21st Century Astronomy
6th Edition
ISBN: 9780393428063
Author: Kay
Publisher: NORTON
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Chapter 3, Problem 32QP
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Find the number of time larger the Venus in the sky at the thinnest crescent than at the gibbous phases and find number of times closer to Earth at the phase of thinnest crescent than at the gibbous phases.
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Chapter 3 Solutions
21st Century Astronomy
Ch. 3.1 - Prob. 3.1ACYUCh. 3.1 - Prob. 3.1BCYUCh. 3.2 - Prob. 3.2CYUCh. 3.3 - Prob. 3.3CYUCh. 3.4 - Prob. 3.4CYUCh. 3 - Prob. 1QPCh. 3 - Prob. 2QPCh. 3 - Prob. 3QPCh. 3 - Prob. 4QPCh. 3 - Prob. 5QP
Ch. 3 - Prob. 6QPCh. 3 - Prob. 7QPCh. 3 - Prob. 8QPCh. 3 - Prob. 9QPCh. 3 - Prob. 10QPCh. 3 - Prob. 11QPCh. 3 - Prob. 12QPCh. 3 - Prob. 13QPCh. 3 - Prob. 14QPCh. 3 - Prob. 15QPCh. 3 - Prob. 16QPCh. 3 - Prob. 17QPCh. 3 - Prob. 18QPCh. 3 - Prob. 19QPCh. 3 - Prob. 20QPCh. 3 - Prob. 21QPCh. 3 - Prob. 22QPCh. 3 - Prob. 23QPCh. 3 - Prob. 24QPCh. 3 - Prob. 25QPCh. 3 - Prob. 26QPCh. 3 - Prob. 27QPCh. 3 - Prob. 28QPCh. 3 - Prob. 29QPCh. 3 - Prob. 30QPCh. 3 - Prob. 31QPCh. 3 - Prob. 32QPCh. 3 - Prob. 33QPCh. 3 - Prob. 34QPCh. 3 - Prob. 35QPCh. 3 - Prob. 36QPCh. 3 - Prob. 37QPCh. 3 - Prob. 38QPCh. 3 - Prob. 39QPCh. 3 - Prob. 40QPCh. 3 - Prob. 41QPCh. 3 - Prob. 42QPCh. 3 - Prob. 43QPCh. 3 - Prob. 44QPCh. 3 - Prob. 45QP
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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_forwardGive me the right answer please and thank you, take your timeCalculate the amount of time it takes for light reflected off the surface of a distant planet to reach us.1. Sunlight takes about 8.3 minutes to travel from the Sun to Earth. What is the Sun-Earth distance in AU? (Give your answer rounded to the nearest AU).2.Light is reflected off the surface of a planet 5.2 AU away from us. How long does it take this light to reach us from the planet? Give your answer in minutes, rounded to exactly one decimal place.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_forward
- Direction: Use your knowledge about solving equations to work out to complete the table below. Show your solution with proper units. R° (meters) T R° / T° { (meters) / Planet Average Times of Radius of Revolution (seconds) (seconds) } Planet's Orbit (Planet's year) R T (seconds) (meters) Mercury 5.7869 x 10:0 7.605 x 10 Venus 1.081 x 101 1.941 x 107 Earth 1.4996 x 10" 3.156 x 10 Mars 2.280 x 101 5.936 x 10 Jupiter 7.783 x 10" 3.743 x 10 Saturn 1.426 x 10 9.296 x 10arrow_forwardFirst we will do a simulation of Eratosthenes measurement of the size of our planet over 2000 years ago, but we will do it on a hypothetical planet. At one location on that planet at noon, the light from its star comes in directly overhead. At another town 200 miles due South, the shadows cast by the towers in that town indicate that the sunlight is coming in at an angle of approximately 10 degrees from directly overhead. Using this information (and the fact that there are 360 degrees in a circle and that the diameter of a circle is equal to the circumstances divided by pi), please show your work to explain what the circumstances and diameter of this new planet isarrow_forwardThe angle between two lines drawn from a point on Earth to two opposite sides of the Moon make an angle of 0.5 degrees. If you do the same thing for the two opposite ends of Andromeda (as shown above), you find an angle of 5 degrees. Let's assume Andromeda and the Moon are equally far away from our location on Earth (of course that's wrong, but how are we supposed to know?) - then how much larger would the diameter of Andromeda be (as indicated by the arrows at the top), compared to the diameter of the Moon? Pick the answer that's closest to what you get under this hypothetical assumption: A. Equal Diameter B. Twice C. Five times D. Ten timesarrow_forward
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