Solution Manual For
An introduction to modern astrophysics Pearson New International Edition by Carroll B.W, Ostlie D.A
Chapter 1-304. Preparing for Business ExamsPreparing for business exams involves mastering both conceptual understanding and practical application. Students are encouraged to study theories and
frameworks but also to
CHAPTER 1
The Celestial Sphere
scenario.Multiple Choice Questions (MCQs): Some law exams, particularly in foundational subjects like contract law or tort law, may feature MCQs designed to test knowledge of legal terminology
1.1 From Fig. 1.7, Earth makes S/P˚ orbits about the Sun during the time required for another planet to make
S/P orbits. If that other planet is a superior planet then Earth must make one extra trip around the Sun to
overtake it, hence
S S
= + 1.
P P
Similarly, for an inferior planet, that planet must make the extra trip, or
S S
=
P P + 1. ˚
Rearrangement gives Eq. (1.1).
1.2 For an inferior planet at greatest elongation, the positions of Earth (E), the planet (P ), and the Sun (S) form
a right triangle (∠EPS = 90○). Thus cos(∠PES) = EP/ES.
From Fig. S1.1, the time required for a superior planet to go from opposition (point P1) to quadrature (P2) can
be combined with its sidereal period (from Eq. 1.1) to find the angle ∠P1SP2. In the same time interval Earth
will have moved through the angle ∠E1SE2. Since P1, E1, and S form a straight line, the angle ∠P2SE2 =
∠E1SE2 — ∠P1SP2. Now, using the right triangle at quadrature, P2S/E2S = 1/ cos(∠P2SE2).
S
P1
Figure S1.1: The relationship between synodic and sidereal periods for superior planets, as discussed in Problem 1.2.
1.3 (a) PVenus = 224.7 d, PMars = 687.0 d
(b) Pluto. It travels the smallest fraction of its orbit before being ―lapped‖ by Earth.
1.4 Vernal equinox: a = 0h, δ = 0○
Summer solstice: a = 6h, δ = 23.5○
Autumnal equinox: a = 12h, δ = 0○
Winter solstice: a = 18h, δ = —23.5○
1
,2 Chapter 1 The Celestial Sphere
1.5 (a) (90○ — 42○) + 23.5○ = 71.5○
(b) (90○ — 42○) — 23.5○ = 24.5○
1.6 (a) 90○ — L < δ < 90○
(b) L > 66.5○
(c) Strictly speaking, only at L = ±90○. The Sun will move along the horizon at these latitudes.
1.7 (a) Both the year 2000 and the year 2004 were leap years, so each had 366 days. Therefore, the number of
days between January 1, 2000 and January 1, 2006 is 2192 days. From January 1, 2006 to July 14, 2006
there are 194 days. Finally, from noon on July 14, 2006 to 16:15 UT is 4.25 hours, or 0.177 days. Thus,
July 14, 2006 at 16:15 UT is JD 2453931.177.
scenario.Multiple Choice4. Preparing for Business ExamsPreparing for business exams involves mastering both conceptual understanding and practical application.
Students are encouraged to study theories and frameworks but also to Questions (MCQs): Some law exams, particularly in foundational subjects like contract law or tort
law, may feature MCQs designed to test knowledge of legal terminology
(b) MJD 53930.677.
1.8 (a) Δa = 9m53.55s = 2.4731○, Δδ = 2○9r16.2rr = 2.1545○. From Eq. (1.8), Δθ = 2.435○.
(b) d = r Δθ = 1.7 × 1015 m = 11,400 AU.
1.9 (a) From Eqs. (1.2) and (1.3), Δa = 0.193628○ = 0.774512m and Δδ = —0.044211○ = —2.65266r. This
gives the 2010.0 precessed coordinates as a = 14h30m29.4s, δ = —62○43r25.26rr.
(b) From Eqs. (1.6) and (1.7), Δa = —5.46s and Δδ = 7.984rr.
(c) Precession makes the largest contribution.
1.10 In January the Sun is at a right ascension of approximately 19h. This implies that a right ascension of roughly
7h is crossing the meridian at midnight. With about 14 hours of darkness this would imply observations of
objects between right ascensions of 0 h and 14 h would be crossing the meridian during the course of the
night (sunset to sunrise).
1.11 Using the identities, cos(90○ — t) = sin t and sin(90○ — t) = cos t , together with the small-angle approxima-
tions cos Δθ ≈ 1 and sin Δθ ≈ 1, the expression immediately reduces to
sin(δ + Δδ) = sin δ + Δθ cos δ cos θ.
Using the identity sin(a + b) = sin a cos b + cos a sin b, the expression now becomes
sin δ cos Δδ + cos δ sin Δδ = sin δ + Δθ cos δ cos θ.
Assuming that cos Δδ ≈ 1 and sin Δδ ≈ Δδ, Eq. (1.7) is obtained.
,CHAPTER 2
Celestial Mechanics
scenario.Multiple Choice Questions (MCQs): Some law exams, particularly in foundational subjects like contract law or tort law, may feature MCQs designed to test knowledge of legal terminology
4. Preparing for Business ExamsPreparing for business exams involves mastering both conceptual understanding and practical application. Students are encouraged to study theories and frameworks but also to
2.1 From Fig. 2.4, note that
r 2 = (x — ae)2 + y2 and r r2 = (x + ae)2 + y2.
Substituting Eq. (2.1) into the second expression gives
r = 2a — (x + ae)2 + y2
which is now substituted into the first expression. After some rearrangement,
x2 y2
+ = 1.
a2 a2(1 — e2)
Finally, from Eq. (2.2),
x2 y2
+ = 1.
a2 b2
2.2 The area integral in Cartesian coordinates is given by
, 2 2 a b 1—x /a a
2b
A= dy dx = a2 — x2 dx = vab.
a b a a
2.3 (a) From Eq. (2.3) the radial velocity is given by
dr a(1 — e2) dθ
vr = = e sin θ . (S2.1)
dt (1 + e cos θ )2 dt
Using Eqs. (2.31) and (2.32)
dθ 2 dA L
. = =
dt dt µr 2 r2
The angular momentum can be written in terms of the orbital period by integrating Kepler’s second law.
If we further substitute A = vab and b = a(1 — e2)1/2 then
L = 2µva2(1 — e2)1/2/P.
Substituting L and r into the expression for dθ/dt gives
dθ 2v(1 e cos θ )2
=
.dt
P(1 — e2)3/2
This can now be used in Eq. (S2.1), which simplifies to
2vae sin θ
vr = .
P(1 — e2)1/2
Similarly, for the transverse velocity
dθ 2va(1 + e cos θ)
vθ = r .
dt (1 — e2)1/2P
3
, 4 Chapter 2 Celestial Mechanics
(b) Equation (2.36) follows directly from v2 = v2 + v2, Eq. (2.37) (Kepler’s third law), and Eq. (2.3).
r θ
2.4 The total energy of the orbiting bodies is given by
1 1 m1m2
E = m v2 + m v2 — G
1 1 2 2
2 2 r
where r = |r2 — r1|. Now,
m2 m1
v1 = r˙1 = — r and v2 = r˙2 = r˙.
m + m2 m + m2
Finally, using M = m1 + m2, µ = m1m2/ (m1 + m2), and m1m2 = µM , we obtain Eq. (2.25).
2.5 Following a procedure similar to Problem 2.4,
L = m1r1 × v1 + m2r2 × v2
m2 m2
= m1 — r× — v
m1 + m2 m1 + m2
+m2 m1 m1
r× v
m1 + m2 m1 + m2
= µr × v = r × p
scenario.Multiple Choice Questions (MCQs): Some law exams, particularly in foundational subjects like contract law or tort law, may feature MCQs designed to test knowledge of legal terminology
2.6 (a) The total orbital angular momentum of the Sun–Jupiter system is given by Eq. (2.30). Referring to the
data in Appendicies A and C, M⊙ = 1.989 × 1030 kg, MJ = 1.899 × 1027 kg, M = MJ + M⊙ =
1.991 × 1030 kg, and µ = MJ M⊙/ (MJ + M⊙) = 1.897 × 1027 kg. Furthermore, e = 0.0489,
a = 5.2044 AU = 7.786 × 1011 m. Substituting,
q
Ltotal orbit = µ GM a 1 — e 2 = 1.927 × 1043 kg m2 s—1.
(b) The distance of the Sun from the center of mass is a=
⊙ µa/M⊙=7.426 × 108 m. The Sun’s orbital
speed is v⊙ = 2va⊙/PJ =12.46 m s—1, where PJ =3.743 × 108 s is the system’s orbital period.
Thus, for an assumed circular orbit,
LSun orbit = M⊙a⊙v⊙ = 1.840 × 1040 kg m2 s—1.
(c) The distance of Jupiter from the center of mass is aJ = µa/MJ = 7.778 × 1011 m, and its orbital
speed is vJ = 2vaJ /PJ = 1.306 × 104 m s—1. Again assuming a circular orbit,
LJupiter orbit = MJ aJ vJ = 1.929 × 1043 kg m2 s—1.
This is in good agreement with
Ltotal orbit — LSun orbit = 1.925 × 1043 kg m2 s—1.
(d) The moment of inertia of the Sun is approximately
2
2
I ~ M R ~ 3.85 × 10 kg m
5
and the moment of inertia of Jupiter is approximately
4. Preparing for Business ExamsPreparing for business exams involves mastering both conceptual understanding and practical application. Students are encouraged to study theories and frameworks but also to
2
I ~ M R2 ~ 3.62 × 1042 kg m2.
5
An introduction to modern astrophysics Pearson New International Edition by Carroll B.W, Ostlie D.A
Chapter 1-304. Preparing for Business ExamsPreparing for business exams involves mastering both conceptual understanding and practical application. Students are encouraged to study theories and
frameworks but also to
CHAPTER 1
The Celestial Sphere
scenario.Multiple Choice Questions (MCQs): Some law exams, particularly in foundational subjects like contract law or tort law, may feature MCQs designed to test knowledge of legal terminology
1.1 From Fig. 1.7, Earth makes S/P˚ orbits about the Sun during the time required for another planet to make
S/P orbits. If that other planet is a superior planet then Earth must make one extra trip around the Sun to
overtake it, hence
S S
= + 1.
P P
Similarly, for an inferior planet, that planet must make the extra trip, or
S S
=
P P + 1. ˚
Rearrangement gives Eq. (1.1).
1.2 For an inferior planet at greatest elongation, the positions of Earth (E), the planet (P ), and the Sun (S) form
a right triangle (∠EPS = 90○). Thus cos(∠PES) = EP/ES.
From Fig. S1.1, the time required for a superior planet to go from opposition (point P1) to quadrature (P2) can
be combined with its sidereal period (from Eq. 1.1) to find the angle ∠P1SP2. In the same time interval Earth
will have moved through the angle ∠E1SE2. Since P1, E1, and S form a straight line, the angle ∠P2SE2 =
∠E1SE2 — ∠P1SP2. Now, using the right triangle at quadrature, P2S/E2S = 1/ cos(∠P2SE2).
S
P1
Figure S1.1: The relationship between synodic and sidereal periods for superior planets, as discussed in Problem 1.2.
1.3 (a) PVenus = 224.7 d, PMars = 687.0 d
(b) Pluto. It travels the smallest fraction of its orbit before being ―lapped‖ by Earth.
1.4 Vernal equinox: a = 0h, δ = 0○
Summer solstice: a = 6h, δ = 23.5○
Autumnal equinox: a = 12h, δ = 0○
Winter solstice: a = 18h, δ = —23.5○
1
,2 Chapter 1 The Celestial Sphere
1.5 (a) (90○ — 42○) + 23.5○ = 71.5○
(b) (90○ — 42○) — 23.5○ = 24.5○
1.6 (a) 90○ — L < δ < 90○
(b) L > 66.5○
(c) Strictly speaking, only at L = ±90○. The Sun will move along the horizon at these latitudes.
1.7 (a) Both the year 2000 and the year 2004 were leap years, so each had 366 days. Therefore, the number of
days between January 1, 2000 and January 1, 2006 is 2192 days. From January 1, 2006 to July 14, 2006
there are 194 days. Finally, from noon on July 14, 2006 to 16:15 UT is 4.25 hours, or 0.177 days. Thus,
July 14, 2006 at 16:15 UT is JD 2453931.177.
scenario.Multiple Choice4. Preparing for Business ExamsPreparing for business exams involves mastering both conceptual understanding and practical application.
Students are encouraged to study theories and frameworks but also to Questions (MCQs): Some law exams, particularly in foundational subjects like contract law or tort
law, may feature MCQs designed to test knowledge of legal terminology
(b) MJD 53930.677.
1.8 (a) Δa = 9m53.55s = 2.4731○, Δδ = 2○9r16.2rr = 2.1545○. From Eq. (1.8), Δθ = 2.435○.
(b) d = r Δθ = 1.7 × 1015 m = 11,400 AU.
1.9 (a) From Eqs. (1.2) and (1.3), Δa = 0.193628○ = 0.774512m and Δδ = —0.044211○ = —2.65266r. This
gives the 2010.0 precessed coordinates as a = 14h30m29.4s, δ = —62○43r25.26rr.
(b) From Eqs. (1.6) and (1.7), Δa = —5.46s and Δδ = 7.984rr.
(c) Precession makes the largest contribution.
1.10 In January the Sun is at a right ascension of approximately 19h. This implies that a right ascension of roughly
7h is crossing the meridian at midnight. With about 14 hours of darkness this would imply observations of
objects between right ascensions of 0 h and 14 h would be crossing the meridian during the course of the
night (sunset to sunrise).
1.11 Using the identities, cos(90○ — t) = sin t and sin(90○ — t) = cos t , together with the small-angle approxima-
tions cos Δθ ≈ 1 and sin Δθ ≈ 1, the expression immediately reduces to
sin(δ + Δδ) = sin δ + Δθ cos δ cos θ.
Using the identity sin(a + b) = sin a cos b + cos a sin b, the expression now becomes
sin δ cos Δδ + cos δ sin Δδ = sin δ + Δθ cos δ cos θ.
Assuming that cos Δδ ≈ 1 and sin Δδ ≈ Δδ, Eq. (1.7) is obtained.
,CHAPTER 2
Celestial Mechanics
scenario.Multiple Choice Questions (MCQs): Some law exams, particularly in foundational subjects like contract law or tort law, may feature MCQs designed to test knowledge of legal terminology
4. Preparing for Business ExamsPreparing for business exams involves mastering both conceptual understanding and practical application. Students are encouraged to study theories and frameworks but also to
2.1 From Fig. 2.4, note that
r 2 = (x — ae)2 + y2 and r r2 = (x + ae)2 + y2.
Substituting Eq. (2.1) into the second expression gives
r = 2a — (x + ae)2 + y2
which is now substituted into the first expression. After some rearrangement,
x2 y2
+ = 1.
a2 a2(1 — e2)
Finally, from Eq. (2.2),
x2 y2
+ = 1.
a2 b2
2.2 The area integral in Cartesian coordinates is given by
, 2 2 a b 1—x /a a
2b
A= dy dx = a2 — x2 dx = vab.
a b a a
2.3 (a) From Eq. (2.3) the radial velocity is given by
dr a(1 — e2) dθ
vr = = e sin θ . (S2.1)
dt (1 + e cos θ )2 dt
Using Eqs. (2.31) and (2.32)
dθ 2 dA L
. = =
dt dt µr 2 r2
The angular momentum can be written in terms of the orbital period by integrating Kepler’s second law.
If we further substitute A = vab and b = a(1 — e2)1/2 then
L = 2µva2(1 — e2)1/2/P.
Substituting L and r into the expression for dθ/dt gives
dθ 2v(1 e cos θ )2
=
.dt
P(1 — e2)3/2
This can now be used in Eq. (S2.1), which simplifies to
2vae sin θ
vr = .
P(1 — e2)1/2
Similarly, for the transverse velocity
dθ 2va(1 + e cos θ)
vθ = r .
dt (1 — e2)1/2P
3
, 4 Chapter 2 Celestial Mechanics
(b) Equation (2.36) follows directly from v2 = v2 + v2, Eq. (2.37) (Kepler’s third law), and Eq. (2.3).
r θ
2.4 The total energy of the orbiting bodies is given by
1 1 m1m2
E = m v2 + m v2 — G
1 1 2 2
2 2 r
where r = |r2 — r1|. Now,
m2 m1
v1 = r˙1 = — r and v2 = r˙2 = r˙.
m + m2 m + m2
Finally, using M = m1 + m2, µ = m1m2/ (m1 + m2), and m1m2 = µM , we obtain Eq. (2.25).
2.5 Following a procedure similar to Problem 2.4,
L = m1r1 × v1 + m2r2 × v2
m2 m2
= m1 — r× — v
m1 + m2 m1 + m2
+m2 m1 m1
r× v
m1 + m2 m1 + m2
= µr × v = r × p
scenario.Multiple Choice Questions (MCQs): Some law exams, particularly in foundational subjects like contract law or tort law, may feature MCQs designed to test knowledge of legal terminology
2.6 (a) The total orbital angular momentum of the Sun–Jupiter system is given by Eq. (2.30). Referring to the
data in Appendicies A and C, M⊙ = 1.989 × 1030 kg, MJ = 1.899 × 1027 kg, M = MJ + M⊙ =
1.991 × 1030 kg, and µ = MJ M⊙/ (MJ + M⊙) = 1.897 × 1027 kg. Furthermore, e = 0.0489,
a = 5.2044 AU = 7.786 × 1011 m. Substituting,
q
Ltotal orbit = µ GM a 1 — e 2 = 1.927 × 1043 kg m2 s—1.
(b) The distance of the Sun from the center of mass is a=
⊙ µa/M⊙=7.426 × 108 m. The Sun’s orbital
speed is v⊙ = 2va⊙/PJ =12.46 m s—1, where PJ =3.743 × 108 s is the system’s orbital period.
Thus, for an assumed circular orbit,
LSun orbit = M⊙a⊙v⊙ = 1.840 × 1040 kg m2 s—1.
(c) The distance of Jupiter from the center of mass is aJ = µa/MJ = 7.778 × 1011 m, and its orbital
speed is vJ = 2vaJ /PJ = 1.306 × 104 m s—1. Again assuming a circular orbit,
LJupiter orbit = MJ aJ vJ = 1.929 × 1043 kg m2 s—1.
This is in good agreement with
Ltotal orbit — LSun orbit = 1.925 × 1043 kg m2 s—1.
(d) The moment of inertia of the Sun is approximately
2
2
I ~ M R ~ 3.85 × 10 kg m
5
and the moment of inertia of Jupiter is approximately
4. Preparing for Business ExamsPreparing for business exams involves mastering both conceptual understanding and practical application. Students are encouraged to study theories and frameworks but also to
2
I ~ M R2 ~ 3.62 × 1042 kg m2.
5
