Download PHYS-633: Problem set #0 Solutions

PHYS-633: Problem set #0
Solutions
.
1. Angles, magnitudes, inverse square law:
a. How far from the Earth would the Sun have to be moved so that its apparent angular
diameter would be 1 arc second? (Express your answer in cm, solar radii, and AU.)
First convert arcsec to radians: 1 arcsec= π/(180×60×60)= 4.8×10−6 radians.
Then
D=
2R
2R
=
= 4.1 × 105 R = 2.9 × 1016 cm = 1900 AU .
α
4.8 × 10−6
(1)
Another way to look at it is to recall the actual angular diameter of the
Sun is about ∼ 0.5o ≈ 1800 arcsec. Thus to get to 1 arcsec, the Sun would
have to move away by a factor 1800, or to 1800 AU.
b. How far away would a Frisbee of diameter 30 cm have to be to subtend the same
angle?
By above,
D=
30
= 6.3 × 107 cm = 63 km .
4.8 × 10−6
(2)
c. At the distance you calculated in (a), what would the Sun’s flux here on Earth be
(i.e. what would the solar constant be)?
The actual solar constant is F = L /4πau2= 1.4 × 106 erg/s/cm2 . If we now
move the Sun 1800 times further away, then by the inverse-square law, the
flux (which is what we mean by the solar constant) would decrease by a
factor 1/18002 . Thus
F,new =
1.4 × 106
= 0.43 erg/s/cm2
18002
(3)
d. What would the Sun’s apparent magnitude be? (Use m = -26.7 for the actual Sun,
the one that’s at 1 AU.)
–2–
The difference in apparent magnitude between two stars is just 2.5 times
the log of the ratio of the flux. Remembering that a lower flux gives a larger
magnitude (i.e. dimmer stars have bigger m), we have
m,new = m + 2.5 log(F /F , new) = −26.7 + 2.5 log(18002 ) = −10.4 .
(4)
Remembering that the brightest stars are around magnitude zero, we see
that the sun would still be a very bright star, about 10,000 times brighter
than the brightest actual star! (Since m=-10 is 10 magnitudes brighter
than m=0, and each difference of 5 in magnitude represents a factor 100 in
brightness, so 1002 =10,000).
2. The star Dschubba (δ Sco) has a parallax p = 8 mas:
Assuming it is a spherical blackbody with radius R = 7.5 R and surface temperature
T = 28, 000 K, compute Dshuba’s
a. Luminosity, in erg/s, and L ;
By Stefan-Boltzmann law:
L = 4πR2 σT 4 = 1.2 × 1038 erg/s = 30, 000 L
(5)
b. Absolute magnitude:
By definition of absolute magnitude normalized to the sun’s value Msun =
+4.8, we have
M = 4.8 − 2.5 log(30, 000) = −6.4 .
(6)
c. Apparent magnitude;
A parallax of α = 8 mas 0.008 arcsec implies a distance of D = 1/0.008 =
125 pc. Thus the apparent magnitude is
m = M + 5 log(D/10pc) = −6.4 + 5 log(12.5) = −0.9 .
(7)
d. Distance modulus;
m − M = 5 log(12.5) = 5.5
(8)
e. Radiant flux at the star’s surface, in CGS, and relative to surface flux of the Sun;
–3–
F (R) =
L
30, 000 × 4 × 1033 erg/s
=
= 3.4 × 1013 erg/cm2 /s .
4πR2
4π(7.5 × 7 × 1010 )2 cm2
(9)
Relative to the flux at sun’s surface, we have
F (R)
L/L
30000
=
=
= 533 .
2
F (R )
(R/R )
(7.5)2
(10)
f. Radiant flux at the Earth’s surface, and the ratio of this to the solar irradiance;
F =
30, 000 4 × 1033 erg/s
L
=
= 6.8 × 10−5 erg/cm2 /s .
4πD2
4π(1253 × 1018 )2 cm2
(11)
Relative to the solar flux at earth’s orbit, we have
30000
F
L/L
=
= 4.5 × 10−11 .
=
2
F
(D/au)
(125 206, 000)2
(12)
g. Peak wavelength λmax .
By Wien’s displacement law,
λmax =
2.9 × 106 nm K
2.9 × 106
=
nm ≈ 100nm .
T
2.8 × 104
(13)
3. Parallax of Mars:
In 1672, an international effort was made to measure the parallax angle of Mars at
opposition, when it was on the opposite side of the Earth from the Sun, and thus closest
to Earth.
a. Consider two observers at the same longitude but one at latitude of 45 degrees
North and the other at 45 degrees South. Work out the physical separation s between
the observers given the radius of Earth is RE ≈ 6400 km.
Viewed from the center of the Earth, the two observers at ±45o are separated
by 90o , thus forming a right angle. The radius lines to each observer thus
form the two lengths of an isosceles triangle with the observers separated
by the base, with length
√
s = 2Re = 8.8 × 103 km .
(14)
–4–
b. If the parallax angle measured is 25 arcsec, what is the distance to Mars? Give your
answer in both km and AU.
D=
s
8.8 × 103 km
=
= 7.3 × 107 km = 0.48 au .
α
25arcsec/206000 (arcsec/radian)
(15)
4. Galaxies: distance, magnitude, and solid angle:
a. What is the apparent magnitude of a galaxy that contains 1011 stars identical to the
Sun (i.e., assume its luminosity is equal to 1011 L ) if it’s at a distance of 5 million
parsecs?
Using eqn. (14) in DocOnotes1-stars.pdf, the apparent magnitude m of an
object with luminosity L at a distance D is given by
m = 4.8 − 2.5 log(L/L ) + 5 log(D/10 pc) = 4.8 − 2.5 × 11 + 5 × (5 + log(5)) = +5.7 ,
(16)
with the latter equality plugging in the above numbers to give m ≈ +3.
b. What is the flux of this galaxy here on the Earth, in cgs units?
1011 4 × 1033 erg/s
L
=
= 1.3 × 10−7 erg/cm2 /s
F =
2
6
18
2
2
4πD
4π (5 × 10 3 × 10 ) cm
(17)
c. If the galaxy is circular in shape, as seen from the Earth, and has a diameter of
50,000 pc, what is its apparent angular diameter?
α=
50 × 103 pc
s
=
= 10−2 rad = 2060 arcsec = 34 arcmin = 0.57 o
D
5 × 106 pc
(18)
d. What solid angle does it subtend (in steradians and in square arc seconds)?
Ω = (π/4)20602 arcsec2 = 3.2 × 106 arcsec2
(19)
Ω = (π/4) α2 = 3 × 10−4 ster
(20)
e. How does the galaxy’s surface brightness (energy/time/area/solid angle) compare to
the Sun’s (express this as a ratio)?
–5–
Assuming both the galaxy and the sun emit isotropically from their surface,
then the ratio of surface brightness is just given by the ratio of their surface
flux,
2
2
Lgal /4πRgal
7 × 1010
Igal
11
= 10
=
= 2.2 × 10−14 .
(21)
2
I
L /4πR
5 × 104 3 × 1018
5. Equilibrium temperature of blackbody near a star:
A spherical blackbody of radius a is at a distance d from a star with an effective temperature Teff and radius R.
a. Assuming the blackbody has a rotation that over time exposes all parts of its surface
to the star’s light, compute its equilibrium temperature T . How does this depend on the
blackbody’s radius a?
4
Noting that the stellar luminosity is L = σTeff
4πR2 , we can write the star’s
4
(R/d)2 . The blackbody
radiative flux at distance d as F (d) = L/4πd2 = σTeff
intercepts an energy πa2 F (d), which in equilibrium it must re-radiate over its
entire spherical surface area 4πa2 . The equilibrium blackbody temperature
T is given by equating this absorbed and emitted energy:
2
R
4
πa2 = σT 4 4πa2 ,
(22)
σTeff
d
which can be readily solved to give
r
T = Teff
R
.
2d
(23)
Note that the blackbody radius a has cancelled, and so this equilibrium
temperature is independent of the absorbing body’s size.
b. Derive an expression for the ratio, λmax,bb /λmax,∗ , of the peak wavelength of emission
for the blackbody to that of the star.
r
Teff
2d
λmax,bb
=
=
.
(24)
λmax,∗
T
R
c. Compute values for parts (a) and (b) in the case of a blackbody at 1 AU from the
sun.
Recalling that au/R ≈ 215, with T ≈ 5800 K, we find
5800
T (1 au) = √
K = 280 K .
430
(25)
–6–
λmax,bb √
= 430 = 20.7 → λmax,bb ≈ 20.7 × 500 nm = 10.4 µm .
λmax,
(26)
6. Equilibrium Temperature of Earth:
a. Modeling earth as a blackbody illuminated by the sun, compute its equilibrium temperature, and compare this to the temperature on a moderate spring day in Delaware.
From part c. of previous problem,
r
√
R
Te = T
≈ 5800 430 ≈ 280 K.
2 au
(27)
On moderate spring day in Delaware, temperature is a bit higher that this,
ca. 20 C (68 F). But given the approximations, this is pretty close to the
characteristic temperature computed for a simple blackbody!
b. According to http://en.wikipedia.org/wiki/Earth, Earth’s “albedo”, meaning
the fraction of received light that is absorbed is only ca. 0.367, with the rest (63.3% )
being refected by, e.g. clouds, snow, etc., without contributing any heat to Earth. So
now redo the calculation in (a) reducing the solar input energy by this amount.
If only a fraction 0.367 of Sun’s luminosity is actually absorbed by Earth,
then the equilibrium temperature should be reduced by a factor 0.3671/4 =
0.78, reducing the above equilibrium temperature now to Te = 219 K =
-54 C.
c. Which result seems more “reasonable”? Briefly discuss what other physics might be
important to include to understand the actual surface temperature of Earth.
This apparently more realistic model thus seems to give a temperature that
is much lower than the typical temperature of the actual Earth.
The key piece of physics missing is the “greenhouse effect”, which blocks the
re-radiation of solar energy, forcing the surface of the Earth to be warmer
than it would otherwise be, much like a blanket at night keeps our skin at
a higher temperature than it would otherwise be.
Bottom line: the greenhouse effect and the albedo effect roughly cancel ,
making the simple blackbody temperature in part (a) come out about right!
7. Brightness of the full moon:
a. Given the Sun’s radius R and distance to Earth ae , compute the Sun’s solid angle
on the sky, both in steradians, and as a fraction of the full sky’s 4π steradians.
–7–
The sun’s solid angle at earth is given by
Z 1
p
dµ = 1 − µ = 1 − 1 − (R /ae )2 ≈ (1/2)(R /ae )2 ≈ 1.1 × 10−5 (28)
Ω /2π =
µ
Ω ≈ 6.8 × 10−5 ster
(29)
Ω /4π ≈ 5.5 × 10−6
(30)
b. By coincidence, the moon covers about this same fraction. Use this fact together
with the moon’s albedo of 0.07 (meaning the moon reflects back 7% of the sunlight that
hits its surface) to estimate how much dimmer the full moon appears in optical light
compared to sunlight on Earth.
Since the solar flux declines as (R /ae )2 , then a perfectly reflecting moon
would have its surface brightness reduced by this fraction. With an albedo
of 0.07, this surface brightness would be reduced by that extra factor. Finally, since the solid angles are the same, the ratio of surface brightness is
also the ratio of apparent brightness.
2
R
Fmoon
= 0.07
= 1.6 × 10−6
(31)
F
ae
c. Given that the Sun’s apparent magnitude is m = −26.7, use your answer for part
(c) to estimate the apparent magnitude of the full moon in optical light.
The moon’s apparent magnitude is
mmoon = −26.7 − 2.5 log(Fmoon /F ) = −12.2
(32)
d. Assume that the sunlight that is not reflected is re-radiated as if the moon were a
blackbody. Compute the average equilibrium temperature of the moon.
Surface flux of the moon’s thermal emission is given by:
2
2
2
0.93 4πR
σT4
0.93 R
σT4
0.93πRmoon
L
=
=
2
4πa2e 4πRmoon
16πa2e
4a2e
r
R
0.25
= 0.93
T ≈ 0.048 T ≈ 273 K
2ae
4
Fem = σTmoon
=
Tmoon
(33)
(34)
–8–
e. Estimate the peak wavelength at which the moon emits this blackbody radiation, comparing it to the peak wavelength of the Sun’s spectrum. What part of the electromagnetic
spectrum does this correspond to?
λmax,moon = λmax,
T
Tmoon
≈
500 nm
≈ 10, 400 nm = 10.4 microns
0.048
(35)
This is in the infrared.
f. For a filter that is centered broadly on this wavelength, what is difference in the
apparent magnitude of the moon compared to the optical magnitude computed in part
(c).
∆m = 2.5 log[fmoon,vis /fmoon,IR ] = 2.5 log[0.07/(.93/4)] ≈ −1.3
(36)
8. Absorption by coal dust.
Imagine you’re a coal miner working under a 1000-watt lamp that is at distance of 10
meter away from your work location.
a. Assuming the lamp emits its light isotropically, what is the flux of light on your
workspace, in watt/m2 ? (Assume the coal mine walls are perfect absorbers, i.e with
zero albedo).
Since the walls absorb all indirect light, the local flux is just due to the
direct light, for which the flux decreases following the inverse-square law:
F (r) =
L
1000
=
= 0.8 watt/m2 .
2
4πr
4π × 102
(37)
b. Now suppose there is a cave-in that fills the mine with black, spherical coal dust
particles of diameter 0.1 mm, and with a uniform number density of 20 particles per
cubic centimeter. What is the optical depth between you and the lamp?
The cross section of dust of diameter d = 0.1 mm is σ = (π/4)d2 = 7.8−5 cm2 .
The optical depth is given in terms of number density n = 20/cm3 , cross
section σ, and distance D = 1000 cm,
τ = n σ D = 20 × 7.8−5 × 103 = 1.56 .
c. What is now the flux on your workspace?
(38)
–9–
Applying the exponential attenuation by optical depth, we find
Fabs = F (r)e−τ = 0.8 × e−1.56 = 0.17 watt/m2 .
(39)
d. What distance had this flux value before the cave in?
Solving the inverse-square law for distance, we find
r
r
L
103
=
= 21.8 m .
r=
4πF
4π0.17
(40)
9. Absorption by interstellar dust.
Imagine you’re an astronomer observing a star with luminosity 1000L that is at a
distance of 10 parsec.
a. What is the flux of light you observe? Give your answer in both watt/m2 and
L /parsec2 .
Again, the flux decreases following the inverse-square law:
F (r) =
L
1000
426
=
= 0.8 L /parsec2 = 0.8
= 3.5 × 10−7 watt/m2 . (41)
2
2
4πr
4π × 10
(316 )2
b. Now assume the space between you and the star contains spherical dust particles of
diameter 1 micron and number density of 6000 particles per cubic kilometer. Assuming
the dust absorbs or scatters light in proportion to its geometric cross-section, what is
the optical depth between you and the star?
Again, the cross section of dust of diameter d = 1 µm is σ = (π/4)d2 =
7.8−9 cm2 . The optical depth is given in terms of number density n =
6000/km3 = 6−12 cm−3 , cross section σ, and distance D = 10 pc = 319 cm,
τ = n σ D = 6−12 × 7.8−9 × 319 = 1.4 .
(42)
c. What is now the flux you observe?
Again, applying the exponential attenuation of optical depth, we find
Fobs = F (r)e−τ = 3.5 × 10−7 × e−1.4 = 8.6 × 10−8 watt/m2 .
(43)
d. If you knew the star’s luminosity but didn’t know about the dust, what distance would
you infer for the star based on your observed flux??
– 10 –
Again, inverting the inverse-square law to infer a distance in terms of the
known luminosity and the observed flux, you infer
s
r
L
103 426
r=
=
= 617 m = 20pc .
(44)
4πFobs
4π8.6−8
e. What is the change in the apparent magitude m of this star resulting from the dust
absorption.
Using the definition of apparent magnitude, we have
∆m = 2.5 log(F/Fobs ) = 2.5 log(e1.4 ) = 2.5 × 1.4 × log(e) = 1.5 .
(45)
f. Finally, generalize your analysis to derive an expression for how the extinction magnitude A, defined as the change in apparent magnitude m due to absorption, depends
on the absorption optical depth τ to any star.
Again, applying the magnitude definition,
A ≡ ∆m = 2.5 log(eτ ) = 2.5τ log(e) = 1.08 τ .
(46)