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Problem Set 3 for Astro 320
Read chapter 7, start chapter 8.1.
Due Oct. 5, 2011, by 5 pm, to my office CCIS 2-109.
Problem 1: The star Mizar in the Big Dipper was the first binary system to be observed
(Benedetto Castelli asked Galileo to observe it in 1617, presumably to confirm his observations of this double star)–though not the first where orbital motion was observed. The
parallax angle to Mizar is 4.2 × 10−2 ” and the angular separation between the two is 14.4
seconds of arc. The orbital period of the system is at least 3000 years. Assume that Mizar
A and B move in circular orbits, that the plane of the orbit is perpendicular to our line of
sight and that the distances between each star and the center of mass are equal.
a) Find the distance between the two stars. Express your answer in parsecs and
in AU.
First we find the distance from us to Mizar; use d = 1/p, get 23.8 pc. Now, we use θ = x/d
(small angle formula) to find x (d=23.8 pc, θ=14.4”). To find it in parsecs, change θ into
radians, θ = 7.02 × 10−5 radians, so x = 1.7 × 10−3 pc. In AU, we can just leave θ in
arcseconds, so x = 343 AU.
b) Find an upper limit on the mass of each star. Express your answer in solar
mass units.
We use P 2 = a3 /M (in solar & Earth units), knowing that for similar masses, we must
interpret this as M =total mass, a is the distance between the 2 bodies (e.g. the sum
of their semimajor axes). Since P > 3000 years, we can write (3000 years)2 < a3 /M , or
M < (343AU )3 /(3000years)2 = 4.48 M . As the two stars have equal distances from the
center of mass, their masses are equal, and both less than 2.24 M .
c) Edward Pickering discovered that Mizar A is actually a binary system in 1889
with a spectroscope, the first such discovery. The spectroscopic period is 20.5
days. Find the separation between the two stars in the Mizar A system. Express
your answer in AU.
Using M = 2.24 M for the total mass, we apply P 2 = a3 /M to get a = (2.24∗(20.5/365)2 )1/3 =
0.19 AU. Since the mass of the binary is an upper limit, the separation of 0.19 AU is also an
upper limit.
d) Mizar B is also a spectroscopic binary with an orbital period of 182.3 days.
Find their separation.
Similarly, Mizar B’s separation will be less than a < (2.24 ∗ (182.3/365)2 )1/3 = 0.82 AU.
e) The star Alcor has a parallax angle of 4.0 × 10−2 ” and is separated from Mizar
by 12 minutes of arc, as seen from Earth. Find the distance between Alcor and
the Mizar star system.
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Alcor’s parallax angle indicates a distance of 25.0 pc, so it is farther away from us (in the
radial direction) by 1.2 pc than Mizar. Their physical separation on the plane of the sky can
be computed with the small angle approximation as x = dθ = 23.8 ∗ (12 ∗ 60/206265) = 0.08
√
pc. So the total distance between them is 0.082 + 1.22 = 1.2 pc.
f ) It has been suggested that the errors in measuring the parallax angles for
Alcor and Mizar may be large enough that their separation is only 0.27 light
years. Suppose this is true and that Alcor’s mass is 1.6 M . Find the period for
Alcor to make a full orbit. Comment on why it is difficult to test the hypothesis
that Alcor and Mizar may be gravitationally bound.
0.27 light years is 0.08 pc (so this question assumes that their distances from Earth are
actually identical). We can use P 2 = a3 /M again, writing a = 0.08 pc = 1.71 × 104 AU and
M < 6.1 M . Thus P >903,000 years. This is so long that Mizar and Alcor won’t make a
substantial fraction of an orbit in the history of astronomy.
Problem 2: C & O, problem 7.6.
a: We get the ratio of stellar masses from the ratio of velocities: m1 /m2 = v2r /v1r , so the
ratio of Star A’s mass to Star B’s is 22.4/5.4 = 4.1.
b: Assuming i = 90 deg and circular orbits, the maximum radial velocities give the orbital
velocities of the stars, and thus the distance they traverse in 6.31 years. A goes at 5.4 × 103
m/s for 6.31 yrs∗3.16 × 107 s/yr= 1.08 × 1012 m; this is 2π times its radius, so a = 1.71 × 1011
m, or 1.14 AU. Similarly, B’s semimajor axis is 4.74 AU. The semimajor axis of the system
is the sum of these, so M = a3 /P 2 = 5.1 M .
c: Ma + Mb = 5.1 M , while Ma /Mb = 4.1, so 4.1Mb + Mb = 5.1M ; Mb = 1.0 M , and
Ma = 4.1 M .
d: For the smaller, rs = v(tb − ta )/2, where v = va + vb = 27.8 km/s. So rs = 27.8 ×
103 m/s(0.58d ∗ 86400s/d)/2 = 7.0 × 108 m, which is equal to one solar radius. The larger
star’s radius can be measured as rl = v(tc − ta )/2, where tc − ta = 1.22 days, so rl = 1.5 × 109
m, or 2.1 solar radii.
e: To find the ratio of the effective temperatures, we use (Bo − Bp )/(Bo − Bs ) = Ts4 /Tl4 .
The brightness ratios can be determined using F1 /F2 = 100(m2 −m1 )/5 . mo − mp for the
primary eclipse = -3.8 magnitudes, giving a flux ratio of Bp /Bo = 0.0302, while for the
secondary eclipse it is -0.04 magnitudes, giving Bs /Bo = 0.964. Thus Bo − Bp = 0.9698Bo ,
Bo − Bs = 0.036Bo , and (Bo − Bp )/(Bo − Bs ) = 26.94. The fourth root of this is 2.28,
which gives the temperature difference between the two stars. (Note that without further
information we can’t distinguish which star, A or B, is hotter.)
Problem 3: You observe a 1.5% drop in the light from a star. You will need data
from Appendices A, C, and G (note there are several tables within App. G).
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a) If the star is a main-sequence star of spectral type G, like our Sun, what is
the radius of the transiting body? Compare this to the radius of Jupiter.
The 1.5% drop indicates that 1.5% of the flux is removed; this requires blocking 1.5% of the
area (assuming that the transiting object is not itself providing significant flux, which would
add too many parameters to the problem). Thus the transiting object has an area of 0.015
√
that of the star, and a radius 0.015 = 0.12 that of the star. If the star is a main sequence
G star like our Sun, its radius will be 0.12 times R , or 8.4 × 107 m. Jupiter’s radius is 11.2
Earth radii, or 7.1 × 107 m; so the transiter is slightly (1.2 times) larger than Jupiter.
b) If the star is a main-sequence star of spectral type M6, what is the radius of
the transiting body? Again, compare to the radius of Jupiter.
For an M6 main-sequence star, the radius is 0.24 R , so the transiter has a radius of 2 × 107
m, only 0.28 times Jupiter’s radius.
c) If the star is a giant star of spectral type M6, what is the radius of the
transiting body? Compare this to the radius of Jupiter and to the Sun.
The M6 giant has a radius of 116 R , so the transiter has a radius of 9.7 × 109 R . This is
137 times Jupiter’s radius, or 13.9 times the Sun’s radius–this is certainly not a planet!
Problem 4: You have a spectrograph capable of measuring radial velocities to 1
m/s. Again use data from Appendix G.
a) What is the change in radial velocity of the star due to a planet of 10 times
Earth’s mass, orbiting a 1 M star in an orbit of 1 year, with an inclination of
90 degrees? Can you detect this?
The velocity of the planet is 2πa/P = 2π ∗ 1.5e11m/(3.16 × 107 s) = 3.0 × 104 m/s. The ratio
of the star’s velocity to the planet’s is v? /vp = mp /m? , so v? = 10 ∗ MEarth ∗ vp /M = 0.90
m/s. So no, we can’t quite detect this.
b) What is the amplitude of the spectroscopic signal if the inclination is 45
degrees?
In this case, the velocities are the same, but what we measure is the radial velocity, vr =
v sin(i), or 0.64 m/s.
c) What is the amplitude of the signal if the star is an M5 main-sequence star,
and the planet is a 3-Earth-mass planet with orbit of 10 days?
The semimajor axis of the planet’s orbit is a = (M ∗ P 2 )1/3 = (0.21M ∗ (10/365)2 )1/3 =
0.054 AU. So its velocity is 5.9 × 104 m/s. By the method above, the star’s velocity is
v? = 3 ∗ MEarth ∗ vp /(0.21 ∗ M ) = 2.5 m/s. (This could be detected by our instrument.)
d) Use an answer from the previous problem set, and App. G, to find the orbital
period which a planet would have to have around an M5 main-sequence star in
order to have temperatures allowing liquid water
on the surface (aim for 300 K).
q
We use Tp = (R?2 T?2 /(4a2 ))1/4 , first solving for a =
3
R?2 T?4 /(4 ∗ Tp4 ) = 1.1 × 1010 m, or 0.076
AU. Thus we can find the period as P = a3/2 /M 1/2 = 0.0763/2 /(0.21)1/2 = 0.045 years, or
16.5 days.
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