Holton, James R. An Introduction to Dynamic Meteorology
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ANSWERS TO SELECTED PROBLEMS
Chapter 1
1. α ≈ (
2
a sin 2φ)/(2g) 2. 35,830 km 3. 0.35 N kg
−1
4. 556 m 5. 1.46 cm to the right 6. 6.25 m s
−1
7. Upward force = 2 × 10
3
N; greater for westward travel 8. 7.8 m eastward
10. 1 rad s
−1
, 1.22 rad s
−1
, 0.17 J 12. 5.536 km, 5.070 km 13. 3˚C
14. 7.987 km 17. 5.187 km
Chapter 2
1. −2hPa/3h 2. −2˚C/h 5. 1.02 cm s
−1
7. 0.725 kg m
−3
8. 135Jkg
−1
,16.4ms
−1
11. 287 s
Chapter 3
1. −1mkm
−1
(−0.23ms
−1
) 2. −10
−3
ms
−2
3. 940 hPa 5. V
grad
/V
g
= 2
7. North, R
t
= 250 km; west, R
t
= 500 km; south, R
t
→∞; east R
t
= 500 km
8. North, 10.5 m s
−1
; west and east, 12.1 m s
−1
; south, 15 m s
−1
10. 25ms
−1
, 34˚ 11. −1.5˚Ch
−1
12. 44 h
13. 238 km to the left
15. North, 9.4 m s
−1
; west and east 11.5 m s
−1
; south, 15 m s
−1
17. 7.66 K
18. −852 km 20. 2 × 10
−5
s
−1
21. 110% 23. 0.96 cm s
−1
509
January 27, 2004 14:0 Elsevier/AID aid
510 answers to selected problems
Chapter 4
1. −2 ×10
7
m
2
s
−1
, −2 ×10
−5
s
−1
2. −5.5ms
−1
(antiyclonic) 3.−2 × 10
−5
s
−1
4. 9.4 ×10
−5
s
−1
, −8.4 ×10
−6
s
−1
5. 0 (annulus), 5 ×10
−5
s
−1
(inner cylinder)
7. −7.2m
2
s
−2
9. h(r) = H +
2
r
2
/(2g)
12. 10
−5
s
−1
, −1.29 ×10
−5
s
−1
, −1546 km
14. 0.71˚ equatorward, 138.6 km upward
Chapter 5
4. 293 s, 1.7 cm 5. 3.21 × 10
3
kg m
−1
s
−1
(mixed layer) 6. De = 14.05 m
8. w
max
= 9.6 × 10
−3
mm s
−1
, v
max
= 0.75 mm s
−1
9. 155 days
10. K
m
= 11 m
2
s
−1
Chapter 6
3. 2.84 ×10
−6
s
−1
4. ω(p) =−(k
2
c
2
p
0
/f
0
π)sin(πp/p
0
) cos[k(x − ct)]
5. k
2
= σ [(f
0
π)/(σp
0
)]
2
8. W
0
=−cf
0
U
0
π/(σp
0
)
9. ω(p) = (Vp
0
/f
0
π)[k
2
(c − U) + β]sin(πp/p
0
) cos[k(x − ct)]
10. c = U −β[k
2
+ (f
2
0
π
2
)/(σp
2
0
]
−1
11. 0.047 Pa s
−1
Chapter 7
2. B
r
= A cos kx
0
, B
i
=−A sin kx
0
, c
r
= ν/k, c
i
= α/k
4. u
= p
(
c − u
)
(
γ p
)
,ρ
= p
(
c −
u
)
2
7. u
= h
(
c − u
)
H
9. ρ
0
u
w
=−ρ
0
m
|
A
|
2
(
2k
)
12. u
=
0.66ms
−1
sin
(
kx + mz
)
,w
=−
0.75ms
−1
sin
(
kx + mz
)
16. −24.3ms
−1
17. −0.633 cm s
−1
Chapter 8
1. 6.04 ×10
4
s (about 17 hrs.)
2. ψ
3
=−ψ
1
= Acos
k(x − c
r
t)
ω
2
=−2kf
0
βAsin
[
k(x −c
r
t)
]
/
σδp(k
2
+ 2λ
2
)
3. ω
2
= δpf
−1
0
λβ
√
3 −1
Asin
[
k
(
x − c
r
t
)
]
/3
6. ψ
1
lags ψ
3
by 65.5˚ 7.
|
B
|
=
A
√
2 −1
1/2
= 6.44 × 10
6
m
2
s
−1
9. 2.38 ×10
9
Jm
−2
13. v
T
is independent of .
Chapter 9
5. 344 K 6. 1.42 m s
−1
8. r = 135.7km
January 27, 2004 14:0 Elsevier/AID aid
answers to selected problems 511
Chapter 10
3. 134ms
−1
8. 6.25 ×10
5
kg s
−2
11. 0.245 cm s
−1
Chapter 11
1. 9.6 ×10
−8
ms
−1
2. 2u
0
l
2
>β
Chapter 12
1. 6697 m 2. 87.6 m s
−1
,48.7ms
−1
,28.0ms
−1
6. |
|=200 m
2
s
−2
, |w
|=1.57 mm s
−1
7. 3.92 × 10
−3
kg m
−1
s
−1
8. ˆw =ˆv(−imy)
ν
2
N
2
exp
−β|m|y
2
(2N
2
)
9. ±6.14˚
Chapter 13
5. 1.3%, 5%, 19%, 59% 6. 1.5% per time step
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January 24, 2004 11:45 Elsevier/AID aid
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