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Astronomy Assignment #6: Newton's Law of Gravity and Structure of the Milky Way Your Name______________________________________ Your Class Meeting Time __________________________ This assignment is due on Friday, 16 Oct and Thursday, 15 Oct Submit this cover sheet with your assignment. Complete the assigned problems from the text listed below and address the Instructor Assigned Topic. Mathematical problems may be hand written. Write out the problem, show your work in solving the problem and state your answer in a complete sentence. Failure to complete all three of these tasks will result in less than full credit awarded. Read the following sections in Astronomy Notes: Chapter 5 Newton's Law of Gravity Introduction Universal Law of Gravity o Characteristics of Gravity---why it is the most important force in astronomy. Inverse Square Law Gravitational Acceleration o Measuring the Mass of the Earth Read the following sections in Astronomy Notes: Chapter14: The Interstellar Medium and the Milky Way Introduction Interstellar Medium (ISM) o Dust o Gas Galactic Structure The Milky Way: our galaxy o Period-Luminosity Relation for Variable Stars o Our Location o Our Motion o Deriving the Galactic Mass from the Rotation Curve Answer the following Review Questions from Nick Strobel’s AstronomyNotes: Chapter 5: Newton's Law of Gravity 1. What things does gravity depend on? Gravity depends on three things. First, the strength of the gravitational attraction between two objects depends directly on the mass of each object, or equivalently on the product of the masses of each object. Second, the strength of the gravitational attraction between two objects depends inversely on the separation of the two objects. In particular, the strength of gravity follows an inverse square relation so that the strength of the gravitational attraction between two bodies depends 1 on 2 . Finally, the strength of the gravitational attraction depends on the intrinsic strength of d gravity as expressed through the Universal Gravitational Constant G. G has a very small value of 6.6710-11 in mks (meters-kilogram-seconds) units and illustrates that gravity is an intrinsically weak force. So weak, that for gravity to be “felt” between two objects, at least one of the objects must be astronomical in mass. 2. How does gravity vary with distance between objects and with respect to what do you measure the distances? The strength of the gravitational attraction between two objects depends inversely on the separation of the two objects. In particular, the strength of gravity follows an inverse square relation so that the 1 strength of the gravitational attraction between two bodies depends on 2 . The distance d is d measured from the center of one object to the center of the other object so long as they are spherical in shape. 3. What would happen to the Earth's orbit if the Sun suddenly turned into a black hole (of the same mass)? Why? If the Sun suddenly turned into a black hole with the same mass as the Sun, the Earth would not feel any difference in the gravitational attraction to the now-black-hole-Sun. The gravitational attraction between any two objects depends only on the mass of each object and the separatin distance from their centers – not on the radius of either object. The Earth would get quite cold since black holes do not emit any radiation (i.e. heat). However, the Earth’s orbit would remain unchanged. 4. Why is gravity called an ``inverse square law''? Gravity called an ``inverse square law'' because its strength depends inversely on the square of the 1 distance between the two objects. In mathematical lingo FGravity 2 . To illustrate, if the d separation between two objects is increased by a factor of two, the strength of their mutual gravitational attraction will decrease by a factor of four. I the separation between two objects is decreased by a factor of five, the strength of their mutual gravitational attraction will increase by a factor of 25. 5. What is the difference between a simple inverse relation and an inverse square relation? The difference between a simple inverse relation and an inverse square relation is that a simple inverse relation decreases much more slowly than an inverse square relation. For example, in a simple inverse relation doubling the distance would simply halve the strength. However, with an inverse square relation doubling the distance would decrease the strength by a factor of four. Inverse square relations rapidly weaken with distance. 6. If the Earth was 3 A.U. from the Sun (instead of 1 A.U.), would the gravity force between the Earth and the Sun be less or more than it is now? By how many times? If the Earth was 3 A.U. from the Sun (instead of 1 A.U.), the gravity force between the Earth and the Sun be less than it is now by 32 = 9 times? 7. If Mercury was 0.2 A.U. from the Sun (instead of 0.4 A.U.), would the gravity force between Mercury and the Sun be less or more than it is now? By how many times? If Mercury was 0.2 A.U. from the Sun (instead of 0.4 A.U.), the gravity force between Mercury and the Sun be more than it is now by 4 times. 8. Why do astronauts in orbit around the Earth feel ``weightless'' even though the Earth's gravity is still very much present? Astronauts in orbit around the Earth are in free fall as they orbit the Earth. They feel weightless because they are falling towards the Earth. They don’t hit the Earth because their horizontal velocity (i.e. orbital velocity) is so large that as they fall they move over the “edge” of the Earth. Answer the following Review Questions from Nick Strobel’s AstronomyNotes: Chapter 14: The Interstellar Medium and the Milky Way. 1. What is the interstellar medium composed of? The interstellar medium is composed of gas and dust. The gas is primarily hydrogen and helium, with trace amounts of other gases. The dust is composed of microscopic (mostly) grains of ices (H2O, CO2, NH3, CH4) , silicate grains (SiO4), Carbon grains, and metals (Fe, Ni) 2. How does dust make stars appear redder than they actually are? Dust makes stars appear redder than they are through a process called interstellar reddening. Interstellar reddening is the process whereby light passing through an interstellar dust cloud is separated into two components. The long wavelength red light component is allowed to pass through the interstellar cloud because the wavelength of red light is longer than the typical diameter of the dust grains, while the bluelight component is scattered (i.e. reflected) in all directions because its wavelength is about the same size as the diameter of the dust grains. Thus much of the blue light from stars behind an interstellar dust cloud is removed from the beam causing the star to look redder than it really is. The front surface of the cloud will look blue from the scattered blue star light and astronomers call this a reflection nebula. The star appears redder than it really is because the blue component of the light is scattered out of the beam. 3. How does dust cause the extinction of starlight? Dust causes the extinction of star light by two processes. If the dust grains are very small, on the order of the wavelength of light, then the light is scattered out of the beam reducing the beams intensity. If the dust grains are larger than the wavelength of light, then the dust grains absorb the star light directly and heat up. The heated dust grains will then re-emit the light energy at infrared wavelengths in all directions. A similar effect can be seen in terrestrial clouds. If the cloud particles are small, then they reflect the sunlight and appear very bright. Rain clouds, however, have much larger cloud particles and these larger ccloud particles absorb the sunlight and appear dark gray in color. 4. Where is the dust thought to come from? Dust grains form from the material in the outflows of stellar winds. Typically giant stars, planetary nebula and supernovas create dust particles from their outflows. 5. What are the characteristics of the gaseous part of the ISM? Is the gas all at the same temperature and density? How do you know? The presence of interstellar gas can be seen when you look at the spectral lines of a binary star system and you see narrow spectral lines that do not move among the broad stellar spectral lines that shift as the two stars orbit each other. The narrow lines are from much colder gas in the interstellar medium between us and the binary system. The gaseous part of the ISM is composed primarily of hydrogen (90%), Helium (10%) and trace amounts of other gasses (e.g. CO, H2O, NH3…). The hydrogen appears in three forms in the ISM; neutral atomic hydrogen (HI), ionized hydrogen atoms (HII) and molecular hydrogen (H2). The ionized hydrogen emits light in the visible band as the electrons recombine with the protons and the neutral atomic and molecular hydrogen emits light in the radio band of the electromagnetic spectrum. Our galaxy, the Milky Way, has about 3 billion solar masses of H I gas with about 70% of it further out in the Galaxy than the Sun. Most of the H I gas is in disk component of our galaxy and is located within 720 light years from the mid-plane of the disk. Distribution of HI in the Galaxy In our galaxy the H II regions are distributed in a spiral pattern associated with the very luminous new O, B and A stars that define the spiral arms. Molecular hydrogen H2 does not produce radio emission. Fortunately, there is evidence of a correlation between the amount of CO and H2, so the easily detected CO radio emission lines (at 2.6 and 1.3 mm) are used to infer the amount of H2. The CO emission is caused by H2 molecules colliding with the CO molecules. An increase in the density of the H2 gas results in more collisions with the CO molecules and an increase in the HII & GMC’s in the plane of the Milky Way CO emission. There is some evidence that indicates that 90% of the H2 is locked up in 5000 giant molecular clouds with masses greater than 105 solar masses and diameters greater than 65 light years. The largest ones, with diameters greater than 160 light years, have more than a million solar masses and make up 50% of the total molecular mass. The Milky Way has about 2.5 billion solar masses of molecular gas with about 70% of it in a ring extending from 13,000 to 26,000 light years from the center. Not much molecular gas is located at 4,900 to 9,800 light years from center but about 15% of the total molecular gas mass is located close to galactic center within 4,900 light years from the center. Most of the molecular clouds are clumped in the spiral arms of the disk and stay within 390 light years of the disk mid-plane. 6. How is the 21-cm line radiation produced? 21-cm line radiation is produced when an electron in excited atomic hydrogen radiatively decays to the ground state by emitting a photon of 21-cm wavelength in the radios portion of the electromagnetic spectrum. Solitary hydrogen has only two allowed spins states (for reasons that only nature knows). Spin is analogous to the spinning of a child’s top although it is quantum mechanical interpretation is less intuitive. An electron and proton process a property a physicist calls “spin”. The electron and proton of the hydrogen atom can spin in opposite directions – called anti-parallel spins – or the electron and proton can spin in the same direction – called parallel spins. The state of parallel spins is just ever so slightly higher in energy – thus parallel spins represent an excited state of the hydrogen atom. The lower energy antiparallel spins represent the lower ground state. The energy difference between these two states is so slight that a hydrogen atom can be excited simply by a collision with another atom. These collisions are common in interstellar clouds of gas and dust. So there are a significant fraction of hydrogen atoms in an interstellar cloud that are in this parallel spin excited state. The excited state, however, is unstable. Given a short time, the excited hydrogen atom will decay back to the ground state by radiating a low energy 21-cm wavelength photon - so called radiative decay.. These 21-cm photons reside in the radio portion of the electromagnetic spectrum. Although it is impossible to detect one 21-cm photon from a cloud of interstellar gas, it is quite simple with a radio telescope to detect the many many billions of 21-cm photons that a single cloud can emit. The detection of 21-cm radiation by radio telescopes along with a measurement of its Doppler-shifted velocity is a cornerstone of modern astronomy and has been used to determine our place in the galaxy and properties of the galaxy. 7. Why is the 21-cm line radiation so important for determining galactic structure and mass? The 21-cm line radiation is so important for determining galactic structure and mass because the 21-cm photons can travel vast distances even through tick interstellar clouds of gas and dust that visual photons cannot pass through. Radio telescopes are extraordinarily sensitive due to their very large diameters and very sensitive electron detectors and amplifiers. 21-cm radio photons from neutral hydrogen clouds can be detected even if they originate from the very opposite side of the galaxy. 8. How is the 21-cm line radiation used to determine galactic structure and mass? The distance of individual neutral hydrogen clouds in the galaxy’s disk can be determined using their radial velocities (that are determined from their Doppler shifts). Thus, astronomers can determine both the distance to these clouds of hydrogen and their velocity. With the hydrogen cloud’s velocity and distance an astronomer can assemble a galactic rotation curve from which the orbital velocity and orbital radius of hydrogen clouds around the center of the galaxy can be deduced. With these two pieces of information the orbital velocity equation can be used to determine the mass of the galaxy within each clouds orbital radius. 9. What is the name for our galaxy and what kind of galaxy is it? Our galaxy is called the Milky Way Galaxy. It is a barred spiral galaxy that consists of a large (100,000 ly diameter) disk of stars and gas and dust around a central bulge and bar ( galactic bars are still pretty mysterious) all embedded in a larger (diameter about 150,000 ly) oblate spheroidal halo of dark matter, globular clusters, and halo stars (very rare). 10. How big is our galaxy? How many stars are in it and how do we know? It is a barred spiral galaxy that consists of a large (100,000 ly diameter) disk of stars and gas and dust around a central bulge and bar ( galactic bars are still pretty mysterious) all embedded in a larger (diameter about 150,000 ly) oblate spheroidal halo of dark matter, globular clusters, and halo stars (very rare). We cannot count the stars in our galaxy because we cannot see them all due to the dust in the disk of the galaxy that limits our visibility to within 10,000 light years of the Sun. However, using the galactic rotation curve derived from 21-cm radiation from interstellar neutral hydrogen clouds astronomers can calculate the mass of our galaxy. From this mass, astronomers can estimate the number of stars. For example, the mass of the Milky Way within the Sun’s orbital radius is about 1041 kilograms. If this mass was solely portioned into sun-like stars, there would be about 1011 sun-like stars – about 100 billion stars. However, we know that the most common stars in the Milky Way galaxy are not sun-like starsbut K & M main sequence stars whose masses are a few tenths of the Sun’s mass. So, the number of K & M main sequence stars equivalent to 100 billion sun-like stars is about 300 billion stars. The number of about 300 billion stars in roughly the number of stars astronomers believe reside in the Milky Way galaxy. 11. How are Cepheids and RR-Lyrae stars considered to be standard candles? How can you find their luminosity? Cepheids and RR-Lyrae variable stars are standard candles because their luminosities are either known “a priori” or can be determined. The luminosity of these objects must first be established by directly measuring their distance by some means such as stellar parallax or some other method. Once the luminosity a these stars are measured (by comparing their apparent magnitudes and their distance) and once the luminosity is established as constant for all objects of that class, then the luminosity of one of these standard candles can be safely assumed when it is seen in another location. 12. How can you use the period-luminosity relation to find distances? The luminosity of a Cepheid is found to correlate very closely to its period of oscillation; the longer the oscillation period the greater the peak luminosity of the star. Thus the luminosity of a Cepheid can be determined by measuring its period of oscillation. Form the period of oscillation, the absolute magnitude can be determined. For example, if the period of oscillation of a Type I Cepheid variable was 10 days, then its absolute magnitude would be about 3.9. If you measured the apparent magnitude of the Cepheid at peak brightness, you could then find its distance using the distance modulus (m-M). For example if the Type I Cepheid with the 10 day period had a peak apparent magnitude of 9 (you need a telescope to see it), the distance modulus would be (m-M) = 6.1 and the distance is given by mM 5 Dpc 10 5 166 pc. 13. Where are we in the galaxy and how do you know? How can the distribution of globular clusters tell you about our place in the Galaxy? Our place in the Milky Way galaxy is defined by comparing our position to the center of the distribution of globular clusters. Globular clusters are not confined to the disk of our galaxy and therefore are visible beyond the 10,000 ly visibility limit imposed by dust within the Milky Way’s disk. We can see globular clusters many tens of thousands of light years away and measure their distances using the standard candles (Cepheid Variables and RR Lyrae) that they contain in abundance. Thus, we know that the Sun is approximately 9,000 parsecs from the center of the globular cluster distribution. This center astronomers had previously assumed and now know represents the center of the Milky Way galaxy. Thus globular clusters were used to define where the center of the galaxy is and our distance from it. 14. What are the four basic components of our galaxy? The four basic components of our galaxy are the bulge/bar/core, the disk, the halo, and the dark matter envelope. Component Composition Dimensions Bulge is about 5 kpc in diameter. The supermassive Old (Population II) and new black hole is located at the galactic center within 10 (Population I) stars, a Bulge/Bar/Core light days of the center. The bar is a complex structure supermassive black hole (about 4 that is not well understood and is around 9 kpc long million solar masses) inclined relative to our position in the galaxy 36 kpc in diameter and 1 kpc in thickness. The gas Old (Population II) and new layer is about 250 pc thick and is centered along the (Population I) stars with gas and galactic equator. The Sun is about 9kpc from the Disk dust concentrated into a thinner galactic center situated between the two dominant gas layer. spiral arms of the galaxy in the smaller Orion Spur spiral arm fragment. Globular Clusters, Halo Stars An oblate spheroid about 50 kpc in diameter along its Halo (Both Population II stars) equatorial plane. Dark Matter The composition of dark matter is as mysterious today as it was when it was first recognized in the second half of the 1900’s. Astronomers have no idea what dark matter is. Most models of the galaxy suggest that dark matter is distributed principally within the galactic halo.