(Introduction to Modern Astrophysics, 2e Bradley Carroll, Dale Ostlie)
(Solution Manual, For Complete File, Download link at the end of this File)
CHAPTER 1
The Celestial Sphere
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
D C 1:
P˚ P
Similarly, for an inferior planet, that planet must make the extra trip, or
S S
D C 1:
P P˚
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 D 90ı). Thus cos.∠PES/ D 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 ∠P2 SE2 D
∠E1 SE2 ∠P1 SP2 . Now, using the right triangle at quadrature, P2 S =E2S D 1= cos.∠P2 SE2 /.
E1 S
P1
E2
P2
Figure S1.1: The relationship between synodic and sidereal periods for superior planets, as discussed in Problem 1.2.
1.3 (a) PVenus D 224:7 d, PMars D 687:0 d
(b) Pluto. It travels the smallest fraction of its orbit before being “lapped” by Earth.
1.4 Vernal equinox: ˛ D 0h , ı D 0ı
Summer solstice: ˛ D 6h , ı D 23:5ı
Autumnal equinox: ˛ D 12h , ı D 0ı
Winter solstice: ˛ D 18h , ı D 23:5ı
1
,2 Chapter 1 The Celestial Sphere
1.5 (a) .90ı 42ı / C 23:5ı D 71:5ı
(b) .90ı 42ı / 23:5ı D 24:5ı
1.6 (a) 90ı L < ı < 90ı
(b) L > 66:5ı
(c) Strictly speaking, only at L D ˙90ı . 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.
(b) MJD 53930.677.
1.8 (a) ˛ D 9m 53:55s D 2:4731ı, ı D 2ı 90 16:200 D 2:1545ı. From Eq. (1.8), D 2:435ı.
(b) d D r D 1:7 1015 m D 11,400 AU.
1.9 (a) From Eqs. (1.2) and (1.3), ˛ D 0:193628ı D 0:774512m and ı D 0:044211ı D 2:652660 . This
gives the 2010.0 precessed coordinates as ˛ D 14h 30m 29:4s , ı D 62ı 430 25:2600 .
(b) From Eqs. (1.6) and (1.7), ˛ D 5:46s and ı D 7:98400 .
(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.90ı t/ D sin t and sin.90ı t/ D cos t, together with the small-angle approxima-
tions cos 1 and sin 1, the expression immediately reduces to
sin.ı C ı/ D sin ı C cos ı cos :
Using the identity sin.a C b/ D sin a cos b C cos a sin b, the expression now becomes
sin ı cos ı C cos ı sin ı D sin ı C cos ı cos :
Assuming that cos ı 1 and sin ı ı, Eq. (1.7) is obtained.
, CHAPTER 2
Celestial Mechanics
2.1 From Fig. 2.4, note that
2
r 2 D .x ae/2 C y 2 and r 0 D .x C ae/2 C y 2 :
Substituting Eq. (2.1) into the second expression gives
q
r D 2a .x C ae/2 C y 2
which is now substituted into the first expression. After some rearrangement,
x2 y2
2
C 2 D 1:
a a .1 e 2 /
Finally, from Eq. (2.2),
x2 y2
C D 1:
a2 b2
2.2 The area integral in Cartesian coordinates is given by
Z a Z b p1x2 =a2 Z
2b a p 2
AD p dy dx D a x 2 dx D ab:
a b 1x 2 =a2 a a
2.3 (a) From Eq. (2.3) the radial velocity is given by
dr a.1 e 2 / d
vr D D 2
e sin : (S2.1)
dt .1 C e cos / dt
Using Eqs. (2.31) and (2.32)
d 2 dA L
D 2 D :
dt r dt r 2
The angular momentum can be written in terms of the orbital period by integrating Kepler’s second law.
If we further substitute A D ab and b D a.1 e 2 /1=2 then
L D 2 a2.1 e 2/1=2 =P:
Substituting L and r into the expression for d=dt gives
d 2.1 C e cos /2
D :
dt P .1 e 2 /3=2
This can now be used in Eq. (S2.1), which simplifies to
2 ae sin
vr D :
P .1 e 2 /1=2
Similarly, for the transverse velocity
d 2 a.1 C e cos /
v D r D :
dt .1 e 2/1=2 P
3
, 4 Chapter 2 Celestial Mechanics
(b) Equation (2.36) follows directly from v 2 D vr2 C v2 , Eq. (2.37) (Kepler’s third law), and Eq. (2.3).
2.4 The total energy of the orbiting bodies is given by
1 1 m1 m2
ED m1 v12 C m2 v22 G
2 2 r
where r D jr2 r1 j. Now,
m2 m1
v1 D rP1 D rP and v2 D rP2 D rP :
m1 C m2 m1 C m2
Finally, using M D m1 C m2 , D m1 m2 = .m1 C m2 /, and m1 m2 D M , we obtain Eq. (2.25).
2.5 Following a procedure similar to Problem 2.4,
L D m1 r1 v1 C m2 r2 v2
m2 m2
D m1 r v
m1 C m2 m1 C m2
m1 m1
Cm2 r v
m1 C m2 m1 C m2
D r v D r p
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, Mˇ D 1:989 1030 kg, MJ D 1:899 1027 kg, M D MJ C Mˇ D
1:991 1030 kg, and D MJ Mˇ = .MJ C Mˇ / D 1:897 1027 kg. Furthermore, e D 0:0489,
a D 5:2044 AU D 7:786 1011 m. Substituting,
q
Ltotal orbit D GM a 1 e 2 D 1:927 1043 kg m2 s1 :
(b) The distance of the Sun from the center of mass is aˇ D a=Mˇ D 7:426 108 m. The Sun’s orbital
speed is vˇ D 2 aˇ=PJ D 12:46 m s1 , where PJ D 3:743 108 s is the system’s orbital period.
Thus, for an assumed circular orbit,
LSun orbit D Mˇ aˇ vˇ D 1:840 1040 kg m2 s1 :
(c) The distance of Jupiter from the center of mass is aJ D a=MJ D 7:778 1011 m, and its orbital
speed is vJ D 2 aJ =PJ D 1:306 104 m s1 . Again assuming a circular orbit,
LJupiter orbit D MJ aJ vJ D 1:929 1043 kg m2 s1 :
This is in good agreement with
Ltotal orbit LSun orbit D 1:925 1043 kg m2 s1 :
(d) The moment of inertia of the Sun is approximately
2
Iˇ Mˇ R2ˇ 3:85 1047 kg m2
5
and the moment of inertia of Jupiter is approximately
2
IJ MJ R2J 3:62 1042 kg m2 :
5
(Solution Manual, For Complete File, Download link at the end of this File)
CHAPTER 1
The Celestial Sphere
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
D C 1:
P˚ P
Similarly, for an inferior planet, that planet must make the extra trip, or
S S
D C 1:
P P˚
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 D 90ı). Thus cos.∠PES/ D 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 ∠P2 SE2 D
∠E1 SE2 ∠P1 SP2 . Now, using the right triangle at quadrature, P2 S =E2S D 1= cos.∠P2 SE2 /.
E1 S
P1
E2
P2
Figure S1.1: The relationship between synodic and sidereal periods for superior planets, as discussed in Problem 1.2.
1.3 (a) PVenus D 224:7 d, PMars D 687:0 d
(b) Pluto. It travels the smallest fraction of its orbit before being “lapped” by Earth.
1.4 Vernal equinox: ˛ D 0h , ı D 0ı
Summer solstice: ˛ D 6h , ı D 23:5ı
Autumnal equinox: ˛ D 12h , ı D 0ı
Winter solstice: ˛ D 18h , ı D 23:5ı
1
,2 Chapter 1 The Celestial Sphere
1.5 (a) .90ı 42ı / C 23:5ı D 71:5ı
(b) .90ı 42ı / 23:5ı D 24:5ı
1.6 (a) 90ı L < ı < 90ı
(b) L > 66:5ı
(c) Strictly speaking, only at L D ˙90ı . 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.
(b) MJD 53930.677.
1.8 (a) ˛ D 9m 53:55s D 2:4731ı, ı D 2ı 90 16:200 D 2:1545ı. From Eq. (1.8), D 2:435ı.
(b) d D r D 1:7 1015 m D 11,400 AU.
1.9 (a) From Eqs. (1.2) and (1.3), ˛ D 0:193628ı D 0:774512m and ı D 0:044211ı D 2:652660 . This
gives the 2010.0 precessed coordinates as ˛ D 14h 30m 29:4s , ı D 62ı 430 25:2600 .
(b) From Eqs. (1.6) and (1.7), ˛ D 5:46s and ı D 7:98400 .
(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.90ı t/ D sin t and sin.90ı t/ D cos t, together with the small-angle approxima-
tions cos 1 and sin 1, the expression immediately reduces to
sin.ı C ı/ D sin ı C cos ı cos :
Using the identity sin.a C b/ D sin a cos b C cos a sin b, the expression now becomes
sin ı cos ı C cos ı sin ı D sin ı C cos ı cos :
Assuming that cos ı 1 and sin ı ı, Eq. (1.7) is obtained.
, CHAPTER 2
Celestial Mechanics
2.1 From Fig. 2.4, note that
2
r 2 D .x ae/2 C y 2 and r 0 D .x C ae/2 C y 2 :
Substituting Eq. (2.1) into the second expression gives
q
r D 2a .x C ae/2 C y 2
which is now substituted into the first expression. After some rearrangement,
x2 y2
2
C 2 D 1:
a a .1 e 2 /
Finally, from Eq. (2.2),
x2 y2
C D 1:
a2 b2
2.2 The area integral in Cartesian coordinates is given by
Z a Z b p1x2 =a2 Z
2b a p 2
AD p dy dx D a x 2 dx D ab:
a b 1x 2 =a2 a a
2.3 (a) From Eq. (2.3) the radial velocity is given by
dr a.1 e 2 / d
vr D D 2
e sin : (S2.1)
dt .1 C e cos / dt
Using Eqs. (2.31) and (2.32)
d 2 dA L
D 2 D :
dt r dt r 2
The angular momentum can be written in terms of the orbital period by integrating Kepler’s second law.
If we further substitute A D ab and b D a.1 e 2 /1=2 then
L D 2 a2.1 e 2/1=2 =P:
Substituting L and r into the expression for d=dt gives
d 2.1 C e cos /2
D :
dt P .1 e 2 /3=2
This can now be used in Eq. (S2.1), which simplifies to
2 ae sin
vr D :
P .1 e 2 /1=2
Similarly, for the transverse velocity
d 2 a.1 C e cos /
v D r D :
dt .1 e 2/1=2 P
3
, 4 Chapter 2 Celestial Mechanics
(b) Equation (2.36) follows directly from v 2 D vr2 C v2 , Eq. (2.37) (Kepler’s third law), and Eq. (2.3).
2.4 The total energy of the orbiting bodies is given by
1 1 m1 m2
ED m1 v12 C m2 v22 G
2 2 r
where r D jr2 r1 j. Now,
m2 m1
v1 D rP1 D rP and v2 D rP2 D rP :
m1 C m2 m1 C m2
Finally, using M D m1 C m2 , D m1 m2 = .m1 C m2 /, and m1 m2 D M , we obtain Eq. (2.25).
2.5 Following a procedure similar to Problem 2.4,
L D m1 r1 v1 C m2 r2 v2
m2 m2
D m1 r v
m1 C m2 m1 C m2
m1 m1
Cm2 r v
m1 C m2 m1 C m2
D r v D r p
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, Mˇ D 1:989 1030 kg, MJ D 1:899 1027 kg, M D MJ C Mˇ D
1:991 1030 kg, and D MJ Mˇ = .MJ C Mˇ / D 1:897 1027 kg. Furthermore, e D 0:0489,
a D 5:2044 AU D 7:786 1011 m. Substituting,
q
Ltotal orbit D GM a 1 e 2 D 1:927 1043 kg m2 s1 :
(b) The distance of the Sun from the center of mass is aˇ D a=Mˇ D 7:426 108 m. The Sun’s orbital
speed is vˇ D 2 aˇ=PJ D 12:46 m s1 , where PJ D 3:743 108 s is the system’s orbital period.
Thus, for an assumed circular orbit,
LSun orbit D Mˇ aˇ vˇ D 1:840 1040 kg m2 s1 :
(c) The distance of Jupiter from the center of mass is aJ D a=MJ D 7:778 1011 m, and its orbital
speed is vJ D 2 aJ =PJ D 1:306 104 m s1 . Again assuming a circular orbit,
LJupiter orbit D MJ aJ vJ D 1:929 1043 kg m2 s1 :
This is in good agreement with
Ltotal orbit LSun orbit D 1:925 1043 kg m2 s1 :
(d) The moment of inertia of the Sun is approximately
2
Iˇ Mˇ R2ˇ 3:85 1047 kg m2
5
and the moment of inertia of Jupiter is approximately
2
IJ MJ R2J 3:62 1042 kg m2 :
5
