Stars and Galaxies
9th Edition
ISBN: 9781305120785
Author: Michael A. Seeds, Dana Backman
Publisher: Cengage Learning
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Chapter 3, Problem 13P
To determine
The extend of prominence on the Sun’s disk during solar eclipses, whether the prominences can be seen if we use protective glasses.
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Part 3
1. The diameter of the Sun is 1,391,400 km. The diameter of the Moon is 3,474.8 km. Find
the ratio, r= Dsa/Dsvan between the sizes.
2. From the point of view of an obs erver on Eanth (consider the Earth as a point-like object),
during the eclipse, the Moon covers the Sun exactly. Sketch a picture to illustrate this
fact. Use a nuler to get a straight line. Your drawing does not need to be in scale.
3. The Sun is 1 Astronomical Unit (AU) away from the Earth. Find the distance between the
Earth and the Moon in AU's using the ratio of similar triangles. Show your work.
DEM=
AU.
Convert this to kilometers. Use 1 AU = 149,600,000 km.
DEM =
km.
I am trying to plot the ground tracks of an orbit. But I am having a problem with finding the longitude. The equation for the longitude is shown in the image. Is the Theta GMST initially zero because the greenwich meridian points to the Aries point (x-axis). How do you calculate alpha or vernal equinox? I saw a formula for alpha which is alpha = arctan(ry/rx), but the formula was for Right Ascension angle. Is the right ascension angle the same as vernal equinox. If not, then what is the formula for vernal equinox.
Next you will (1) convert your measurement of the semi-major axis from arcseconds to AU, (2) convert your measurement of the period from days to years, and (3) calculate the mass of the planet using Newton's form of Kepler's Third Law.
Use Stellarium to find the distance to the planet when Skynet took any of your images, in AU. Answer: 4.322 AU
Use this equation to determine a conversion factor from 1 arcsecond to AU at the planet's distance. You will need to convert ? = 1 arcsecond to degrees first. Answer: 2.096e-5 AU
(2 x 3.14 x 4.322 x (.000278/360) = 2.096e-5)
Next, use this number to convert your measurement of the moon's orbital semi-major axis from arcseconds to AU.
A) Calculate a in AU.
B) Convert your measurement of the moon's orbital period from days to years.
C) By Newton's form of Kepler's third law, calculate the mass of the planet.
D) Finally, convert the planet's mass to Earth masses: 1 solar mass = 333,000 Earth masses.
Chapter 3 Solutions
Stars and Galaxies
Ch. 3 - Prob. 1RQCh. 3 - Prob. 2RQCh. 3 - Prob. 3RQCh. 3 - Prob. 4RQCh. 3 - Prob. 5RQCh. 3 - Prob. 6RQCh. 3 - Prob. 7RQCh. 3 - Prob. 8RQCh. 3 - Prob. 9RQCh. 3 - Prob. 10RQ
Ch. 3 - Prob. 11RQCh. 3 - Prob. 12RQCh. 3 - Prob. 13RQCh. 3 - Why do solar eclipses happen only at new moon? Why...Ch. 3 - Prob. 15RQCh. 3 - Prob. 16RQCh. 3 - What is the angular diameter of the Moon in the...Ch. 3 - Why cant the Moon be eclipsed when it is halfway...Ch. 3 - Prob. 19RQCh. 3 - Prob. 20RQCh. 3 - Prob. 21RQCh. 3 - Prob. 22RQCh. 3 - Pretend the Moons orbit around Earth is a perfect...Ch. 3 - Prob. 2PCh. 3 - Prob. 3PCh. 3 - Prob. 4PCh. 3 - Prob. 5PCh. 3 - Prob. 6PCh. 3 - Prob. 7PCh. 3 - Prob. 8PCh. 3 - Prob. 9PCh. 3 - Prob. 10PCh. 3 - Prob. 11PCh. 3 - Examine the list of upcoming lunar eclipses in...Ch. 3 - Prob. 13PCh. 3 - Prob. 14PCh. 3 - Prob. 15PCh. 3 - When will the eclipse described in Problem 15 next...Ch. 3 - Prob. 17PCh. 3 - Prob. 18PCh. 3 - Prob. 1LLCh. 3 - Prob. 2LLCh. 3 - Prob. 3LLCh. 3 - Prob. 4LLCh. 3 - Prob. 5LLCh. 3 - Prob. 6LLCh. 3 - Prob. 7LLCh. 3 - Prob. 9LLCh. 3 - Prob. 10LLCh. 3 - Prob. 11LLCh. 3 - Prob. 12LL
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- 1. These images were taken six months apart, first when Earth was as far to one side of Alpha Centauri as it can get and again when Earth was as far to the other side of Alpha Centauri as it can get. Consequently, the baseline between the two observing positions is how many AU across? Answer: 1.7 arcsec USE 1.7 arcsec NOT 2.946 2. First, convert this to kilometers using your measurement of how many kilometers are in an AU. 3. Now convert the baseline to kilometers using the true value for the number of kilometers in an AU. 4. Calculate the distance to Alpha Centauri using parallax and the true baseline in kilometers. 5. Google and record the true value. 6. Calculate your percent error 7. Discuss significant sources of errorarrow_forwardIn countries at far northern latitudes, the winter months tend to be so cloudy that astronomical observations are nearly impossible. Why can’t good observations of the stars be made at those places during the summer months?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
- 1. These images were taken six months apart, first when Earth was as far to one side of Alpha Centauri as it can get and again when Earth was as far to the other side of Alpha Centauri as it can get. Consequently, the baseline between the two observing positions is how many AU across? Answer: 1.7 arcsec 2. First, convert this to kilometers using your measurement of how many kilometers are in an AU. 3. Now convert the baseline to kilometers using the true value for the number of kilometers in an AU. 4. Calculate the distance to Alpha Centauri using parallax and the true baseline in kilometers. 5. Google and record the true value. 6. Calculate your percent error 7. Discuss significant sources of errorarrow_forwardThe nearest star to our sun is Proxima Centauri, at a distance of 4.3 light-years from the Sun. A light-year is the distance that light travels in one year (365 days). How far away, in kilometers, is Proxima Centauri from the Sun?Express your answer using two significant figures.arrow_forward1. a) Why is it necessary to have a long baseline when using triangulation to measure the distances to objects in space? b) Of all visible objects in the celestial sphere, which appears to move the least? Why?arrow_forward
- Calculate the total amount of radiative energy per second intercepted by Mars from the Sun using the flux of radiation from the Sun at Mars' orbital radius. Flux of radiation from the Sun at Mars' orbital radius is 597 W m-2. The luminosity of the Sun Ls = 3.8×1026 W. Mars orbits at a distance of 2.25×1011 m (1.5 AU) from the Sun. Note: Consider carefully the cross-sectional area Mars presents to the outwards flow of radiative energy when answering this question.arrow_forwardBased on what you've learn on the impact of the Earth-Sun distance on the seasons, what can you say about the the cause of the seasons? (Give ALL correct answers, i.e., B, AC, BCD...)A) Earth's axis is tilted an an angle of 23.5 degrees compared to a line perpendicular (straight up and down) to its orbit, which is the main cause for the seasons.B) Earth's speed varies in its orbit around the Sun, giving us summer when Earth is moving fastest and winter when Earth is moving slowest.C) The Earth-Sun distance play a major role in creating seasons on Earth.D) The tilt of Earth's axis causes the Northern Hemisphere to be closer to the Sun than the southern hemisphere in summer, and vice versa in winter. E) The tilt of Earth's axis causes different portions of the Earth to receive more or less direct sunlight at different times of year.F) Earth's Northern Hemisphere is always tilted away from the Sun at an angle of 23.5 degrees.G) Earth's Northern Hemisphere is always tilted toward the Sun…arrow_forwardThe 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_forward
- As we discuss in class, the radius of the Earth is approximately 6370 km. Theradius of the Sun, on the other hand, is approximately 700,000 km. The Sun is located,on average, one astronomical unit (1 au) from the Earth. Imagine that you stand near Mansueto Library, at the corner of 57th and Ellis.Mansueto’s dome is 35 feet (10.7 meters) high. Let’s imagine we put a model of theSun inside the dome, such that it just fits — that is, the model Sun’s diameter is 35 feet The nearest star to the Solar System outside of the Sun is Proxima Centauri,which is approximately 4.2 light years away. Given the scale model outlined above,how far would a model Proxima Centauri be placed from you? Give your answer inmiles and kmarrow_forwardA distant galaxy has an apparent magnitude of 13 and is 5,000 kpc away. What is its absolute magnitude? (Round your answer to at least one decimal place.) The difference in absolute magnitude between two objects viewed from the same distance is related to their fluxes by the flux-magnitude relation. FA = 2.51(MB - MA) FB How does the absolute magnitude of this galaxy compare to the Milky Way (M = -21)? F, distant galaxy FMilky Wayarrow_forwardUsing a 8-m reflector telescope on Mars, what is the maximum distance we could measure using stellar parallax?arrow_forward
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