ASTR 321-Fall 2015 –

ASTR 321-Fall 2015
1) Imagine that a planet formed in a place where all the elements were originally in gas of solar relative
elemental composition. When gas cooled, some compounds condensed to form solids and others were stranded
in the gas phase. Only the solids could get together and form the planet. All of the carbon was bonded to
oxygen to form CO gas. Fe (iron) condensed as pure metal and did not form oxides. All of the O left after
forming 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 of
An easy way to do this problem is just to consider what happens with a clump of gas that starts with ten oxygen
atoms along with H, C, Mg, Si and Fe in solar proportion (see below). All the C forms CO and the remaining
oxygen 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/H
mass 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. The
plot is for atom ratios and to convert to mass ratios you must take into account molecular weights. For example
the 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 more
massive than H. For this problem you can use the following relative masses: H=1, C=12, O=16, Mg=24, Si=28
and Fe=56.
Relative Abundances (Atomic) of the Elements in the Sun
Note: it is difficult to accurately read C and O from the plot so make log oxygen =7 and log carbon = 6.7
2) The following is a list of all of the stable isotopes (heavier than Cd 110) of the elements cadmium (Cd, 48
protons), indium (In, 49 protons), tin (Sb, 50 protons) and antimony (Sb, 51 protons).
cadmium 110Cd, 111Cd, 112Cd, 113Cd, 114Cd, 116Cd
indium 113In, 115In
tin 112Sn, 114Sn, 115Sn, 116Sn, 117Sn, 118Sn, 119Sn, 120Sn, 122Sn, 124Sn
antimony 121Sb,123Sb
All of the other isotopes of these elements are radioactive with lifetimes that are shorter than the time-scale of
the slow (S) neutron capture process. Starting with 110Cd mark the location of each of the listed isotopes on the
following plot of neutron number VS proton number and draw a line that shows the path of the S process
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from 110Cd to 121Sb. The S process goes to the right until a short-lived radioactive isotope is reached. When
this isotope ß decays (emits an electron aka a beta particle) the product is the next heavier element with one
more 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 S
process__________? Hint: If the isotopes are plotted is correctly then isotopes with the same mass (# of
neutrons 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 greatest
distance from the Sun is 3 AU and closest point is 1AU. For this problem, assume that the Earth has a perfectly
circular orbit at 1AU (almost true) at that the asteroid has a circular orbit 3 AU from the Sun. How long does
the rock take to reach 1 AU (Answer in years)? If the timing is just right, the rock will impact Earth. What is
the rock’s impact speed on Earth (km/s)? The Earth’s escape velocity is 11.2 km/s. For this problem you will
have to determine the difference between rock and Earth speeds at 1AU and then take into account the effect of
Earth’s gravitational field. The KE/mass of impact (0.5 V2
impact) will equal the KE of the approach velocity
(0.5 V2) plus the KE of the escape speed (0.5 V2
4) Use a diagram showing the Sun and the asteroid and Earth orbits (1 and 3AU) to show the relative positions
of Earth and asteroid when the rock was blasted off the asteroid. Place Earth at the bottom of its orbit when it
gets hit by the rock.
5) The plot shows how the radial velocity of
Pollux (a 2 solar mass star) varies over time. The
sinusoidally varying doppler shift of the star’s
spectral lines indicate the presence of a planet
orbiting the star in a circular orbit. Assume that
the planet’s orbit is viewed edge on and use
Kepler’s 3rd law to determine the distance
between the planet (in AU) and the star and the
mass of the planet (in Jupiter masses). The
vertical axis is the difference between the
measured velocity and the average velocity.
Positive values indicate the star is moving away
and negative values indicate it is coming towards
you. The orbit velocity is half of the difference
between the peak and valley on this plot.

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