Astronomy Today (9th Edition)
9th Edition
ISBN: 9780134450278
Author: Eric Chaisson, Steve McMillan
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Videos
Question
Chapter 2, Problem 1P
To determine
The distance that corresponds to the angle of 1 arc minute
(a) The Moon
(b) The Sun
(c) Saturn
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Mars' 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?
(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.
Calculate the angle of (a) 10 (degree) (b) 1′ (minute of arc or arcmin) and (c) 1″(second of arc or arc second) in radians. Use 3600=2π rad, 10 =60′ and1′ = 60 ″
Chapter 2 Solutions
Astronomy Today (9th Edition)
Ch. 2 - Prob. 1DCh. 2 - Prob. 2DCh. 2 - Prob. 3DCh. 2 - Prob. 4DCh. 2 - Prob. 5DCh. 2 - Prob. 6DCh. 2 - Prob. 7DCh. 2 - Prob. 8DCh. 2 - Prob. 9DCh. 2 - Prob. 10D
Ch. 2 - Prob. 11DCh. 2 - Prob. 12DCh. 2 - Prob. 13DCh. 2 - Prob. 14DCh. 2 - Prob. 15DCh. 2 - Prob. 1MCCh. 2 - Prob. 2MCCh. 2 - Prob. 3MCCh. 2 - Prob. 4MCCh. 2 - Prob. 5MCCh. 2 - Prob. 6MCCh. 2 - Prob. 7MCCh. 2 - Prob. 8MCCh. 2 - Prob. 9MCCh. 2 - Prob. 10MCCh. 2 - Prob. 1PCh. 2 - Prob. 2PCh. 2 - Prob. 3PCh. 2 - Prob. 4PCh. 2 - Prob. 5PCh. 2 - Prob. 6PCh. 2 - Prob. 7PCh. 2 - Prob. 8P
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
- Consider the Earth's orbit around the Sun to be circular with radius R = 9.30 x 107 mi and it takes 365 days to complete one revolution. What is the distance Earth traveled for one revolution (circumference of a circle is 2??2πR )?arrow_forwardIn a distant star system there are many inhabitable planets. One of these planets is named Qomar. Qomar is 3.2 AU's from its star and takes 6.5 Earth years to go around its star once. There is another planet in the same star system called Ferenginar. Ferenginar is 0.9 AUs from the star. What is the length of a Ferengi year (on Ferenginar) in terms of Earth years?arrow_forwardTycho Brahe’s observations of the stars and planets were accurate to about 1 arc minute. To what distance does this angle correspond at the distance of (a) the Moon; (b) the Sun; and (c) Saturn (at closest approach)?arrow_forward
- Newton’s law of gravitation and the formula for centripetal acceleration can be used to show that: T^2=(4π^2/Gms)R^3 where G is the universal constant of gravitation and MS is the mass of the Sun. Take logarithms to base 10 of both sides of the equation to complete the expression for 2 lg T.2 lg T = ……………… × lg R + ……………………arrow_forwardI measured the angular separation of Jupiter's moons in arcminutes/arcseconds and converted this angle to radians. The conversion factor for degrees to radians was 57.3 degrees per radian. Which unit of measurement is larger? 1 degree or 1 radian?arrow_forward3. The Moon has a period of 27.3 days and a mean distance of 3.90×105 km from the center of Earth. a. Use Kepler's laws to find the period of a satellite in orbit 6.70x103 km from the center of Earth. b. How far above Earth's surface is this satellite?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_forwardThe Halley’s Comet regularly passes by the earth on its tour around the sun (at the time of Jesus’ birth itwas something different, most probably). The semi-major axis of the elliptical path is 17.8 AU(astronomical unit = 150·109 m). Halley’s last visit at our earth was in 1985. Are you going to experience the next visit?arrow_forwardUsing the m, = 1.99 x 1030 kg and the mę = 5.98 x 1024 kg and the distance between them as 1.00 AU (astronomical unit), what would be the orbital period of an object orbiting the sun at a distance of 2.12 AU? (please give answer in years) Number Unitsarrow_forward
- (to two decimal places): (what is ‘h’?) Eccentricity of earth orbit is 0.0167 µ(sun) = 1.32712E+11 km^3/s^ semimajor axis of Earth orbit = 1.49598E+08 need to figure out what ‘h’ is. a) Calculate the speed of the earth around the sun at aphelion? (29.29 KM/S) b) At perihelion? (30.29 KM/S)arrow_forwardAccording to Lunar Laser Ranging experiment the average distance LM from the Earth to the Moon is approximately 3.92 x 105 km. The Moon orbits the Earth and completes one revolution relative to the stars in approximately 27.5 days (a sidereal month). Calculate the orbital velocity of the Moon in m/s. Answer: Choose...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_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Kepler's Three Laws Explained; Author: PhysicsHigh;https://www.youtube.com/watch?v=kyR6EO_RMKE;License: Standard YouTube License, CC-BY