long the trip will take in years as well as in seconds.arrow_forwardParallaxes are measured in fractions of an arcsecond. One arcsecond equals 1/60 arcmin; an arcminute is, in turn, 1/60th of a degree (°). To get some idea of how big 1° is, go outside at night and find the Big Dipper. The two pointer stars at the ends of the bowl are 5.5° apart. The two stars across the top of the bowl are 10° apart. (Ten degrees is also about the width of your fist when held at arm’s length and projected against the sky.) Mizar, the second star from the end of the Big Dipper’s handle, appears double. The fainter star, Alcor, is about 12 arcmin from Mizar. For comparison, the diameter of the full moon is about 30 arcmin. The belt of Orion is about 3° long. Keeping all this in mind, why did it take until 1838 to make parallax measurements for even the nearest stars?arrow_forwardThe nearest star to our sun is Proxima Centauri, at a distance of 4.3 light-years from the Sun. A light-year is the distance that light travels in one year (365 days). How far away, in kilometers, is Proxima Centauri from the Sun?Express your answer using two significant figures.arrow_forward
- Give me the right answer please and thank you, take your timeCalculate the amount of time it takes for light reflected off the surface of a distant planet to reach us.1. Sunlight takes about 8.3 minutes to travel from the Sun to Earth. What is the Sun-Earth distance in AU? (Give your answer rounded to the nearest AU).2.Light is reflected off the surface of a planet 5.2 AU away from us. How long does it take this light to reach us from the planet? Give your answer in minutes, rounded to exactly one decimal place.arrow_forwardWhen we look at a particular star, we are seeing it as it was 371 years ago. How far away from us (in meters) is the star? Take a year to be 365.25 days.arrow_forwardThe Sun’s actual diameter is about 1,400,000 kilometers. How many “Earth diameters” is this? Given your 3-inch Earth, how large (i.e what diameter) of a ball would you need to represent the Sun? The average Earth–Sun distance is about 149,600,000 km. To represent this distance to scale, how far away would you have to place your 3-inch Earth from your Sun?arrow_forward
- Earth is about 150 million kilometers from the Sun, and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Determine the apparent brightness that we would measure for the Sun if we were located 12 times Earth's distance from the Sun.arrow_forwardUse the equation E = mc^2 where E is energy in Joules (J), m is mass in kilograms (kg) and c is the speed of light 3 x 10^8 m/s to answer the following: a) One ton of TNT releases 4.18 gigajoules of energy. The metric prefix giga means billion. a) How much mass would be required to release an equivalent amount of energy? b) How much energy (J) is equivalent to 1 kilogram of mass?arrow_forward
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