Download AST 341 - Homework I - Solutions

AST 341 - Homework I - Solutions
TA: Marina von Steinkirch, [email protected]
State University of New York at Stony Brook
September 13, 2010
1
Halley’s Orbit. (1 point)
Halley’s orbital period (P)
Halley’s eccentricity (e)
76 years = 2.4 × 109 s
0.9673
a) What is the semimajor axis (a)? When calculating in Astronomic Units
(AU) and years, instead of meters and seconds (SI), this problem is very
simple. The third Kepler’s law states the relation between period (in
years) of the orbit and the average distance (ā, in AU),
P 2 = ā3 ,
On the other hand, the relation to the semimajor axis is
P 2 ∝ ka3 ,
with a proportionality constant k that depends on the units. When averaging the distance over the eccentric anomaly, one has ā = a, therefore
we can apply directly the third law,
P 2 = ka3 .
(1.1)
Since the proportionality k is the same for all bodies orbiting around the
sun, we have
P2
P2
k = 3 = 3Earth = constant.
a
aEarth
Since PEarth = 1 = PEarth , we have k = 1 and equation 1.1 gives
a = 17.83 AU = 2.6 × 1012 m .
1
b) Estimate the mass of the Sun. In the same fashion as the problem 2.31
of the textbook, we have
M =
4π 2 a3
.
G P2
(1.2)
Now we need to put everything in SI since we want to find the mass in
kilograms,
M =
4π 2
(2.6 × 1012 )3
∼ 2 × 1030 kg .
6.673 × 10−11 (2.4 × 109 )2
c) Calculate the perihelion and the aphelion. From equation 2.3 of the textbook, we have
r=
a(1 − e2 )
.
1 + e cos θ
(1.3)
In perihelion, θ = 0o and
rp = a(1 − e) = 17.83 × (1 − 0.9673) = 0.58AU = 8.68 × 1010 m.
In aphelion, θ = 180o and
ra = a(1 + e) = 17.83 × (1 + 0.9673) = 35.08AU = 5.25 × 1012 m.
d) Determine the orbital speed (v) at perihelion, aphelion, the semiminor axis (b).
The orbital speed at perihelion (vp ) is given by the equation 2.33 in the
the textbook
GM
GM 1 + e vp2 =
(1 + e) =
(1.4)
rp
a
1−e
Plugging in the values, in SI units,
vp2 =
6.673 × 10−11 × 2 × 1030 1.9673 2
= 2.98 × 109 (m/s) ,
8.68 × 1010
0.0327
vp = 5.45 × 104 m/s.
The orbital speed at aphelion (va ) is given by equation 2.34 in the the
textbook,
va2 =
GM
GM 1 − e (1 − e) =
.
ra
a
1+e
Plugging in the values, in SI units,
va2 =
6.673 × 10−11 × 2 × 1030 0.0327 2
= 8.24 × 105 (m/s) ,
5.25 × 1012
1.9673
2
(1.5)
va = 907.73 m/s.
The orbital speed at the semiminor axis (vb ) is given by using
b2 = a2 (1 − e2 )
(1.6)
on any of the previous equation.
vb = 7.04 × 103 m/s.
e) Compare the kinetic energy (K) of the comet at perihelion and aphelion.
Kp
=
Ka
2
1
2
2 µvp
1
2
2 µva
=
vp2
∼ 3860
va2
Earth as a Blackbody. (1 point)
Internal heat Flow (Fi )
Albedo (a)
Earth’s actual Temperature
Luminosity of Sun (L ) at 1 AU
Distance (d)
0.0075 W/m2
39%
280 K
3.839 × 1026 W
1 AU = 1.496 × 1011 m
First of all, the intensity from the sun that reaches the outer atmosphere of
the Earth is the solar constant or solar irradiance, and it can be easily calculate
as in the example 3.2.1. From the equation 3.2 of the the textbook ,
F =
L
,
4πr2
(2.1)
we have
F = 1365 W m−2 .
However, we need to take account the albedo in this intensity of flux,
Iabs = (1 − a)F = 0.61 × 1365 = 832.65 W m−2 .
Now, we calculate the total power absorbed and the the total power radiated,
in and from the surface. The sunlight strikes the Earth on just one side, therefore
the total power being absorbed is
2
Pabs = Iabs × πREarth
,
while the amount of power radiated away is
2
Prad = Irad × 4πREarth
,
3
where Irad is the intensity being emitted. This is given by the Stefan-Boltzmann
law for a blackbody, where T is the temperature of the blackbody and σ =
5.67 × 10−8 W/m2 K 4 a constant,
Irad = σT 4 .
(2.2)
Finally, the thermal equilibrium is given by
2
2
Prad = Pabs −→ Iabs × πREarth
= Irad × 4πREarth
,
resulting in
Iabs
832.65
=
= 208.2 W m−2 ,
4
4
and putting back in 2.2, we find
Irad =
T4 =
208.2
= 246K.
5.67 × 10−8
This value is different of the actual value 280K due the fact that the temperature on Earth is influenced by the presence of an atmosphere, as we can see
from effects such as the greenhouse.. One secondary explanation is that Earth
is not a perfect blackbody. A last observation is that we did not use the value
for the internal flux in problems of this kind.
For future references, we can use the equation for the equilibrium temperature:
(1 − a)1/4
Tequilibrium = 280K
.
(2.3)
d1/2
3
Wien Law for a Blackbody. (0.5 point)
We derive Wien law from the equation of the blackbody spectrum (the Planck
function given by the equations 3.22 and 3.24 of the textbook), in terms of
wavelength of the radiation of the blackbody,
Bλ (T, λ) =
1
8πhc
1
2hc2
=
,
hc
hc
λ5 e λkT
λ5 e λkT
−1
−1
(3.1)
or its description in terms of frequency,
Bν (T, λ) =
2hv 3
1
8πhν 3
1
=
.
hc
hc
2
c e λkT − 1
c e λkT − 1
(3.2)
Making use of any of them, the derivation starts by doing
∂B
= 0.
∂λ
hc
Defining a dimensionless variable x = λkT
for the first equation or x =
for the second, we obtain a equation of the form
xex = 5(ex − 1) or xex = 3(ex − 1).
4
hν
kT
These are the Lambert’s log Functions and can only be solved numerically.
Putting in some program such as Mathematica, we obtain x = 4.96... and x =
2.82..., respectively. Adjusting units on Kelvin, putting meters to nanometers
(1 nm = 10−9 m) and the values of the constants c, h and k, we finally have
λmax T '
4
hc 1
= 2.9 × 10−3 mK.
k x
A model of the star Dschubba as a Blackbody.
(2 points)
Surface Temperature of Dschubba(T )
Radius (r)
Distance from Earth (d)
1 pc
28,000 K
5.16 × 109 m
123 pc
3.086 × 1016
a) Find the Luminosity. From equation 2.1 we have
L = F × 4πr2 = F × 4π × (5.16 × 109 )2 .
Approximating Dschubba as a blackbody, then the radiant flux at the
surface can be calculated from the temperature from 2.2,
F = σ × T 4 = 5.67 × 10−8 × 28, 0004 = 3.5 × 1010 W m−2 .
plugging it in the first equation, we have the bolometric luminosity (luminosity integrated over all wavelengths) of Dschubba,
L = 1.17 × 1031 = 30, 374 × L .
b) Find the Absolute Bolometric magnitude (M ). From equation 3.8 of
the textbook, we can find the absolute bolometric magnitude of the star
from its flux and from comparing it to the absolute magnitude and the
luminosity of the Sun,
L M = M − 2.5 log10
.
(4.1)
L
Luminosity bolometric of Sun (L )
Absolute bolometric magnitude of Sun (M )
Resulting in
M = −6.47
5
3.839 × 1026 W
+4.74
c) Find the Aparent bolometric magnitude. From equation 3.6 of the
textbook
d ,
(4.2)
m − M = 5 log10 (d) − 5 = 5 log10
10 pc
we have
m = −1.02.
d) Find the Distance modulus. From the same 3.6, the distance modulus
is
m − M = 5.45.
e) Find the Radiant flux at the star’s surface. From 2.1 we make r =
R,
1.17 × 1031
F =
4π × (5.16 × 109 )2 = 3.5 × 1010 W m−2
Another way of finding the same result is If we approximate Dschubba as
a blackbody, then the radiant flux at the surface can be calculated from
the temperature from 2.2,
F = σ × T 4 = 5.67 × 10−8 × 28, 0004 = 3.5 × 1010 W m−2 .
Note that we had already found this result in item a.
f ) Find the Radiant flux at Earth’s surface, compare with the solar irradiance.
Now we do r = d and let us assume that the flux received at the Earth’s
surface is not attenuated by its atmosphere. We then have
F =
L
1.17 × 1031
=
= 6.46 × 10−8 W m−2 .
2
4πd
4π(123 × 3.086 × 1016 )2
g) Peak wavelength λmax . Using equation 3.15 of the textbook, the Wiens
displacement law gives
λmax =
5
0.00289
= 103.2 nm.
T
The Planck Function Bλ . (1.5 points)
a) Show that the Rayleigh-Jeans equation → the Planck function for enough large λ.
The Plank spectrum function, equation 3.1 or 3.22 of the textbook
Bλ (T, λ) =
2hc2
1
,
hc
λ5 e λkT
−1
(5.1)
can be writte as a first-order Taylor expansion of the exponential for λ hc
hc
hc
hc
kT λ − 1 =
kT or 1 kT λ as e
kT λ , resulting in
Bλ (T ) =
6
2hc2 1
.
hc
λ5 kT
λ
From equation 3.20, the Rayleigh-Jeans equation is
2ckT
,
λ4
Bλ '
which is exactly the previous result. We clearly see that it is a classical
result since the h was canceled.
b) Plot the Planck Function and the Rayleigh-Jeans for the Sun.
6
Blackbody radiation and the Stefan-Boltzmann
constant. (2 points)
a) Integrate over all λ to obtain the total luminosity of a blackbody model star.
hc
in equation 3.20 and then integrate it on the whole
Let us make A = kT
space,
Z ∞
1
1
Lλ = 8π 2 R2 hc2
dλ.
5
A/λ
λ e
−1
0
Now we make the change of variable λ =
2
2
2
∞
Z
Lλ = 8π R hc
0
Lλ = 8π 2 R2 hc2 ×
0
kT 4
hc
with
du
dλ
= − λa2 , obtaining
1 du 1
.
(A/u)3 eu − 1 A
R∞
From the straigthfoward integral
A
u
u3 du
eu −1
×
=
π4
15 ,
we finally have
π4
2 π5 k4
=
4πR2 T 4
15
15 h3 c2
b) Compare to the Stefan-Boltzmann equation and show σ. The StefanBoltzmann equation for the luminosity (from 2.2 and 3.17 in the textbook)
is
L = σ4πR2 T 4 .
(6.1)
Comparing to item a, we have
σobt =
2 π5 k4
,
15 h3 c2
which is exactly the answer.
c) Compare the value of this expression to the value in Appendix A.
Comparing to the appendix A of the textbook,
σobt
σ
5.6 × 10
2 π 5 k4
15 h3 c2
−2
−8
7
Wm
K −4
7
UBV Colors to Shaula star. (2 points)
Surface Temperature(T )
Visible color (V )
U-B
B-V
Parallax angle
22,000K
1.62, blue-white
-0.90
0.23
0.0046400
a) Estimate U − B and B − V . Compare to the measured values. From
the equation 3.33 of the textbook we have
!
R
Fλ dλ
U − B = −2.5 log10 R
+ CU −B ,
(7.1)
Fλ dλ
and we can approximate these functions to the Planck functions, whose
depends on λ and T . We write, for example,
U − B = −2.5 log10
B
365 ∆λU
B440 ∆λB
+ CU −B .
(7.2)
Substituting the values for each bandwidth, from page 75, and making use
U
B
V
B=365 nm
B= 440 nm
B=550 nm
∆λ= 68 nm
∆λ=98 nm
∆λ=89 nm
of CU −B = −0.87 and CB−V = 0.65, we find
U − B = −1.08,
and
B − V = −1.36.
8