FOUNDATIONS OF ASTRONOMY (LL)-W/MINDTAP
14th Edition
ISBN: 9780357000502
Author: Seeds
Publisher: CENGAGE L
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Chapter 15, Problem 10RQ
To determine
The reason behind the fact that why the spiral arms cannot be physically connected structure and what will happen when they are physically connected.
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Why couldn't spiral arms be physically connected structures? What would happen to them?
A galaxy's rotation curve is a measure of the orbital speed of stars as a function of distance
from the galaxy's centre. The fact that rotation curves are primarily flat at large galactocen-
tric distances (vrot(r) ~ constant) is the most common example of why astronomer's believe
dark matter exists. Let's work out why!
Assuming that each star in a given galaxy has a circular orbit, we know that the accelera-
tion due to gravity felt by each star is due to the mass enclosed within its orbital radius r and
equal to v?/r. Here, ve is the circular orbit velocity of the star. (a) Show that the expected
relationship between ve and r due to the stellar halo (p(r) xr-3.5) does not produce a flat
rotation curve. (b) Show that a p(r) ∞ r¯² density profile successfully produces a flat ro-
tation curve and must therefore be the general profile that dark matter follows in our galaxy.
Observations indicate that each galaxy contains a supermassive black hole at its center. These black holes can be hundreds of
thousands to billions of times more massive than the Sun. Astronomers estimate the size of such black holes using
multiple methods.
One method, using the orbits of stars around the black hole, is an application of Kepler's third law. The mass of the black hole
can be found by using the given equation, where a is the semi-major axis in astronomical units, P is the period in years, and k is
a constant with a value of 1 Mo X year²/ AU³.
a³
M = k-
p²
What is the mass of a supermassive black hole if a star orbits it with a semimajor axis of 959 AU and a period of 13.3 years?
mass:
Another method measures the speed of gas moving past the black hole. In the given equation, v is the velocity of the gas (in
kilometers per second), r is the distance of the gas cloud from the black hole (in kilometers), and G is Newton's gravitational
constant. In this equation, G = 1.33 ×…
Chapter 15 Solutions
FOUNDATIONS OF ASTRONOMY (LL)-W/MINDTAP
Ch. 15 - What evidence can you give that we live in a...Ch. 15 - Prob. 2RQCh. 15 - Why didnt astronomers before Shapley realize how...Ch. 15 - Prob. 4RQCh. 15 - Prob. 5RQCh. 15 - Prob. 6RQCh. 15 - Which parts of a spiral galaxy comprise the...Ch. 15 - Prob. 8RQCh. 15 - Prob. 9RQCh. 15 - Prob. 10RQ
Ch. 15 - Prob. 11RQCh. 15 - Prob. 12RQCh. 15 - Prob. 13RQCh. 15 - Prob. 14RQCh. 15 - Prob. 15RQCh. 15 - Prob. 16RQCh. 15 - Prob. 17RQCh. 15 - Prob. 18RQCh. 15 - Prob. 19RQCh. 15 - Prob. 20RQCh. 15 - Prob. 21RQCh. 15 - Prob. 22RQCh. 15 - Prob. 23RQCh. 15 - Prob. 24RQCh. 15 - Prob. 25RQCh. 15 - Prob. 26RQCh. 15 - Rank these objects from oldest to youngest the...Ch. 15 - What evidence contradicts the top-down hypothesis...Ch. 15 - Prob. 29RQCh. 15 - The story of a process makes the facts easier to...Ch. 15 - Prob. 1PCh. 15 - Prob. 2PCh. 15 - Prob. 3PCh. 15 - Prob. 4PCh. 15 - Prob. 5PCh. 15 - Prob. 6PCh. 15 - Prob. 7PCh. 15 - Prob. 8PCh. 15 - If the Sun is 4.6 billion years old, how many...Ch. 15 - Prob. 10PCh. 15 - Prob. 11PCh. 15 - Prob. 12PCh. 15 - Prob. 13PCh. 15 - Prob. 14PCh. 15 - Prob. 15PCh. 15 - Prob. 1SOPCh. 15 - Prob. 2SOPCh. 15 - Prob. 2LTLCh. 15 - Prob. 3LTLCh. 15 - Prob. 4LTLCh. 15 - Prob. 5LTL
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- The rate at which a nebular cloud rotates increases as the cloud collapses to form systems of stars and planets. Consider a small segment of a nebular cloud with a mass m of 1.9 x 102" kg, tangential velocity vinitial equal to 6.8 km s-1 located at an orbital distance rinitial = 2.5 x 10* km. After the cloud collapses, the same small segment is located at an orbital distance rinal = 3.2 x 10° km. Calculate the change of the rotational velocity, Ao, for the cloud segment, assuming perfectly circular orbits. Perform your work and report your solution using two significant figures. Δω- 16605 rad s-!arrow_forwardThe Tully-Fischer method relies on being able to relate the mass of a galaxy to its rotation velocity. Stars in the outer-most regions of the Milky Way galaxy, located at a distance of 50 kpc from the galactic centre, are observed to orbit at a speed vrot determine the mass in the Milky Way that lies interior to 50 kpc. Express your answer in units of the Solar mass. 250 km s-1. Using Kepler's 3rd Law,arrow_forwardIndicate whether the following statements are true or false. (Select T-True, F-False. If the first is T and the rest F, enter TFFFFF). A) If we find an O type star in our galaxy, it must be in the disk. B) The nearest large spiral Galaxy, similar in size to the Milky Way, is the Andromeda Galaxy (M31). It is located about 2 million light years from Earth. C) The disk of the Milky Way galaxy is about 100,000 light years in diameter. D) On very large scales, matter in the Universe is distributed in clumps and voids. E) Distances to most stars in the Milky Way are measured by parallax. F) RR Lyrae and Cepheid variable stars are used to measure the distance to nearby galaxies.arrow_forward
- The Tully-Fischer method relies on being able to relate the mass of a galaxy to its rotation velocity. Stars in the outer-most regions of the Milky Way galaxy, located at a distance of 50 kpc from the galactic centre, are observed to orbit at a speed vrot = 250 km s−1. Using Kepler’s 3rd Law, determine the mass in the Milky Way that lies interior to 50 kpc. Express your answer in units of the Solar mass.arrow_forwardThe Plummer sphere is a model for some star clusters, and it has the gravitational potential as given in the provided image. M and rp are both constant. What is the density profile ρ(r)? When the Plummer sphere is observed from a large distance along the z axis, what is its surface density, Σ(R), where R is the distance from the center?arrow_forwardThe very first “image" of a black hole, at the centre of galaxy M87, was recently taken by the Event Horizon Telescope (EHT). More accurately, EHT imaged radio emission from the disc of gas that orbits the black hole with a lack of emission from the centre being attributed to the black hole. This image was only possible because EHT is not a single radio telescope, but is in fact a network of telescopes from around the world that take advantage of something known as interferometry. Interferometry is a method for combining the light from multiple telescopes, which results in an image that could have been taken by a telescope that has a diameter equal to the distance between the telescopes referred to as the "“baseline"-rather than the size of each individual telescope. EHT in particular combines observations from several Very Long Baseline Interferometry (VLBI) stations in order to achieve a high angular resolution. (a) Given that the "baseline" of EHT is effectively the diameter of the…arrow_forward
- The very first "image" of a black hole, at the centre of galaxy M87, was recently taken by the Event Horizon Telescope (EHT). More accurately, EHT imaged radio emission from the disc of gas that orbits the black hole with a lack of emission from the centre being attributed to the black hole. This image was only possible because EHT is not a single radio telescope, but is in fact a network of telescopes from around the world that take advantage of something known as interferometry. Interferometry is a method for combining the light from multiple telescopes, which results in an image that could have been taken by a telescope that has a diameter equal to the distance between the telescopes-referred to as the “baseline"-rather than the size of each individual telescope. EHT in particular combines observations from several Very Long Baseline Interferometry (VLBI) stations in order to achieve a high angular resolution. (a) Given that the "baseline" of EHT is effectively the diameter of the…arrow_forwardThe rate at which a nebular cloud rotates increases as the cloud collapses to form systems of stars and planets. Consider a small segment of a nebular cloud with a mass m of 1.9 × 102" kg, tangential velocity vinitial equal to 6.8 km s- located at an orbital distance rinitial = 2.5 × 10“ km. After the cloud collapses, the same small segment is located at an orbital distance rinal = 3.2 x 10° km. Calculate the change of the rotational velocity, A®, for the cloud segment, assuming perfectly circular orbits. Perform your work and report your solution using two significant figures. 5.0 x10-6 Ao = rad s-1 Incorrectarrow_forwardH5. A star with mass 1.05 M has a luminosity of 4.49 × 1026 W and effective temperature of 5700 K. It dims to 4.42 × 1026 W every 1.39 Earth days due to a transiting exoplanet. The duration of the transit reveals that the exoplanet orbits at a distance of 0.0617 AU. Based on this information, calculate the radius of the planet (expressed in Jupiter radii) and the minimum inclination of its orbit to our line of sight. Follow up observations of the star in part reveal that a spectral feature with a rest wavelength of 656 nm is redshifted by 1.41×10−3 nm with the same period as the observed transit. Assuming a circular orbit what can be inferred about the planet’s mass (expressed in Jupiter masses)?arrow_forward
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