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 1, Problem 19Q
To determine
The minimum diameter of the object on the moon using the telescope witha clear view for a subtended angle of 2 arcsec.
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Next 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.
1. 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 error
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.
Chapter 1 Solutions
Universe: Stars And Galaxies
Ch. 1 - Prob. 1QCh. 1 - Prob. 2QCh. 1 - Prob. 3QCh. 1 - Prob. 4QCh. 1 - Prob. 5QCh. 1 - Prob. 6QCh. 1 - Prob. 7QCh. 1 - Prob. 8QCh. 1 - Prob. 9QCh. 1 - Prob. 10Q
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- The diameter of the earth’s moon 2159is miles. The diameter of Jupiter is 86881miles. How many of earth’s moons could fit across the diameter of Jupiter? Round your answer to the nearest whole number.arrow_forwardIf 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 2.1105 AU.) (Hint: Use the small-angle formula in Reasoning with Numbers 3-1.)arrow_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
- 1. 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_forwardEAn astronaut arrives on the planet Oceania and climbs to the top of a cliff overlooking the sea. The astronaut's eye is 100 m above the sea level and he observes that the horizon in all directions appears to be at angle of 5 mrad below the local horizontal. What is the radius of the planet Oceania at sea level? How far away is the horizon from the astronaut? 6000 km and 50 km 3600 km and 20 km 2000 km and 40 km 8000 km and 40 kmarrow_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_forward
- The table below presents the semi-major axis (a) and Actual orbital period for all of the major planets in the solar system. Cube for each planet the semi-major axis in Astronomical Units. Then take the square root of this number to get the Calculated orbital period of each planet. Fill in the final row of data for each planet. Table of Data for Kepler’s Third Law: Table of Data for Kepler’s Third Law: Planet aau = Semi-Major Axis (AU) Actual Planet Calculated Planet Period (Yr) Period (Yr) __________ ______________________ ___________ ________________ Mercury 0.39 0.24 Venus 0.72 0.62 Earth 1.00 1.00 Mars 1.52 1.88 Jupiter…arrow_forwardPluto has been hard to measure from Earth because of the atmosphere. In 2007 Young and Buie measured Pluto as having a diameter of 2322 km. In 2015 the New Horizons probe reached pluto and measured it up close and we now know the actual diameter is 2372 km. What was the percent error of the 2007 measurement? Enter your answer as a percent with a negative value if the 2007 measurement was too small and a positive value if it was too large.arrow_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 5 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. ote: 1 pc = 2.1 x 10° AU). times fainter What would its apparent magnitude be? Need Help? Read Itarrow_forward
- A Sense of Proportion: Neptune is about 50,000 km in diameter, and its largest moon, Triton, is about 2700 km in diameter. If you represent the planet with a ball 10 inches in diameter, how big a ball will you need to represent Triton?arrow_forwardNeptune is an average distance of 4.5×10^9 km from the Sun. - How many astronomical units (AU) is Neptune from the Sun? One AU is 1.50×10^8 km. - Estimate the length of the Neptunian year using your answer from part (a).arrow_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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