Foundations of Astronomy, Enhanced
13th Edition
ISBN: 9781305980686
Author: Michael A. Seeds; Dana Backman
Publisher: Cengage Learning US
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Question
Chapter 13, Problem 2DQ
To determine
How the Earth’s day and night look like from the Earth’s surface when the Sun turns to white dwarf and whether the Earth’s day would be bright and still the phases of the Moon could be seen.
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Read this main idea: The sun is the center of our solar system. Choose three details that go with the main idea.
The sun's gravity holds the planets in place. It provides them with heat and light.
The largest stars, called supergiants, are 1,500 times bigger than our sun.
It takes Earth 365 days to orbit the sun. Jupiter takes 12 years!
Our sun is not the largest or hottest star. It is a medium sized yellow star.
Radio telescopes use radio waves to show stars in great detail.
Astronomers long ago and today use star charts to map star locations.
All of the planets in our solar system revolve around one star-our sun.
Stars can be blue, white, yellow, or red. Blue stars are the hottest.
A reflector telescope bounces star light through mirrors.
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).
Part 3
1. The diameter of the Sun is 1,391,400 km. The diameter of the Moon is 3,474.8 km. Find
the ratio, r= Dsa/Dsvan between the sizes.
2. From the point of view of an obs erver on Eanth (consider the Earth as a point-like object),
during the eclipse, the Moon covers the Sun exactly. Sketch a picture to illustrate this
fact. Use a nuler to get a straight line. Your drawing does not need to be in scale.
3. The Sun is 1 Astronomical Unit (AU) away from the Earth. Find the distance between the
Earth and the Moon in AU's using the ratio of similar triangles. Show your work.
DEM=
AU.
Convert this to kilometers. Use 1 AU = 149,600,000 km.
DEM =
km.
Chapter 13 Solutions
Foundations of Astronomy, Enhanced
Ch. 13 - Prob. 1RQCh. 13 - Prob. 2RQCh. 13 - Prob. 3RQCh. 13 - Prob. 4RQCh. 13 - Prob. 5RQCh. 13 - Prob. 6RQCh. 13 - Prob. 7RQCh. 13 - Prob. 8RQCh. 13 - Prob. 9RQCh. 13 - Prob. 10RQ
Ch. 13 - Prob. 11RQCh. 13 - Prob. 12RQCh. 13 - Prob. 13RQCh. 13 - Prob. 14RQCh. 13 - Prob. 15RQCh. 13 - Prob. 16RQCh. 13 - Prob. 17RQCh. 13 - Prob. 18RQCh. 13 - Prob. 19RQCh. 13 - Prob. 20RQCh. 13 - Prob. 21RQCh. 13 - Prob. 22RQCh. 13 - Prob. 23RQCh. 13 - Prob. 24RQCh. 13 - Prob. 25RQCh. 13 - Prob. 26RQCh. 13 - Prob. 1DQCh. 13 - Prob. 2DQCh. 13 - Prob. 3DQCh. 13 - Prob. 4DQCh. 13 - Prob. 5DQCh. 13 - Prob. 1PCh. 13 - Prob. 2PCh. 13 - Prob. 3PCh. 13 - Prob. 4PCh. 13 - Prob. 5PCh. 13 - Prob. 6PCh. 13 - Prob. 7PCh. 13 - Prob. 8PCh. 13 - Add a fourth column to Table 13-1 and write in the...Ch. 13 - Prob. 10PCh. 13 - Prob. 11PCh. 13 - Prob. 12PCh. 13 - Prob. 13PCh. 13 - Prob. 14PCh. 13 - Prob. 15PCh. 13 - Prob. 1LTLCh. 13 - Prob. 2LTLCh. 13 - Prob. 3LTLCh. 13 - Prob. 4LTLCh. 13 - Prob. 5LTL
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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
- Is the Sun an average star? Why or why not?arrow_forwardDo 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_forwardAs seen from Earth, the Sun has an apparent magnitude of about 26.7 . What is the apparent magnitude of the Sun as seen from Saturn, about 10 AU away? (Remember that one AU is the distance from Earth to the Sun and that the brightness decreases as the inverse square of the distance.) Would the Sun still be the brightest star in the sky?arrow_forward
- The figure below shows the radial velocity of a star plotted as a function of time over the course of 20 days. Where is the planet in its orbit around the star when the star's radial velocity is 18 km/s? How do I determine this?arrow_forwardEarth 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_forwardTime From this light curve, we can deduce that... O the star has a high mass exoplanet orbiting it O the star has an exoplanet orbiting it that has an eccentric orbit O the star has an exoplanet orbiting it that has an eccentric orbit O the star has an exoplanet that is not on the same orbital plane as the star L Brightnessarrow_forward
- What is the apparent magnitude of the Sun as seen from Venus at perihelion? What is the apparent magnitude of the sun as seen from Pluto at aphelion?arrow_forwardFrom Earth, the Sun appears brighter than any other star because the Sun is the O newest star O hottest star O largest star O closest stararrow_forwardEarth 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).arrow_forward
- 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).arrow_forwardThe planet Jupiter is more than 5 times as far from the Sun as planet Earth. How does the brightness of the Sun appear at this greater distance?arrow_forward(Okay so its really astronomy) Star A and Star B are both 5 pc away, but Star A is at your zenith and Star B is at your nadir. How much more flux does Earth receive from Star A relative to Star B? Assume there are no interstellar clouds between you and each star.arrow_forward
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