The Solar System
10th Edition
ISBN: 9781337672252
Author: The Solar System
Publisher: Cengage
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Chapter 1, Problem 4SP
To determine
The size of Sun model.
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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…
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?
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?
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?
Chapter 1 Solutions
The Solar System
Ch. 1 - Prob. 1RQCh. 1 - What is the largest dimension of which you have...Ch. 1 - What is the difference between the Solar System,...Ch. 1 - What is the difference between the Moon and a...Ch. 1 - Prob. 5RQCh. 1 - Why are light-years more convenient than miles,...Ch. 1 - Why is it difficult to detect planets orbiting...Ch. 1 - Prob. 8RQCh. 1 - What is the difference between the Milky Way and...Ch. 1 - What are the Milky Way Galaxys spiral arms?
Ch. 1 - Prob. 11RQCh. 1 - Where are you in the Universe? If you had to give...Ch. 1 - Prob. 13RQCh. 1 - Prob. 14RQCh. 1 - Prob. 15RQCh. 1 - How do we know? How does the scientific method...Ch. 1 - The equatorial diameter of Earth is 7928 miles. If...Ch. 1 - Prob. 2PCh. 1 - One astronomical unit (AU) is about 1.5 108 km....Ch. 1 - A typical galaxy is shown on the first page of the...Ch. 1 - Prob. 5PCh. 1 - Venus orbits 0.72 AU from the Sun. What is that...Ch. 1 - Light from the Sun takes 8 minutes to reach Earth....Ch. 1 - The Sun is almost 400 times farther from Earth...Ch. 1 - If the speed of light is 3.0 × 105 km/s, how many...Ch. 1 - Prob. 10PCh. 1 - Prob. 11PCh. 1 - Prob. 12PCh. 1 - How many galaxies like our own would it take if...Ch. 1 - Arrange the following in order of increasing size:...Ch. 1 - Arrange the following in order of increasing...Ch. 1 - Prob. 3SPCh. 1 - Prob. 4SPCh. 1 - Look at the center of Figure 1–4. Approximately...Ch. 1 - Look at Figure 1-6. How can you tell that Mercury...Ch. 1 - Prob. 3LLCh. 1 - Look at Figure 1-9. Would you say that the...Ch. 1 - Prob. 5LLCh. 1 - Prob. 6LL
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- The light a planet receives from the Sun (per square meter of planet surface) decreases with the square of the distance from the Sun. So a planet that is twice as far from the Sun as Earth receives (1/2)2=0.25 times (25%) as much light and a planet that is three times as far from the Sun receives (1/3)2=0.11 times (11%) as much light. How much light is received by the moons of Jupiter and Saturn (compared to Earth), worlds which orbit 5.2 and 9.5 times farther from the Sun than Earth?arrow_forwardWhy is it so hard to see planets around other stars and so easy to see them around our own?arrow_forwardDo the previous problem again, this time using the information that the Sun is 150,000,000 km away. You will get a very large number of km as your answer. To get a better feeling for how the distances compare, try calculating the time it takes light at a speed of 299,338 km/s to travel from the Sun to Earth and from Alpha Centauri to Earth. For Alpha Centauri, figure out how long the trip will take in years as well as in seconds.arrow_forward
- Earth is about 150 million kilometers from the Sun (1 Astronomical Unit, or AU), and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Using these two facts and the inverse square law for light, determine the apparent brightness that we would measure for the Sun if we were located at the following positions. a) At the orbit of Jupiter (780 million km from the Sun).arrow_forwardEarth is about 150 million kilometers from the Sun (1 Astronomical Unit, or AU), and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Using these two facts and the inverse square law for light, determine the apparent brightness that we would measure for the Sun if we were located at the following positions. a) At the orbit of Venus (67 million km from the Sun)arrow_forwardEarth is about 150 million kilometers from the Sun (1 Astronomical Unit, or AU), and the apparent brightness of the Sun in our sky is about 1300 watts/m^2. Using these two facts and the inverse square law for light, determine the apparent brightness that we would measure for the Sun if we were located at the following positions. a) At the orbit of Venus (67 million km from the Sun). b) At the orbit of Jupiter (780 million km from the Sun). c) At the mean distance of Pluto (40 Astronomical Units).arrow_forward
- If there were a planet about 9 times as far from the Sun as Earth is, what would its solar constant be? The solar constant of the Earth is 1350 watts per square meter. This question is not multiple choice. Provide an answer that is in watts per square meter (round to the nearest 10 watts per square meter).arrow_forwardEarth is about 150 million kilometers from the Sun (1 Astronomical Unit, or AU), and the apparent brightness of the Sun in our sky is about 1300 watts/m^2. Using these two facts and the inverse square law for light, determine the apparent brightness that we would measure for the Sun if we were located at the following positions. b) At the orbit of Jupiter (780 million km from the Sun).arrow_forwardIf you get a slope of 15 degrees per day for sunspot motion from Earth, what are the synodic and sidereal periods of rotation?arrow_forward
- The angular separation in degrees of two objects is (physical separation × 360°) / (2 π × distance). If an individual was observing our solar system from Castor at a distance of 7.6 light years. What angular resolution, in arcsecond, is needed to resolve the Sun-Earth system as distinct points of light?arrow_forwardAccording to the solar nebula theory, why is Earths orbit nearly in the plane of the Suns equator?arrow_forwardIf 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_forward
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