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 20Q
To determine
The angular size of the Venus as seen from the earth on April 18, 2006.
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On August 27, 2003, the planet Mars was at a distance of 0.373 AU from Earth. The diameter of Mars is 6788 km. Calculate the angular diameter of Mars, as seen from Earth on August 27, 2003. Give your answer in arcminutes.
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.
A 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?
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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- 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_forwardThe 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_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
- 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…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_forwardThe 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_forward
- 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.arrow_forward- How far (in km) is 3.5 lightyears(ly) – the distance traveled by light in one Earth year? - How much is this same value in parsecs and (C) in astronomical units (AU)? Use 299,732 km/s for the speed of light (c) and 1 year = 365 days. Show your solution and write your answer in both regular notation and scientific notation.arrow_forwardUsing Kepler’s Third Law (r3 = MT2 where M is the mass of the central star) find the orbital radius in astronomical units of this planet. M = 1.5 times the mass of the sun. Remember to convert days to years using 365.25 as the length of a year in days. Key Points to know: - The semimajor axis of the planet in AU is r = 0.0379 AU - The circumference of the orbit is l = 3.562 x 10^10 m - The orbital velocity in m/s is v = 1.874 x 10^5 m/s Questions that need to be answered: - With that orbital velocity, the radius of the orbit in meters, find the centripetal acceleration of our exoplanet: - Knowing the acceleration that our planet experiences, calculate the force that the host star exerts on the planet: - Knowing the force on the planet, the orbital radius, and the mass of the parent star, use the equation for gravitational force to find the mass of our planet (m2). (To get m1 in kg multiply the mass of the star in solar masses by 1.98 x 1030).arrow_forward
- Neptune 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_forwardUse 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_forwardYou are given the following data from observations of an exoplanet: Using Kepler’s Third Law (r3 = MT2 where M is the mass of the central star) find the orbital radius in astronomical units of this planet. M = 1.5 times the mass of the sun. Remember to convert days to years using 365.25 as the length of a year in days. What is the semimajor axis of this planet in AU? - Knowing the orbital radius in both kn and AU, use the value in km to find the circumference of the orbit, then convert that to meters. (Assume the orbit is a perfect circle). - Knowing the orbital circumference and the period in days, convert the days to seconds (multiply by 86,400) and find the orbital velocity in m/s - With that orbital velocity, the radius of the orbit in meters, find the centripetal acceleration of our exoplanet - Knowing the acceleration that our planet experiences, calculate the force that the host star exerts on the planet - Knowing the force on the planet, the orbital radius, and the mass of the…arrow_forward
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