CHAPTER 1
,2 Chapter 1 The Celestial Sphere
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
= + 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
, 1.5 (a) (90○ — 42○) + 23.5○ = 71.5○
(b) (90○ — 42○) — 23.5○ = 24.5○
CHAPTER 1
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.
(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.
,
,2 Chapter 1 The Celestial Sphere
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
= + 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
, 1.5 (a) (90○ — 42○) + 23.5○ = 71.5○
(b) (90○ — 42○) — 23.5○ = 24.5○
CHAPTER 1
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.
(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.
,
