Download Homework #8 Solutions - Department of Physics and Astronomy

Astronomy 204 - Homework #8 Solutions
Problem 12-4: Find the distance in parsecs to a visual binary that consists of stars of absolute
bolometric magnitudes of +5.0 and +2.0. The mean angular separation is 0.005”, and the
observed orbital period is ten years. The stars obey the mass-luminosity relation, equations 125a, b, and c. What assumptions have you used to arrive at your answer?
Note that we do not know the inclination of the plane of the star’s orbits. Let’s assume that the
orbital inclination is 0E(in the plane of the sky).
We next need to figure out the masses of the stars. We can estimate these from their bolometric
magnitudes. These stars are similar to the sun, so it is appropriate to use α = 4.0 in the massluminosity relation.
⎛L ⎞
log10 ⎜ * ⎟ = 1.9 − 0.4M bol*
⎝ L: ⎠
⎛L ⎞
log10 ⎜ * ⎟ = 1.9 − 0.4 ( 2.0 ) = 1.1
⎝ L: ⎠
L*
= 12.6
L:
L* ⎛ M ⎞
=⎜
⎟
L: ⎝ M : ⎠
4
⎛L ⎞
M
= 4 ⎜ * ⎟ = 4 12.6 = 1.9
M:
⎝ L: ⎠
And a mass of 0.9Mu for the MBol =+5.0 star.
Now using Kepler’s Harmonic Law,
3
3
⎛ a" ⎞
( M 1 + M 2 ) P = a = ⎜ '' ⎟ = ( a"d pc )
⎝π ⎠
2
(1.9 + 0.9 )(10 )
d pc
3
2
= ( 0.05 d pc )
1
⎡
⎤
3
280
(
)
⎢⎣
⎥⎦
=
= 131 pc
0.05
3
Problem 12-6: The velocity curves of a double-line spectroscopic binary are observed to be
sinusoidal, with amplitudes of 20 km/s and 60 km/s and a period of 1.5 years. (a) What is the
orbital eccentricity? (b) Which star is more massive and what is the ratio of the stellar masses?
(c) If the orbital inclination is 90º, find the relative semi-major axis (in astronomical units) and
the individual stellar masses (in solar masses).
Since the radial velocity curves are sinusoidal, the orbits must be circular. Eccentricity = 0.
Since the orbits are circular we know that v = 2πR/P. Thus,
⎛ km ⎞
60
⎟
M 1 R2 v2 ⎜⎝
s ⎠
=
= =
=3
M 2 R1 v1 ⎛ km ⎞
⎜ 20
⎟
s ⎠
⎝
Thus, star 1 (with the smaller velocity) is 3
times more massive than star 2. M : M :
V1
R2
For circular orbits the relative semi-major Star 2
axis must be the sum of the two orbital radii.
CM
R1
V2
a = R1 + R2
⎛
1.5 yr ) ⎜ 80
(
P ( v1 + v2 )
⎝
=
=
2π
2π
Now using Kepler’s Harmonic Law:
( M 1 + M 2 ) = ( 3M 2 + M 2 ) = 4M 2
3
a 3 ( 4.02 )
=
= 28.9M :
2
P2
(1.5)
M 1 = 7.2 M :
M 2 = 21.7 M :
km ⎞
⎟
7
s ⎠ ⎛ 3.16 ×10 s ⎞ ⎛ 1AU
⎞
⎜
⎟⎜
⎟ = 4.02 AU
8
yr
⎝
⎠ ⎝ 1.5 × 10 km ⎠
Star 1
Problem 12-9: The star Sirius A has a surface temperature of 10,000 K, a radius 1.8Ru, and
Mbol=1.4; the radius of its white dwarf companion, Sirius B, is 0.01Ru and Mbol=11.5.
(a) What is the ratio of their luminosities?
LSiriusA L1
=
= 100.4( m2 − m1 ) = 100.4(11.5−1.4) = 1.10 ×104
LSiriusB L2
(b) What is the ratio of their effective temperatures?
L = 4π R 2σ T 4
1
⎛ L ⎞4
T ∝⎜ 2 ⎟
⎝R ⎠
1
TSiriusA
TSiriusB
1
2 4
2 4
⎡⎛ L
⎡
⎤
⎞ ⎛ RSiriusB ⎞ ⎤
4 ⎛ 0.01 ⎞
SiriusA
= ⎢⎜
⎟⎜
⎟ ⎥ = ⎢(1.10 × 10 ) ⎜
⎟ ⎥ = 0.76
⎝ 1.8 ⎠ ⎦⎥
⎢⎣⎝ LSiriusB ⎠ ⎝ RSiriusA ⎠ ⎥⎦
⎢⎣
Thus, Sirius B is the hotter star.
(c) If they orbit at i = 90E, which star is eclipsed at primary minimum?
This is asking which eclipse would be deeper – have the greatest loss of light. Since Sirius B is
the hotter star, the eclipse would be deeper when it is covered by Sirius A.
(d) If you photometer can measure magnitudes to an accuracy of .0.001, would you be able to
detect the hypothetical primary eclipse?
⎛ Lno eclipse ⎞
m primary − mno eclipse = 2.5log10 ⎜
⎜ Lprimary ⎟⎟
⎝
⎠
⎛ 1.10 × 104 + 1 ⎞
= 2.5log10 ⎜
⎟
4
⎝ 1.10 ×10 ⎠
1
1
⎛
⎞
⎛
⎞
= 2.5log10 ⎜1 +
= 2.5log10 ⎜1 +
4 ⎟
4 ⎟
⎝ 1.10 × 10 ⎠
⎝ 1.10 ×10 ⎠
= 0.000099 ≈ 0.0001
Thus, our photometer is insufficiently sensitive by a factor of 10 to detect the primary eclipse.
Problem 13-13: Using Figures 13-8 and 13-11, estimate the distance to an M Ib star of apparent
magnitude +1.0.
From Figure 13-8C, an MIb stars has MV = -4.5. If mV = 1.0, then
m − M = 5log10 ( d pc ) − 5
d = 10
m − M +5
5
= 10
1.0 − ( −4.5) + 5
5
= 102.1 = 126 pc
Problem 13-16: Using the HR diagram in Figure 13-7 and the relationship between temperature
and spectral type in Figure 13-6, estimate how many times larger is Betelgeuse than
(a) Antares
Betelgeuse has a spectral type of M2 and M =-5.0. Antares has spectral type M1 and M = -2.6.
Thus, both temperatures are about 3500 K
LAntares
0.4( M Betelgeuse − M Antares )
0.4 ⎡ −5.0 −( −2.6 ) ⎦⎤
= 10
= 10 ⎣
= 0.11
LBetelgeuse
1
1
⎛ L
⎞2 ⎛ T
⎞
RAntares
−2
= ⎜ Antares ⎟ ⎜ Betelgeuse ⎟ = ( 0.11) 2 (1) = 0.33
⎜
⎟
RBetelgeuse ⎝ LBetelgeuse ⎠ ⎝ TAntares ⎠
2
(b) β Crucis
β Crucis is a B0 star implying a temperature of 25,000K and M = -5.0.
Lβ Crucis
LBetelgeuse
= 10
(
0.4 M Betelgeuse − M β Crucis
)
= 10
1
Rβ Crucis
RBetelgeuse
0.4 ⎣⎡ −5.0 − ( −5.0 ) ⎦⎤
= 1.0
2
1
⎛ Lβ Crucis ⎞ 2 ⎛ TBetelgeuse ⎞
−2
=⎜
= (1) 2 ( 7.1) = 0.02
⎜
⎟
⎟
⎜L
⎟ ⎜T
⎟
⎝ Betelgeuse ⎠ ⎝ β Crucis ⎠
(c) α Centauri
α Centauri is a G2 star of M = 4.5.
Lα Centauri
LBetelgeuse
Rα Centauri
RBetelgeuse
= 10
(
0.4 M Betelgeuse − M α Centauri
⎛ Lα Centauri
=⎜
⎜L
⎝ Betelgeuse
1
)
= 10
⎞ 2 ⎛ TBetelgeuse
⎟⎟ ⎜⎜
⎠ ⎝ Tα Centauri
0.4 ⎡⎣ −5.0 − ( 4.5) ⎤⎦
2
= 1.0 × 10−4
1
⎞
−2
−4 2
⎟⎟ = (1.0 × 10 ) (1.7 ) = 0.0035
⎠