Astro 301
5/11/10
1
End-of-Lecture Problems: Answers (Lecture 10–26)
Lecture 10.
The net result of the p-p reaction, can be summarized as 4
×
1
H
→
4
He+
energy
, so 4 H
nuclei are converted to He in each reaction. The energy released is
Δ
E
=
Δ
mc
2
, where
Δ
m
is the mass difference between the 4 H nuclei and the produce He nucleus. The number of
p-p reactions/sec required to sustain the Sun’s luminosity is
N
~
L
/
Δ
E
.
The mass of H converted to He/sec is then 4
m
H
N
. On the other hand, the amount of mass
converted to energy/sec is
Δ
mN
(alternatively, we can get the mass-energy conversion
rate by taking the time derivative of Einstein’s equation
dE/dt
= (
dm/dt
)
c
2
=
L
). We can
now calculate, respectively, the total mass converted to He and the total mass converted
to energy over Sun’s lifetime to date:
M
(H
→
He) ~9
×
10
28
kg (0.04 M
);
M
(
Δ
E
)~6.5
×
10
26
kg (3
×
10
-4
M
).
Use the mass loss rate given in the lecture (~10
–14
M
/yr ) to estimate the total mass lost
due to the solar wind: ~4.6
×
10
–5
M
.
Lecture 11.
The ratio of luminosities is given by
L
Sirius
/
L
= 100
[M(Sun)–M(Sirius)]/5
, where the “M”’s
refer to the absolute magnitudes (NB: NOT masses). Thus, Sirius is about 23
×
more
luminous.
Similarly, the flux ratio is
F
Sirius
/
F
= 100
[m(Sun)–m(Sirius)]/5
, where the “m”’s are the
apparent magnitudes. For the Sun, m = –26.7 (slide 4), hence
F
Sirius
/
F
≈
8
×
10
–11
. Flux is
related to luminosity via the inverse-square law, hence
F
Sirius
/
F
= (
L
Sirius
/
L
)
×
(1
AU/
r
Sirius
)
2
. Solving for the distance,
r
Sirius
≈
5.37
×
10
5
AU or 2.6 pc.
Luminosity, radius, and surface temperature are related by the blackbody equation
L
=
4
π
R
2
σ
T
4
. Recasting in terms of Solar values and solving for the radius,
R
/
R
= (
L
/
L
)
–1/2
×
(
T
/
T
)
–2
. Thus, for Sirius, R
≈
1.65
R
.
Lecture 12.
First, use the small angle formula to estimate the linear diameter of the nebular:
D
=
"
d
206 265
. With
d
= 2000 pc and
θ
= 30
×
60 = 1800”, we have
D
≈
17.5 pc.
The size of the nebula is related to the number of ionizing photons/sec,
Q
, produced by
the central star cluster. Assuming that it is spherical, the Stromgren radius is
R
S
≈
8.7 pc =
2.7
×
10
17
m. Balancing photionizations and recombinations within the nebula gives
Q
=
4
3
R
S
3
n
e
n
H
#
H
. Within the Strömgren sphere, the gas is fully ionized so