Foundations of Astronomy
13th Edition
ISBN: 9781305079151
Author: Michael A. Seeds, Dana Backman
Publisher: Cengage Learning
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Textbook Question
Chapter 1, Problem 6P
Venus orbits 0.72 AU from the Sun. What is that distance in kilometers? (Hint: See Problem 3.)
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Chapter 1 Solutions
Foundations of Astronomy
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 - Why do astronomers now label Pluto a dwarf planet?Ch. 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 - Prob. 10RQ
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 - Prob. 1DQCh. 1 - Prob. 2DQCh. 1 - Prob. 3DQCh. 1 - Prob. 4DQCh. 1 - The equatorial diameter of Earth is 7928 miles. If...Ch. 1 - The equatorial diameter of the Moon is 3476...Ch. 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 - The time of the Cambrian explosion is listed on...Ch. 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 - How long does it take light to cross the diameter...Ch. 1 - The nearest galaxy to our home galaxy is about 2.5...Ch. 1 - How many galaxies like our own would it take if...Ch. 1 - Look at the center of Figure 14. Approximately...Ch. 1 - Look at Figure 1-6. How can you tell that Mercury...Ch. 1 - Prob. 3LTLCh. 1 - Look at Figure 1-12. Would you call the...Ch. 1 - Of the objects listed here, which would be...Ch. 1 - Prob. 6LTL
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- how many seconds are in a mars year that has approx 678 earth days ?arrow_forward1 AU = 1.496 x10^11 marrow_forward(a) Jupiter's third-largest natural satellite, Io, follows an orbit with a semimajor axis of 422,000 km (4.22 ✕ 105 km) and a period of 1.77 Earth days (PIo = 1.77 d). To use Kepler's Third Law, we first must convert Io's orbital semimajor axis to astronomical units. One AU equals 150 million km (1 AU = 1.50 ✕ 108 km). Convert Io's a value to AU and record the result. aIo = AU (b) One Earth year is about 365 days. Convert Io's orbital period to Earth years and record the result. PIo = yr (c) Use the Kepler's Third Law Calculator to calculate Jupiter's mass in solar units. Record the result. MJup(Io) = MSun (d) Based on this result, Jupiter's mass is about that of the Sun. Jupiter has a similar fraction of the Sun's volume. The two objects therefore have rather similar density! In fact, Jupiter has a fairly similar composition as well: most of its mass is in the form of hydrogen and helium.arrow_forward
- A new planet is discovered orbiting a distant star. Observations have confirmed that the planet has a circular orbit with a radius of 12 AU and takes 117 days to orbit the star. Determine the mass of the star. State your answer with appropriate mks units. [NOTE: AU ..stands.for...astronomical unit". It is the average distance between Earth & the Sun. 1 AU≈ 1.496 x 1011 m.] Enter a number with units. I be quite large and your calculator will display the answer as a power of 10. If, as an example, your answer was 8.54 x 1056, you would type "8.54e56" into the answer box (remember to state your units with your answer).]arrow_forwardSuppose, hypothetically, that the Earth orbited the Sun at half its current distance. (That is, at 1/2 AU instead of 1 AU). What would be the length of the year? What else would be different?arrow_forwardMars' period (its "year") was noted by Kepler to be about 687 days (Earth days), which is (687d / 365 d) = 1.88 yr. Determine the distance of Mars from the Sun using the Earth as reference. (The distance of Earth from the Sun is 1.50 x 10 m) !3! Thu TMS IMS TE TES TES TE 2 28 x 10 m TES yr After reading and understanding the concept Gravity, please do the following problems: 1. What keeps a satellite up in its orbit around the Earth?arrow_forward
- The radius of the Earth's orbit is 1.50 1011 m and that of Mars is 2.28 1011 m. The star that this planet orbits is identical to our Sun. What is the orbital period of this planet?In years?arrow_forwardKepler's 1st law says that our Solar System's planets orbit in ellipses around the Sun where the closest distance to the Sun is called perihelion. Suppose I tell you that there is a planet with a perihelion distance of 2 AU and a semi-major axis of 1.5 AU. Does this make physical sense? Explain why or why not.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_forward
- What would be the period of revolution of a hypothetical planet whose circular orbit around the sun has a radius of 1.75 AU? (Hint: 1 AU = 1 Astronomical Unit = 1.5*1011) a) 2.3 yrs b) 1.45 yrs c) 2.9 yrsarrow_forwardthanks. Moon is at the distance 384400 km from Earth and orbits the Earth every ∼28 days. If the radius of the Moon is 1737 km (consider it to be spherical), what is the area of the moon as measured by the observer on Earth? (Hint: Length contraction)arrow_forward2. A planet orbits a star at 2.534 AU once every 41.39 years. What is the planet's velocity? (1 AU 1.496E11 m, 1 year = 365.25 days)arrow_forward
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Kepler's Three Laws Explained; Author: PhysicsHigh;https://www.youtube.com/watch?v=kyR6EO_RMKE;License: Standard YouTube License, CC-BY