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hwk02ans
Ast 4001, 2015 Sept 24
Answers for homework set 2
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1. Calculate the gas constant  in both SI and c.g.s. units for the following types of gas.
Precision of about 1 percent is suitable, so you can pretend that every baryon has the
same effective mass as a free proton.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
Non-ionized hydrogen molecules (Ordinary H2 gas.)
Non-ionized hydrogen atoms (all the molecules have been dissociated)
Fully ionized hydrogen (a plasma composed of protons and electrons) .
Fully ionized helium (He++ plasma, including the electrons).
Non-ionized 56Fe atoms.
Fully ionized 56Fe.
A fully ionized mixture in which 75% of the mass is hydrogen and 25% is helium.
Earth’s atmosphere. Pretend that 80% of the molecules are N2 and 20% are O2 .
Answers: In each case,  = k / m av . Since one-percent accuracy is sufficient,
m av will be good enough if we pretend that each type of atom has mass = A m H
where A = atomic weight rounded off to an integer, and that the electron mass is
practically zero. (Doing it more accurately isn't much harder, but then we'd need
to look up each atomic mass.)
(a) m ( H2 ) = about 3.35 × 10 g , so   4.12 × 10 c.g.s. = 4120 SI units.
(b) atomic hydrogen,   8.25 × 10 c.g.s. = 8250 SI units.
(c) ionized hydrogen, m av = m H / 2:
  1.65 × 10 c.g.s. = 16500 SI units.
(d) ionized helium: nucleus + two electrons for each atom, so m av  m He / 3 
1.333 m H . Therefore   6.19 × 10 c.g.s. = 6190 SI units.
(e) iron atoms:   1.47 × 10 c.g.s. = 147 SI units.
(f) fully ionized iron: each atom becomes nucleus + 26 electrons, so m av  56 m H / 27
 2.07 m H .   3.98 × 10 c.g.s. = 3980 SI units, comparable to H2 .
(g) Imagine a gas sample with mass = 400 m H (for example). In order to have the
specified hydrogen and helium mass fractions, it must consist of
300 H atoms + 25 He atoms + 350 electrons = 675 particles.
Therefore m av  400 m H / 675, so   1.39 × 10 c.g.s. = 13900 SI units.
(h) A sample of 8 nitrogen molecules and 2 oxygen molecules (for example) would
have total mass = 288 m H , so m av  28.8 m H per particle. Thus we find
  2.86 × 10 c.g.s. = 286 SI units. This is smaller than the ionized-hydrogen
value by a factor of almost 60.
4001hwk02 - p2
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2.In order to solve the hydrostatic equation for an atmosphere, dP / dz = g , we need
to have some additional information that relates mass density  ( z ) to the pressure P ( z ) .
For reasons too complicated to explain here, often a power-law relation is surprisingly realistic:
P  Kwhere K and  are constants. The exponent usually has some value between
1.2 and 1.7.
(a) (This was done in class; repeat the steps yourself.) Assuming that P = K  and g is
a constant, solve the hydrostatic equation for P( z ) . The solution will include a constant of
integration, which can be adjusted to suit any particular case. Assuming that it's an ideal gas
so P = T with = constant, convert your solution into a formula for temperature
T ( z ) . Adopt a constant of integration that makes T = T 0 at z = 0.
Solution: The math is routine (albeit a little ugly) and leads to
K {}g z + constant .
But K  is merely P /  = T .
T (z) = T0
Suitably adjusting the constant, we find
{} {g z /  } .
(b) Write a simple expression for the atmosphere's scale height H . Scale height is usually
viewed as a rough order-of-magnitude parameter, and in this case a reasonable choice is
the height z where T = T 0 / 2.
Answer: H  {  } T / g . The leading factor is model-dependent,
but the expression T / g applies to almost any planetary or stellar atmosphere.
(Many authors would omit the 2 in the formula, because H is not precisely defined.)
(c) For atmospheric dynamics a critical quantity is the speed of sound w, defined in this
case by w 2 = P /  (Memorize the P /  part!) Write a simple formula for H as
a function of w and the other main parameters.
Answer: H  {  } w 2 / g .
The w 2 / g factor is most important.
(d) Suppose an object falls from height H , ignoring aerodynamic drag. What speed does
it have when it reaches z = 0, and what does this have to do with part (c) above?
Answer: That speed is comparable to w . With parameters used above, the
elementary-textbook value would be w divided by the square root of  .
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4001hwk02 - p3
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3. A spectroscopic binary star system. It's best to do this problem in solar units.
The conversion factor for velocities is 1 AU / yr = 4.74 km / s .
The figure shows observed Doppler velocity curves, V( t ), for a particular spectroscopic
binary. Since the curves are sinusoidal, evidently it has a circular orbit. Use elementary
physics to analyze the system with accuracy of about 2%. Explain your reasoning.
(a) Estimate the two stars' mass ratio, MB /M A .
Answer: Since the two stars are going around their mutual center of mass,
M A V A = M B V B where each V is a circular speed. (Conservation of
momentum.)
We can see the speeds in the Doppler amplitudes, shown in the figure:
V A  33 km/s and V B  57 km/s . Therefore M B / M A  33 / 57  0.58.
(b) Assume that we see the orbit edge-on (orbit inclination i = 90°). Based on the
velocity amplitudes and the orbital period, calculate the separation between the two
stars' centers. Express your answer in AU.
Answer: First, according to the figure the orbital period is about 80 days.
The relative orbit speed must be 33 + 57 km/s = 90 km/s = about 19 AU/year
= about 4.16 AU in 80 days. This is the circumference of the relative orbit.
Divide by 2 to get the separation between the stars: about 0.66 AU.
(Radii of the two individual circles around the center of mass are r A  0.24 AU
and r B  0.42 AU .)
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4001hwk02 - p4
(c) Deduce the total mass and the individual masses MA and MB , relative to the Sun.
(In other words our unit of mass is Msun.)
Answer: For a relative circular orbit in Solar units we have M (total) = r 3 / P 2 .
Plugging in r = 0.66 AU from part b and P = 0.219 year (i.e., 80 days),
we find M (total) = close to 6.0 M sun .
Apportioning this in a 1.0 : 0.58 ratio based on part a , we find M A  3.8 M sun
and M B  2.2 M sun .
Note: If we did not know that i = 90°, then each mass value would have an
undetermined factor of 1 / sin 3 i . This would be a very substantial uncertainty,
e.g. a factor of almost 3 if i  45°. For most purposes we can deal with this
only by observing a large sample of binary systems and using statistics based on
a spherically-random distribution of i -values. However, a few spectroscopic
binaries are also eclipsing binaries, indicating that i is close to 90°.
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