WebAssign for Seeds' The Solar System
10th Edition
ISBN: 9780357724729
Author: Seeds
Publisher: Cengage Learning US
expand_more
expand_more
format_list_bulleted
Videos
Textbook Question
Chapter 1, Problem 6P
Venus orbits 0.72 AU from the Sun. What is that distance in kilometers? (Hint: See Problem 3.)
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
One year on Planet AAA is 5 time the length of one earth year and it orbits its sun at an
average distance 4.38 AU. Planet ZZZ orbits the same sun at an average distance of
2.53 AU. Determine the length of one year on Planet ZZZ in earth days.
Two planets orbit the same star in circular orbits. One orbits at a distance of 167AU and takes 1.5days to complete an orbit. The second planet orbits at a distance of 4.7 AU. How long does it take the the second planet to complete one orbit? answer in days.
A planet's speed in orbit is given by V = (30 km/s)[(2/r)-(1/a)]0.5 where V is the planet's velocity, r is the distance in AU's from the Sun at that instant, and a is the semimajor axis of its orbit.
Calculate the Earth's velocity in its orbit (assume it is circular):
What is the velocity of Mars at a distance of 1.41 AU from the Sun?
What is the spacecraft's velocity when it is 1 AU from the Sun (after launch from the Earth)?
What additional velocity does the launch burn have to give to the spacecraft? (i.e. What is the difference between the Earth's velocity and the velocity the spacecraft needs to have?)
How fast will the spacecraft be traveling when it reaches Mars?
Does the spacecraft need to gain or lose velocity to go into the same orbit as Mars?
Chapter 1 Solutions
WebAssign for Seeds' 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
Knowledge Booster
Learn more about
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
- 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
- Suppose, 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_forwardThe 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_forward
- Kepler'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_forwardWhat 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_forward
- thanks. 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_forwardMathematically, prove the accuracy of Kepler’s 3rd law by computing and recording the values in the data table below.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Foundations of Astronomy (MindTap Course List)PhysicsISBN:9781337399920Author:Michael A. Seeds, Dana BackmanPublisher:Cengage Learning
Stars and Galaxies (MindTap Course List)PhysicsISBN:9781337399944Author:Michael A. SeedsPublisher:Cengage Learning
Horizons: Exploring the Universe (MindTap Course ...PhysicsISBN:9781305960961Author:Michael A. Seeds, Dana BackmanPublisher:Cengage Learning
AstronomyPhysicsISBN:9781938168284Author:Andrew Fraknoi; David Morrison; Sidney C. WolffPublisher:OpenStax
Foundations of Astronomy (MindTap Course List)
Physics
ISBN:9781337399920
Author:Michael A. Seeds, Dana Backman
Publisher:Cengage Learning
Stars and Galaxies (MindTap Course List)
Physics
ISBN:9781337399944
Author:Michael A. Seeds
Publisher:Cengage Learning
Horizons: Exploring the Universe (MindTap Course ...
Physics
ISBN:9781305960961
Author:Michael A. Seeds, Dana Backman
Publisher:Cengage Learning
Astronomy
Physics
ISBN:9781938168284
Author:Andrew Fraknoi; David Morrison; Sidney C. Wolff
Publisher:OpenStax
Kepler's Three Laws Explained; Author: PhysicsHigh;https://www.youtube.com/watch?v=kyR6EO_RMKE;License: Standard YouTube License, CC-BY