ASTR 321-Fall 2015 – custom papers
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ASTR 321-Fall 20151) Imagine that a planet formed in a place where all the elements were originally in gas of solar relativeelemental composition. When gas cooled, some compounds condensed to form solids and others were strandedin the gas phase. Only the solids could get together and form the planet. All of the carbon was bonded tooxygen to form CO gas. Fe (iron) condensed as pure metal and did not form oxides. All of the O left afterforming CO, formed solids made of MgO, SiO2 and H2O. H2O formed last from O not taken by other elements.No other elements played major roles in making the planet. What fraction of the planet’s mass is composed ofH2O?An easy way to do this problem is just to consider what happens with a clump of gas that starts with ten oxygenatoms along with H, C, Mg, Si and Fe in solar proportion (see below). All the C forms CO and the remainingoxygen makes MgO, SiO2 and H2O. Only a small fraction H condenses to make H2O.The plot below shows the relative atomic abundances of the elements in the Sun. If you wanted to get the O/Hmass ratio from the plot: Log H is ~10.2 and log O is ~7 so the atom O/H ratio is 107/1010.2 ~6.3 x 10-4. Theplot is for atom ratios and to convert to mass ratios you must take into account molecular weights. For examplethe O/H atom ratio of 6.3 x 10-4 gives a 16 times larger O/H mass ratio of 1.0 x 10-2 because O is 16 times moremassive than H. For this problem you can use the following relative masses: H=1, C=12, O=16, Mg=24, Si=28and Fe=56.Relative Abundances (Atomic) of the Elements in the SunNote: it is difficult to accurately read C and O from the plot so make log oxygen =7 and log carbon = 6.72) The following is a list of all of the stable isotopes (heavier than Cd 110) of the elements cadmium (Cd, 48protons), indium (In, 49 protons), tin (Sb, 50 protons) and antimony (Sb, 51 protons).cadmium 110Cd, 111Cd, 112Cd, 113Cd, 114Cd, 116Cdindium 113In, 115Intin 112Sn, 114Sn, 115Sn, 116Sn, 117Sn, 118Sn, 119Sn, 120Sn, 122Sn, 124Snantimony 121Sb,123SbAll of the other isotopes of these elements are radioactive with lifetimes that are shorter than the time-scale ofthe slow (S) neutron capture process. Starting with 110Cd mark the location of each of the listed isotopes on thefollowing plot of neutron number VS proton number and draw a line that shows the path of the S processPage 2/3from 110Cd to 121Sb. The S process goes to the right until a short-lived radioactive isotope is reached. Whenthis isotope ß decays (emits an electron aka a beta particle) the product is the next heavier element with onemore proton and one less neutron. The purely “R” stable isotopes lie to the right of the path of the S process.Circle the isotopes that can only be made by the r process. Which isotope can only be made by the Sprocess__________? Hint: If the isotopes are plotted is correctly then isotopes with the same mass (# ofneutrons plus # of protons) lie on lines with a slopes of –1.3) An impact event on an asteroid propels a rock on a low inclination prograde elliptical orbit whose greatestdistance from the Sun is 3 AU and closest point is 1AU. For this problem, assume that the Earth has a perfectlycircular orbit at 1AU (almost true) at that the asteroid has a circular orbit 3 AU from the Sun. How long doesthe rock take to reach 1 AU (Answer in years)? If the timing is just right, the rock will impact Earth. What isthe rock’s impact speed on Earth (km/s)? The Earth’s escape velocity is 11.2 km/s. For this problem you willhave to determine the difference between rock and Earth speeds at 1AU and then take into account the effect ofEarth’s gravitational field. The KE/mass of impact (0.5 V2impact) will equal the KE of the approach velocity(0.5 V2) plus the KE of the escape speed (0.5 V2esc).4) Use a diagram showing the Sun and the asteroid and Earth orbits (1 and 3AU) to show the relative positionsof Earth and asteroid when the rock was blasted off the asteroid. Place Earth at the bottom of its orbit when itgets hit by the rock.5) The plot shows how the radial velocity ofPollux (a 2 solar mass star) varies over time. Thesinusoidally varying doppler shift of the star’sspectral lines indicate the presence of a planetorbiting the star in a circular orbit. Assume thatthe planet’s orbit is viewed edge on and useKepler’s 3rd law to determine the distancebetween the planet (in AU) and the star and themass of the planet (in Jupiter masses). Thevertical axis is the difference between themeasured velocity and the average velocity.Positive values indicate the star is moving awayand negative values indicate it is coming towardsyou. The orbit velocity is half of the differencebetween the peak and valley on this plot.
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