UNIVERSE LL W/SAPLINGPLUS MULTI SEMESTER
11th Edition
ISBN: 9781319278670
Author: Freedman
Publisher: MAC HIGHER
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Chapter 1, Problem 19Q
To determine
The diameter in kilometers (km) of the smallest crater which is visible from the Moon, when a telescope can give a clear view of the objects and features that subtend an angle of at least 2 arcseconds.
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The smallest detail visible through Earth-based telescopes is about 1 arc second in diameter. What linear size is this on the Moon?
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Chapter 1 Solutions
UNIVERSE LL W/SAPLINGPLUS MULTI SEMESTER
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
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- If you observed the Solar System from the nearest star (distance = 1.3 parsecs), what would the maximum angular separation be between Earth and the Sun? (Note: 1 pc is 2.1105 AU.) (Hint: Use the small-angle formula in Reasoning with Numbers 3-1.)arrow_forwardWhat is the maximum angular diameter of the dwarf planet Ceres when it is closest to Earth? Could Earth-based telescopes detect surface features? Could the Hubble Space Telescope? (Hint: Use the small-angle formula, Eq. 3-1.) (Notes: Ceress average distance from the Sun is 2.8 AU and its diameter is 950 km. The best angular resolution of Earth-based telescopes at visual wavelengths is about 1 arc second and of Hubble about 0.1 arc second.)arrow_forwardEarlier in this chapter, we modeled the solar system with Earth at a distance of about one city block from the Sun. If you were to make a model of the distances in the solar system to match your height, with the Sun at the top of your head and Pluto at your feet, which planet would be near your waist? How far down would the zone of the terrestrial planets reach?arrow_forward
- The smallest detail visible through a Earth-based telescope is about 1.00 arcsecond in diameter. Use the small angle formula to determine to the right number of significant figures the size of the object in meters this would represent on Mars as it is at the closest distance from Earth. (??????? ???????? (arc−seconds)2.06×105=?????? ??????????????? ; ???−????????????−????=5.46×105 ??).arrow_forwardEarth orbits 1 AU from the Sun, and the Oort cloud extends from about 10,000 to 100,000 AU from the Sun. If you represent Earths orbit around the Sun with a paper plate 4 inches in radius, how far away will the inner edge of the Oort Cloud be? The outer edge? Express your answers in miles.arrow_forwardVenus can be as bright as apparent magnitude −4.7 when at a distance of about 1 AU. How many times fainter would Venus look from a distance of 1 pc? What would its apparent magnitude be? Assume Venus has the same illumination phase from your new vantage point. (Hints: Light follows an inverse square law as does gravity, review Section 5-1c; also, review the definition of apparent visual magnitudes, Chapter 2.) (Note: 1 pc = 2.1 × 105 AU.)arrow_forward
- The smallest detail visible through Earth-based telescopes is about 1 arc second in diameter. What linear size is this on the Moon? (Hint: Use the small-angle formula, Eq. 3-1.)arrow_forwardGalileo’s telescope showed him that Venus has a large angular diameter (61 arc seconds) when it is a crescent and a small angular diameter (10 arc seconds) when it is nearly full. Use the small-angle formula to find the ratio of its maximum distance to its minim um distance. Is this ratio compatible with the Ptolemaic shown on the right-hand page of Concept Art 4A? (Hint: See Reasoning with Numbers 3-1.)arrow_forwardThe main advantage of taking images of our planet from high-altitude orbits is that you can see a large fraction of the Earth's surface. The main disadvantage is that there will be too many low-altitude satellites obscuring your view of the Earth high-altitude satellites must launch to the west, which is more costly being at high altitude makes your satellite subject to strict international regulations you can't see as much detail in images taken from high altitudearrow_forward
- A Sense of Proportion: Mercury averages only 0.39 AU from the Sun, Venus 0.72 AU and Mars 1.52 AU. If you built a model solar system and represent the average distance from the Sun to Earth as 10 inches, how far would you place Mercury, Venus and Mars from the Sun?arrow_forwardWhat is an Astronomical Unit (AU) based on?arrow_forwardIf you observed the Solar System from the vantage point of the nearest star, at a distance of 1.3 pc, what would the maximum angular separation be between Earth and the Sun? (Hint: Use the small-angle formula, Eq. 3-1.) (Note: 1 pc = 2.1 105 AU.)arrow_forward
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