21st Century Astronomy
6th Edition
ISBN: 9780393428063
Author: Kay
Publisher: NORTON
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Chapter 2, Problem 42QP
To determine
Find the time it will take the Moon to move a distance equal to its diameter across the sky.
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Chapter 2 Solutions
21st Century Astronomy
Ch. 2.1 - Prob. 2.1ACYUCh. 2.1 - Prob. 2.1BCYUCh. 2.2 - Prob. 2.2CYUCh. 2.3 - Prob. 2.3CYUCh. 2.4 - Prob. 2.4CYUCh. 2.5 - Prob. 2.5CYUCh. 2 - Prob. 1QPCh. 2 - Prob. 2QPCh. 2 - Prob. 3QPCh. 2 - Prob. 4QP
Ch. 2 - Prob. 5QPCh. 2 - Prob. 6QPCh. 2 - Prob. 7QPCh. 2 - Prob. 8QPCh. 2 - Prob. 9QPCh. 2 - Prob. 10QPCh. 2 - Prob. 11QPCh. 2 - Prob. 12QPCh. 2 - Prob. 13QPCh. 2 - Prob. 14QPCh. 2 - Prob. 15QPCh. 2 - Prob. 16QPCh. 2 - Prob. 17QPCh. 2 - Prob. 18QPCh. 2 - Prob. 19QPCh. 2 - Prob. 20QPCh. 2 - Prob. 21QPCh. 2 - Prob. 22QPCh. 2 - Prob. 23QPCh. 2 - Prob. 24QPCh. 2 - Prob. 25QPCh. 2 - Prob. 26QPCh. 2 - Prob. 27QPCh. 2 - Prob. 28QPCh. 2 - Prob. 29QPCh. 2 - Prob. 30QPCh. 2 - Prob. 31QPCh. 2 - Prob. 33QPCh. 2 - Prob. 34QPCh. 2 - Prob. 35QPCh. 2 - Prob. 36QPCh. 2 - Prob. 37QPCh. 2 - Prob. 38QPCh. 2 - Prob. 39QPCh. 2 - Prob. 40QPCh. 2 - Prob. 41QPCh. 2 - Prob. 42QPCh. 2 - Prob. 43QPCh. 2 - Prob. 44QPCh. 2 - Prob. 45QP
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- Use the table to answer questions 13 through 15. A student collects the following data about the Sun, stars, moon, and Earth. Time of Day Sun Visible Moon Visible Stars Visible 5 am Sun Location Near horizon Above horizon Overhead Yes No Yes 10 am No Yes No No 1 pm 5 pm 9 pm Yes No Near horizon No Yes No Not visible No Yes Yes O What research question is the student investigating? A. How long does it take Earth to rotate on its axis? B. Does the moon rotate at a faster rate than Earth does? C. How do the locations of the stars relate to the moon? D. What is the relationship between time of day and seeing objects in the sky?arrow_forwardThe average Earth-Moon distance is 3.84 X 10^5 km, while the Earth-Sun is 1.496 X 10^8 km. Since the radius of the Moon is 1.74 X 10^3 km and that of the Sun is 6.96 X 10^5 km. a) Calculate the angular radius of the Moon and the Sun, qmax, according to the following figure. D Bax R b) Calculate the solid angle of the Moon and the Sun as seen from Earth. (c) Interpret its results; Would this be enough to explain the occurrence of total solar eclipses?arrow_forwardd. Diameter of the Sun I h A d. Standing on the beach at sunset, you extend the tip of your finger at your full arm length from your face, covering the Sun. Upon moving your finger around, you find that only about half of its width is needed to completely cover the Sun's diameter. You measure your finger width to be 0.5 inches. You know your arm length to be 28.0 inches. You have be told that the Sun is approximately 93 million miles away. Use this information to determine the approximate diameter of the Sun, filling in the table below with the proper quantities measure in meters. 1 = ½ finger g = eye level height d = object h = Diameter f= object height from level to top of eye level X = Arm A = angle width length distance from eye of the Sun objectarrow_forward
- Right Ascension and Declination is a coordinate system for objects in the sky, and is analogous to longitude and latitude coordinates, respectively, for objects on Earth. Right ascension (RA) coordinates are given in hours (h), minutes (m), and seconds (s). Declination (DEC) coordinates are given in degrees (°), arcminutes ('), and arcseconds ("). Sirius is the brightest star in the night sky. Its RA and DEC coordinates are 6h 45 m 7.96 s and -16° 44' 78.6". Using unit conversion, find the RA coordinate only in hours and round the coordinate to 5 significant figures.arrow_forwardThe planet Earth has a semi-major axis of a = 1.00 AU and an orbital period of P= 1 sidereal year = 365.25 days = 3.156 x 10^7 s. Compute the orbital periods of bodies orbiting the Sun with each of the following semi-major axes. a) a = 0.1 AU b) a = 10 AU c) a = 100 AU d) a = 1000 AU e) a = 10,000 AU 1 AU = 1.496 x 10^8 km = 1.496 x 10^11 m = 1.496 x 10^13 cm. GM(sun) = 1.327 x 10^20 m^3/s^2 = (Newton's Constant) x (Mass of Sun) %3D %3Darrow_forwarda. Describe the concept of "sphere of influence" and how it is estimated. b. Calculate the SOI for the Moon relative to the Earth. c. Would a single lone star have a computed sphere of influence, as defined in this class, which could be calculated? If no, why not? If yes, how would you do it?arrow_forward
- On the evening of an autumnal equinox day Siddhant noticed that Mars was exactly along the north-south meridian in his sky at the exact moment when the sun was setting. In other words, the Sun and Mars subtended an angle of exactly 90° as measured from the Earth. If the orbital radius of Mars is 1.52 au, What will be the approximate rise time of the mars on the next autumnal equinox day?arrow_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. F/ Earrow_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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