Universe
11th Edition
ISBN: 9781319039448
Author: Robert Geller, Roger Freedman, William J. Kaufmann
Publisher: W. H. Freeman
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Chapter 4, Problem 15Q
To determine
The diagram explaining the reason for the longer sidereal period of Mercury, compared to its synodic period, and the shorter sidereal period of Jupiter, compared to its synodic period.
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Saturn has an angular size of 16”, and an observed Synodic Period of 1.035yrs. Saturn’s moon, Titan orbits the planet with an angular separation of 192”, with a period of 15.9days. From these observations we can determine Saturn’s mass.
Use the Sidereal Period of Saturn above to find the distance to Saturn from the Sun.
The Earth covers about 1° per day in its orbit about the Sun, and the solar
day is slightly longer than the sidereal day. If Earth spun in a retrograde
direction like Venus but it still had the same sidereal period (23 hr 56
min), how long would the solar day be?
The solar day would be
72 hr and
min.
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…
Chapter 4 Solutions
Universe
Ch. 4 - Prob. 1CCCh. 4 - Prob. 2CCCh. 4 - Prob. 3CCCh. 4 - Prob. 4CCCh. 4 - Prob. 5CCCh. 4 - Prob. 6CCCh. 4 - Prob. 7CCCh. 4 - Prob. 8CCCh. 4 - Prob. 9CCCh. 4 - Prob. 10CC
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- In Ptolemys model, how do the epicycles of Mercury and Venus differ from those of Mars, Jupiter, and Saturn?arrow_forwardWhat phases would Venus show if the geocentric model were correct?arrow_forwardI. Directions: Complete the given table by finding the ratio of the planet's time of revolution to its radius. Average Radius of Orbit Times of Planet R3 T2 T?/R3 Revolution Mercury 5.7869 x 1010 7.605 x 106 Venus 1.081 x 1011 1.941 x 107 Earth 1.496 x 1011 3.156 x 107 1. What pattern do you observe in the last column of data? Which law of Kepler's does this seem to support? II. Solve the given problems. Write your solution on the space provided before each number. 1. You wish to put a 1000-kg satellite into a circular orbit 300 km above the earth's surface. Find the following: a) Speed b) Period c) Radial Acceleration Given: Unknown: Formula: Solution: Answer: Given: Unknown: Formula: Solution: Answer: Given: Unknown: Formula: Solution: Answer:arrow_forward
- From the data measured read off the period, P and the orbital radius, a from thegraph for the moon Ganymede . These values will have units of hours for the period P, and Jupiter Diameters (J.D.) for a. Enter your results here:P (period) = _________ hours a (orbital radius) = ________ J.D. After,In order to use Kepler's Third Law, you need to convert the period into years, using: 1 day = 24 hours and 1 year = 365.25 days. The orbital radius must be converted to A.U., using 1050 J.D. = 1 A.U. Enter your converted values here: P (period) = _________ years a (orbital radius) = ________ A.U.arrow_forwardMars is 1.5 times as far away from the Sun as Earth. Earth’s axis is tilted at 23.5o compared to the ecliptic. The axis of Mars is tilted at 25o compared to the ecliptic. The atmosphere on Earth is 100 times as thick as the atmosphere on Mars. Which of the following statements is true? 1.)Mars is so cold that the water there is ice, while Earth does not have any ice 2.)When it is summer in Earth’s northern hemisphere, it is winter on Mars’ southern hemisphere 3.) Earth has seasons, Mars does not 4.) All of the water on Mars is frozen, while Earth has water in solid, liquid and gas formarrow_forward"Aristarchus' Method of Determining the Distance to the Sun" At the first quarter of the Moon (at position Q), the angle (EQS) equals 90° (see the sketch below). Modern observations show that the interval from new Moon (near position N) to first quarter (at Q) is 35 minutes shorter than that from first quarter to full Moon (near position F). Given that the lunar synodic period (the interval between two identical lunar phases) is 29 days and 12.73 hours, estimate the Earth-Sun distance (ES) in terms of the Earth-Moon distance.arrow_forward
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