Astro 301
5/11/10
1
EndofLecture Problems: Answers (Lecture 10–26)
Lecture 10.
The net result of the pp 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
pp 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 massenergy 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 inversesquare 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
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '10
 DrAndrewRobinson
 Astronomy, Energy, Mass, Black hole

Click to edit the document details