Universe
11th Edition
ISBN: 9781319039448
Author: Robert Geller, Roger Freedman, William J. Kaufmann
Publisher: W. H. Freeman
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Chapter 1, Problem 20Q
To determine
The angular size (in arcminutes) of Venus as seen from Earth on April 18, 2006, when it was at a distance of 0.869 au from Earth. Given that the diameter of Venus is 12,104 km.
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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.
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.
Suppose you were given a 3 in diameter ball to represent the Earth and a 1 in diameter ball to represent the Moon. (The actual ratio of Earth diameter to Moon diameter is 3.7 to 1.)
The actual average Earth–Moon distance is about 384,000 kilometers, and Earth’s diameter is about 12,800 kilometers. How many “Earth diameters” is the distance from Earth to the Moon?
Based on your answer to Question 2, what is the correct scaled distance of the Moon, using the 3-inch ball as Earth?
The 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? Give your answer in feet.
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? Give your answer in feet.
Could we use this scale to visualize the solar system instead of just the Earth and Moon? Why or Why…
Chapter 1 Solutions
Universe
Ch. 1 - Prob. 1CCCh. 1 - Prob. 2CCCh. 1 - Prob. 3CCCh. 1 - Prob. 1QCh. 1 - Prob. 2QCh. 1 - Prob. 3QCh. 1 - Prob. 4QCh. 1 - Prob. 5QCh. 1 - Prob. 6QCh. 1 - Prob. 7Q
Ch. 1 - Prob. 8QCh. 1 - Prob. 9QCh. 1 - Prob. 10QCh. 1 - Prob. 11QCh. 1 - Prob. 12QCh. 1 - Prob. 13QCh. 1 - Prob. 14QCh. 1 - Prob. 15QCh. 1 - Prob. 16QCh. 1 - Prob. 17QCh. 1 - Prob. 18QCh. 1 - Prob. 19QCh. 1 - Prob. 20QCh. 1 - Prob. 21QCh. 1 - Prob. 22QCh. 1 - Prob. 23QCh. 1 - Prob. 24QCh. 1 - Prob. 25QCh. 1 - Prob. 26QCh. 1 - Prob. 27QCh. 1 - Prob. 28QCh. 1 - Prob. 29QCh. 1 - Prob. 30QCh. 1 - Prob. 31QCh. 1 - Prob. 32QCh. 1 - Prob. 33QCh. 1 - Prob. 34QCh. 1 - Prob. 35QCh. 1 - Prob. 36QCh. 1 - Prob. 37QCh. 1 - Prob. 38QCh. 1 - Prob. 39QCh. 1 - Prob. 40QCh. 1 - Prob. 41QCh. 1 - Prob. 42QCh. 1 - Prob. 43QCh. 1 - Prob. 44Q
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- The Mars Robotic Lander for which we are making these calculations is designed to return samples of rock from Mars after a long time of collecting samples, exploring the area around the landing site, and making chemical analyses of rocks and dust in the landing area. One synodic period is required for Earth to be in the same place relative to mars as when it landed. Calculate the synodic period (in years) using the following formula: 1/Psyn = (1/PEarth) - (1/PMars) where PEarth is the sidereal period of the Earth (1 year) and PMars is the sidereal period of Mars. If 3/4 of a Martian year was spent collecting samples and exploring the terrain around the landing site, calculate how long the Mars Robotic Lander expedition took!arrow_forwardA planet's speed in orbit is given by V = (30 km/s)[(2/r)-(1/a)]0.5 where V is the planet's velocity, r is the distance in AU's from the Sun at that instant, and a is the semimajor axis of its orbit. Calculate the Earth's velocity in its orbit (assume it is circular): What is the velocity of Mars at a distance of 1.41 AU from the Sun? What is the spacecraft's velocity when it is 1 AU from the Sun (after launch from the Earth)? What additional velocity does the launch burn have to give to the spacecraft? (i.e. What is the difference between the Earth's velocity and the velocity the spacecraft needs to have?) How fast will the spacecraft be traveling when it reaches Mars? Does the spacecraft need to gain or lose velocity to go into the same orbit as Mars?arrow_forwardThe 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 kmarrow_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_forwardDione, a moon of Saturn, has an orbital radius of 377,400 km, and an orbital period of about 2.737 Earth days. Find the orbital period of Rhea, another moon of Saturn, which has an orbital radius of 527,040 km. Find the period in Earth days. Round to the nearest hundredth. Don't worry about putting the unit, just put the answer.arrow_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_forward
- The chart shows the length of time for each planet, in Earth days, to make one complete revolution around the Sun. Orbital Period of Planets iY the Solar System Orbital Period (Earth days) 88 225 365 687 4333 10 759 30 685 60 189 Planet Mercury Venus Earth Mars Jupiter Satum Uranus Neptune Source: NASA Use the data table above to compare the length of a year on Mars and Neptune. (HS-ESS1-4) a. One year on Neptune is almost 100 times longer than a year on Mars. b. One year on these two planets is nearly equal. c. One year on Mars is almost 100 times longer than a year on Neptune. d. One year these two planets is roughly equal to a year on Earth. Use the data table above to determine which of the following statements is TRUE. (HS-ESS1-4) a. There is no relationship between a planet's distance from the Sun and its length of year. b. The closer a planet is to the Sun, the longer the planet's year. c. One year on all planets is about 365 days long. d. The farther away a planet is from the…arrow_forwardAs we discuss in class, the radius of the Earth is approximately 6370 km. Theradius of the Sun, on the other hand, is approximately 700,000 km. The Sun is located,on average, one astronomical unit (1 au) from the Earth. Imagine that you stand near Mansueto Library, at the corner of 57th and Ellis.Mansueto’s dome is 35 feet (10.7 meters) high. Let’s imagine we put a model of theSun inside the dome, such that it just fits — that is, the model Sun’s diameter is 35 feet The nearest star to the Solar System outside of the Sun is Proxima Centauri,which is approximately 4.2 light years away. Given the scale model outlined above,how far would a model Proxima Centauri be placed from you? Give your answer inmiles and kmarrow_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_forward
- Use the ellipse tool to draw Earth’s orbit (e = 0.02, a = 1 AU). What are the closest and farthest distances of Earth from the Sun? Closest to the Sun (perihelion) à __0___ AU Farthest from the Sun (aphelion) à __3.96___ AU How do these distances compare?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_forwardThe angle on the sky between Venus and the Sun is measured to be 46.3° when Venus is at greatest eastern elongation. What is the distance of Venus from the Sun, measured in AU? Choose the answer below that most closely matches your answer. Select one: а. 1.763 AU O b. 0.587 AU Ос. 0.652 AU O d. 0.846 AU Ое. 0.723 AUarrow_forward
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