Universe: Stars And Galaxies
6th Edition
ISBN: 9781319115098
Author: Roger Freedman, Robert Geller, William J. Kaufmann
Publisher: W. H. Freeman
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Chapter 23, Problem 8Q
To determine
The distance up to the galaxy.
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The difference in absolute magnitude between two objects is related to their fluxes by the flux-magnitude relation:
FA / FB = 2.51(MB - MA)
A distant galaxy contains a supernova with an absolute magnitude of -19. If this supernova were placed next to our Sun (M = +4.8) and you observed both of them from the same distance, how much more flux would the supernova emit than the Sun?
Fsupernova / FSun = ?
The figure below shows the spectra of two galaxies A and B.
Looking for ___Mpc
Chapter 23 Solutions
Universe: Stars And Galaxies
Ch. 23 - Prob. 1QCh. 23 - Prob. 2QCh. 23 - Prob. 3QCh. 23 - Prob. 4QCh. 23 - Prob. 5QCh. 23 - Prob. 6QCh. 23 - Prob. 7QCh. 23 - Prob. 8QCh. 23 - Prob. 9QCh. 23 - Prob. 10Q
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- Would you expect to observe every supernova in our own Galaxy? Why or why not?arrow_forwardCan you please help with Part 2 of 2? Thank you.arrow_forwardIf a galaxy contains a supernova that at its brightest has an apparent magnitude of +15, how far away is the galaxy? Assume that the absolute magnitude of the supernova is -14. Hint: Use the magnitude-distance formula: d = 10(mv - My + 5)/5 Mpcarrow_forward
- If a galaxy contains a supernova that at its brightest has an apparent magnitude of +25, how far away is the galaxy? Assume that the absolute magnitude of the supernova is -14. (Hint: use the magnitude - distance formula : d = 10 (Mv-Mv +5)/5 _______ Mpcarrow_forwardIf a galaxy contains a supernova that at its brightest has an apparent magnitude of 16, how far away is the galaxy? Assume that the absolute magnitude of the supernova is −18. (Hint: Use the magnitude-distance formula d = 10(mV − MV + 5)/5.)arrow_forward1arrow_forward
- If a galaxy contains a supernova that at its brightest has an apparent magnitude of +15, how far away is the galaxy? Assume that the absolute magnitude of the supernova is −17. Use the magnitude-distance formula: d = 10(mV − MV + 5)/5 .arrow_forwardNeeds Complete typed solution with 100 % accuracy.arrow_forwardA planetary nebula expanded in radius 0.3 arc seconds in 30 years. Doppler measurements show the nebula is expanding at a rate of 35 km/s. How far away is the nebula in parsecs? First, determine what distance the nebular expanded in parsecs during the time mentioned. Δd = vpc/sTs So we first need to convert the rate into pc/s and the time into seconds: vpc/s = vkm/s (1 pc / 3.09 x 1013km) vpc/s = ? Ts = (Tyr)(365 days/yr)(24 hrs/day)(3600 s/hr) Ts = ? s Δd= vpc/sTs Therefore, Δd = ? pcarrow_forward
- A Type la supernova explodes in a galaxy at a distance of 6.10×107 light-years from Earth. If astronomers detect the light from the supernova today, how many years T have passed since the supernova exploded? T= 2.07 x10 -5 years Given a Hubble constant of 74.3 km/s/Mpc, at what speed v is this galaxy moving away from Earth? v= km/s What is this galaxy's redshift? redshift:arrow_forwardFigure 2 shows the "rotation curve" of NGC 2742. It plots the “radial velocity (V)" (how fast material is moving either toward or away from us) that is measured for objects at different distances (R = radius") from the center of the galaxy. The center of the galaxy is at 0 kpc (kiloparsecs) with a speed of 9 km/sec away from us. (These velocities have been corrected for the observed tilt of the galaxy and represent true orbital velocities of the stars and gas.) 200 100 U4779 -100 As you can see, one side of the galaxy is moving with a negative velocity (spinning toward us), while the other side has a positive velocity (spinning away from us). Using Newton's gravity equation, we will be able to determine the gravitational mass of the entire galaxy and how the mass varies versus distance from the galaxy's center. -200 -8 8 -4 Radius (kpc) Read the following text carefully and follow the instructions: Select five radii spaced evenly from 0-10 kpc across the galaxy. Your selections should…arrow_forwardThe best parallaxes obtained with Hipparcos have an accuracy of 0.001 arcsec. If you want to measure the distance to a star with an accuracy of 10%, its parallax must be 10 times larger than the typical error. How far away can you obtain a distance that is accurate to 10% with Hipparcos data? The disk of our Galaxy is 100,000 light-years in diameter. What fraction of the diameter of the Galaxy’s disk is the distance for which we can measure accurate parallaxes?arrow_forward
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