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Solutions Set 2 for Astro 320
Read chapters 3.1-3.2, 3.6, 5.1, 6.1, and 7.1-7.2.
Problem 1: Comparing stellar magnitudes, radii. Two physically identical stars are
at different distances from us. Star A has parallax 10−3 ”, and radius 108 m. It appears 25
times brighter than Star B.
a) What are the distances to stars A and B?
Star A has d=1/p”=1000 parsecs. Star B is 25 times fainter, so must be 5 times farther
away (inverse square law), 5000 parsecs.
b) Star A has an apparent magnitude of 17.4. What is its absolute magnitude? What
is the absolute magnitude of star B?
m − M = 5 log(d) − 5. For A, M = 17.4 − 5 log(1000) + 5=+7.4. Star B is identical, so also
has M =+7.4.
c) Star C has a (approximately) blackbody spectrum which peaks at the same wavelength as that of star A. Star C has a parallax angle of 10−2 ” and appears to be 104 times
brighter than Star A. What is the radius of Star C?
Star C must be at a distance of d = 1/10−2 = 100 pc. So it’s 10 times closer than star A,
which should make it 100 times brighter were the stars identical. But it’s 100 times brighter
than that, so is 100 times more luminous. Since the peak of its (blackbody-like) spectrum
is in the same place as A’s, the only remaining variable to explain C’s luminosity is its radius. As L = 4πR2 σT 4 , increasing the radius by a factor of 10 gives a factor 100 increase in
luminosity. So C’s radius is 109 m.
d) What is the apparent magnitude of Star C?
Comparing its magnitude to that of A, its apparent magnitude will be m1 = m2 −2.5 log(F1 /F2 ) =
17.4 − 2.5 log(104 ) = 7.4.
Problem 2: How big and bright are Jupiter and Uranus? “Opposition” (in the
context of the outer planets) means that one of the outer planets (e.g. Mars, Jupiter) is
lined up in a straight line with the Earth and Sun, bringing the planet as close as it can
come to the Earth. Jupiter reaches opposition roughly every 13 months. Since its orbit (and
Earth’s) is elliptical, the minimum distance varies between oppositions. On Oct. 29, 2011,
Jupiter reaches opposition, producing an Earth-Jupiter distance of 4.0 AU. You’ll need data
from Appendix C for this problem (B if you have the first edition of the textbook).
a) What was the angular diameter of Jupiter at opposition in 2011?
Jupiter has physical diameter 2 ∗ 11.209 ∗ 6.378 × 106 m=1.43 × 108 m. θ ' x/d =
1
1.43 × 108 m/(4.0 ∗ 1.50 × 1011 m) = 2.38 × 10−4 radians, or 2.38 × 10−4 ∗ 205265=48.8
arcseconds.
b) Jupiter’s apparent magnitude in Oct. 2011 will be -2.9. Assuming circular orbits,
estimate its apparent magnitude when it’s on the far side of the Sun from us (don’t worry
about the Sun being in the way).
Assume its distance is 1 AU (our average distance to the Sun) plus 5.2 AU (Jupiter’s average
distance to the Sun). So Jupiter will be 6.2/4.0=1.55 times as far away. By the inverse
square law, it will be 1/1.552 = 0.416 times as bright. The difference in magnitudes is
-2.5*log(F1 /F2 )=0.95 magnitudes, so it will have an apparent magnitude of -1.95.
c) On Sept. 25, 2011, Uranus will also reach opposition. Use the data in Appendix C
to estimate the distance to Uranus from the Earth (assume circular orbits). What will the
maximum angular diameter of Uranus be?
Uranus’s physical diameter is 2 ∗ 4.0073 ∗ 6.378 × 106 m=5.11 × 107 m. At closest approach,
it is 19.2-1=18.2 AU away, so θ ' x/d = 5.11 × 107 m/(18.2 ∗ 1.50 × 1011 m) = 1.9 × 10−5
radians, or 3.8 arcseconds.
d) Do you think you might be able to resolve its disk with the U of A telescope? Why
or why not (show a calculation)?
The U of A telescope has an aperture of 12 inches (0.33 m), and therefore an angular
resolution (in V band, 550 nm) of θ = 1.22 ∗ (5.5 × 10−7 m/0.33m) = 2.0 × 10−6 radians,
or 0.4”. However, seeing generally sets the minimum resolution achievable with the U of A
scope, and is typically ∼1”, meaning that structures smaller in angular size than 1” cannot
be distinguished from a point. Uranus, at 3.8 arcseconds, is large enough that it should be
resolvable, appearing as a disk in the U of A telescope.
Problem 3: Why is the earth warm? Assume a blackbody-like planet orbits a star in a
circular orbit with radius a. The star has radius R? and effective surface temperature T? .
a) Assuming that energy is conserved, derive an equation for the effective surface temperature of the planet. The temperature should depend only on a, R? , and T? .
For energy conservation, the energy captured by the planet equals the energy the planet
is radiating. So (flux at planet’s orbit from star)*(area planet presents to star)=(flux
from planet’s surface)*(area planet radiates from). The flux from the star at the planet
is L? /(4πa2 ), where a is the radius of the planet’s orbit, and the stellar luminosity is
L? = 4πR?2 σT?4 . The area the planet presents to the star is πRp2 . The flux from the planet
is σTp4 , while its radiating area is 4πRp2 . So the equation becomes (4πR?2 σT?4 /(4πa2 )) ∗
(πRp2 ) = (σTp4 ) ∗ (4πRp2 )). Dividing both sides by σ4πRp2 , we get πR?2 T?4 /(4πa2 ) = Tp4 , or
Tp = (R?2 T?4 /(4a2 ))1/4 .
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b) Treating the Earth as a blackbody and using SI values for the astronomical unit,
solar radius, and solar temperature (T =5770 K), calculate the Earth’s effective surface
temperature in K and in C.
2 4
T /(4a2 ))1/4 = 278 K, or 5 degrees Celsius. (Side
Using the expression above, TEarth = (R
note: the average Earth temperature is currently about 10 degrees higher than this (although
Edmonton’s is very close, ∼3.6 C), due to the greenhouse effect.)
Problem 4: Modelling the star Dschubba. The star Dschubba can be modelled as a
blackbody with surface temperature 28,000 K and radius 5.15 × 109 m. The star is 180 pc
from the Earth. The bolometric correction for Dschubba is BC = −2.47. Calculate:
a) Dschubba’s luminosity as a fraction of the Sun’s luminosity.
Since L = 4πR?2 ∗ σT 4 , LDschubba /L = (RDschubba /R )2 ∗ (TDschubba /T )4 = 3.0 × 104 L .
b) Absolute bolometric magnitude.
We’ll use the Sun’s absolute bolometric magnitude 4.74, M1 = M − 2.5 log(3.0 × 104 ) =
−6.45.
c) Apparent bolometric magnitude.
m = M + 5 log(d/10pc) = −6.45 + 5 log(180/10) = −0.17.
d) Radiant flux at the surface of Dschubba.
F = σT?4 = 3.5 × 1010 W m−2 .
e) Observed flux from Dschubba at Earth’s surface.
We can use its luminosity (from part a, 3 × 104 L ) and distance; F = L/(4πd2 ) = 2.9 × 10−8
W m−2 .
f) Peak wavelength.
Using λmax T = 2.9 × 10−3 m K, and T = 28000 K, λmax = 1.04 × 10−7 m=104 nm.
g) What part of the spectrum is the peak wavelength of Dschubba’s light in? (The
optical spectrum is considered to range from 400 to 700 nm. Look at Fig. 6.25 for the rest.)
Compared to the Sun’s colour, how will this star appear?
The peak of the spectrum is in the UV. This star should appear bluish compared to the Sun.
h) What is the apparent visual magnitude of Dschubba?
We use BC = mbol − V , so V = −0.17 + 2.47 = 2.3.
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