Horizons: Exploring the Universe (MindTap Course List)
14th Edition
ISBN: 9781305960961
Author: Michael A. Seeds, Dana Backman
Publisher: Cengage Learning
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Chapter 15, Problem 1P
If you observed the Solar System from the nearest star (distance = 1.3 parsecs), what would the maximum angular separation be between Earth and the Sun? (Note: 1 pc is
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Thinking about the Scale of the Solar System As we discussed, the radius of the Earth is approximately 6370 km. The Sun, on the other hand, is approximately 700,000 km in radius and located, on average, one astronomical unit (1 au=1.5x108 km) from the Earth. Imagine that you stand near Mansueto Library, at the corner of 57th and Ellis. You hold a standard desk globe, which has a diameter of 12 inches, and you want to build a model of the Sun, Earth, and their separation that keeps all sizes and lengths in proportion to one another. a) How big would the Sun be in this scale model? Give your answer in feet and meters. b) The nearest star to the Solar System outside of the Sun is Proxima Centauri, which is approximately 4.2 light years away (a light year is the distance light travels in one year, or approximately 9.5x1012 km). Given the scale model outlined above, how far would a model Proxima Centauri be placed from you? Give your answer in miles and km.
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If you observed the solar system from the nearest star (distance 1.3 parsecs), what would the...
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If you observed the solar system from the nearest star (distance 1.3 parsecs), what would the maximum
angular separation be between Earth and the sun? Note that 1 pc is 2.1 105 AU. (Hint: Use the small-angle
formula, Chapter 3.)
The International Space Station is about 90 meters across and about 380 kilometers away. One night it appears to be the same angular size as Jupiter. Jupiter is 143,000 km in size. Use S = r x a to figure out how far away Jupiter is in AU. Note 1 AU = 1.5 x 108 km
Chapter 15 Solutions
Horizons: Exploring the Universe (MindTap Course List)
Ch. 15 - What produced the helium now present in the Sun’s...Ch. 15 - What produced the iron and heavier elements like...Ch. 15 - What evidence can you cite that disks of gas and...Ch. 15 - According to the solar nebula theory, why is the...Ch. 15 - Why does the solar nebula theory predict that...Ch. 15 - Prob. 6RQCh. 15 - If you visited another planetary system, would you...Ch. 15 - Why is almost every solid surface in our Solar...Ch. 15 - What is the difference between condensation and...Ch. 15 - Why don’t Terrestrial planets have rings like the...
Ch. 15 - How does the solar nebula theory help you...Ch. 15 - How does the solar nebula theory explain the...Ch. 15 - What does the term differentiated mean when...Ch. 15 - What processes cleared the nebula away and ended...Ch. 15 - Why would astronomically short lifetime of gas and...Ch. 15 - Prob. 16RQCh. 15 - What evidence can you cite that planets orbit...Ch. 15 - Why is the existence of “hot Jupiters” puzzling?...Ch. 15 - How Do We know? The evidence is overwhelming in...Ch. 15 - How Do We know? How can scientists know anything...Ch. 15 - If you could visit another planetary system while...Ch. 15 - Prob. 2DQCh. 15 - If the solar nebula hypothesis is correct, do you...Ch. 15 - If you observed the Solar System from the nearest...Ch. 15 - Prob. 2PCh. 15 - Prob. 3PCh. 15 - Prob. 4PCh. 15 - Prob. 5PCh. 15 - Prob. 6PCh. 15 - Suppose that Earth grew to its present size in 1...Ch. 15 - Prob. 8PCh. 15 - Prob. 9PCh. 15 - Prob. 1LTLCh. 15 - Why do astronomers conclude that the surface of...Ch. 15 - Prob. 3LTL
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- If you observed the Solar System from the vantage point of the nearest star, at a distance of 1.3 pc, what would the maximum angular separation be between Earth and the Sun? (Hint: Use the small-angle formula, Eq. 3-1.) (Note: 1 pc = 2.1 105 AU.)arrow_forwardVenus can be as bright as apparent magnitude -4.7 when at a distance of about 1AU. How many times fainter would Venus look from a distance of 7pc? Assume Venus has the same illumination phase from your new vantage point. (Hint: Recall the inverse square law; also, review the definition of apparent visual magnitudes. Note: 1 pc = 2.1 x 10^5 AU _______ times fainter what would it's apparently magnitude be? _______arrow_forward1. These images were taken six months apart, first when Earth was as far to one side of Alpha Centauri as it can get and again when Earth was as far to the other side of Alpha Centauri as it can get. Consequently, the baseline between the two observing positions is how many AU across? Answer: 1.7 arcsec 2. First, convert this to kilometers using your measurement of how many kilometers are in an AU. 3. Now convert the baseline to kilometers using the true value for the number of kilometers in an AU. 4. Calculate the distance to Alpha Centauri using parallax and the true baseline in kilometers. 5. Google and record the true value. 6. Calculate your percent error 7. Discuss significant sources of errorarrow_forward
- Use Kepler's 3rd Law and the small angle approximation. a) An object is located in the solar system at a distance from the Sun equal to 41 AU's . What is the objects orbital period? b) An object seen in a telescope has an angular diameter equivalent to 41 (in units of arc seconds). What is its linear diameter if the object is 250 million km from you? Draw a labeled diagram of this situation.arrow_forward1. These images were taken six months apart, first when Earth was as far to one side of Alpha Centauri as it can get and again when Earth was as far to the other side of Alpha Centauri as it can get. Consequently, the baseline between the two observing positions is how many AU across? Answer: 1.7 arcsec USE 1.7 arcsec NOT 2.946 2. First, convert this to kilometers using your measurement of how many kilometers are in an AU. 3. Now convert the baseline to kilometers using the true value for the number of kilometers in an AU. 4. Calculate the distance to Alpha Centauri using parallax and the true baseline in kilometers. 5. Google and record the true value. 6. Calculate your percent error 7. Discuss significant sources of errorarrow_forwardPart 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.arrow_forward
- Use this light curve of a star with a transiting exoplanet to answer the following. If the exoplanet is orbiting a star identical to our own Sun, what is its average orbital distance, in AU? What is the period in years of the transiting exoplanet? Use this light curve of a star with a transiting exoplanet to answer the following questions. Brightness 0 V V V B 5 10 15 20 Time (months) 25 30 35arrow_forwardNext you will (1) convert your measurement of the semi-major axis from arcseconds to AU, (2) convert your measurement of the period from days to years, and (3) calculate the mass of the planet using Newton's form of Kepler's Third Law. Use Stellarium to find the distance to the planet when Skynet took any of your images, in AU. Answer: 4.322 AU Use this equation to determine a conversion factor from 1 arcsecond to AU at the planet's distance. You will need to convert ? = 1 arcsecond to degrees first. Answer: 2.096e-5 AU (2 x 3.14 x 4.322 x (.000278/360) = 2.096e-5) Next, use this number to convert your measurement of the moon's orbital semi-major axis from arcseconds to AU. A) Calculate a in AU. B) Convert your measurement of the moon's orbital period from days to years. C) By Newton's form of Kepler's third law, calculate the mass of the planet. D) Finally, convert the planet's mass to Earth masses: 1 solar mass = 333,000 Earth masses.arrow_forwardConsidering that Earth experiences an average intensity of sunlight of 1330 W/m? and is at a distance from the Sun of 1.0 AU = 150 million km, and considering that that the apparent magnitude of the Sun as seen from Earth is m = -26.7, (a) then how far from the Sun would a distant Kuiper- belt world need to be in order for the apparent magnitude of the Sun to be m = -11.0 as seen on that world? Give your answer in AU. (b) What would be the orbital period of this world? Give your answer in Earth years.arrow_forward
- (a) The distance to a star is approximately 7.80 1018 m. If this star were to burn out today, in how many years would we see it disappear? years (b) How long does it take for sunlight to reach Saturn?arrow_forwardConsider the attached light curve for a transiting planet observed by the Kepler mission. If the host star is identical to the sun, what is the radius of this planet? Give your answer in terms of the radius of Jupiter. Brightness of Star Residual Flux 0.99 0.98 0.97 0.006 0.002 0.000 -8-881 -0.06 -0.04 -0.02 0.00 Time (days) → 0.02 0.04 0.06arrow_forwardVenus can be as bright as apparent magnitude −4.7 when at a distance of about 1 AU. How many times fainter would Venus look from a distance of 7 pc? Assume Venus has the same illumination phase from your new vantage point. (Hints: Recall the inverse square law; also, review the definition of apparent visual magnitudes. Note: 1 pc = 2.1 ✕ 105 AU). [fill in the blank] times fainter What would its apparent magnitude be?arrow_forward
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