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tiL12214_appe_643-698.indd Page 671 09/10/10 8:37 AM user-f463 Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefile

m _
14.15. Use the equation provided in the question to solve for the speed ρ = ∴ m = ρV

at which the galaxies are moving apart from each other. V
g
_ 27 3
_ v = H 0 d m = 5.52 (1.0868 × 10 cm )
km
3
20

_ cm
s

H 0 = _ g cm 3
_
km
27 _
Mly _ = 5.52(1.0868 × 10 ) ×
20

s

= 2 Mly cm 3 1
d = 2 Mly Mly 27
v = ? = 6.00 × 10 g
km
_ Mly

_

= (20)(2) s _ × Determine the mass in kilograms.
Mly 1
1 kg
3 )
_ 27 ( _
km

= 40 = 6.00 × 10 g
s 1 × 10 g
kg
27
_ _ )
6.00 × 10
=
× g (

1 × 10 3 g
CHAPTER 15 24
= 6.00 × 10 kg
15.1. The volume is determined by using the formula for the volume

15.3. This problem requires a conversion ratio to be developed to

of a sphere and the relationship between the radius and

convert actual units into model units. Then the actual distances
diameter of a circle.
are multiplied by the conversion ratio to obtain the model
4
D = 6.787 × 10 km dimensions.
V = ?
4 _ 3 4 _ D _ 3 Determine the conversion ratio (CR) from actual units to
V = πr and r = 2D ∴ V = π ( )
3 3 2 model units:
Convert diameter from kilometers to meters. 8
1.0 × 10 km actual = 1.0 cm model
8
3
1 × 10 m
4
7
1.0 cm model
1.0 × 10 km actual

6.787 × 10 km ( _ ) = 6.787 × 10 m __ = __

1 km 8 8
1.0 × 10 km actual 1 × 10 km actual
4 _ __ ) 3 __
7
6.787 × 10 m
1.0 cm model
V = π ( 1 =

3 2 8
1.0 × 10 km actual
4 _ 7 3 _
−8 cm model

= π(3.394 × 10 m) CR = 1.0 × 10
3
km actual
4 _ 22 3

= π(3.910 × 10 m ) Determine the model dimension of Sun to Neptune in
3
23
= 1.638 × 10 m 3 centimeters.
9
d actual = 4.497 × 10 km
15.2. Density is the mass divided by the volume; hence, this
−8 cm model
relationship can be used to determine the mass. First, the CR = 1.0 × 10 _

volume needs to be determined from the diameter. km actual
d model = ?
4
D = 1.2756 × 10 km d model = d actual (CR)
V = ? 9 −8 cm model
_
(
)
4 _ 3 4 _ D _ 3 = 4.497 × 10 km actual 1.0 × 10
V = πr and r = 2D ∴ V = π ( ) km actual
3 3 2
9
−8
cm model
= 4.497 × 10 (1.0 × 10 ) km actual _

Convert diameter from kilometers to centimeters to match km actual
1
density units. = 4.5 × 10 cm model
= 45 cm
5
1 × 10 cm
9
4
1.2756 × 10 km ( _ ) = 1.2756 × 10 cm

1 km 15.4. Determine the conversion ratio (CR) from actual units to
4 _ __ 3 model units:
9
1.2756 × 10 cm

V = π ( )

3 2
4 _ 8 3 1.0 AU actual = 25 cm model

= π(6.3780 × 10 cm) _ _
3 =
1.0 AU actual
25 cm model

4 _ 26 3 1.0 AU actual 1.0 AU actual
= π(2.5945 × 10 cm ) _

3 1 =
25 cm model

27
= 1.0868 × 10 cm 3 1.0 AU actual
cm model

CR = 25 _

Determine the mass of Earth in grams. AU actual
g
_ Determine the model dimension of Sun to Mercury in
ρ Earth = 5.52

cm 3 centimeters.
27
V = 1.0868 × 10 cm 3
d actual = 0.38 Au d model = d actual (CR)
m = ?
(
)
cm model
cm model

CR = 25 _ = 0.38 AU actual 25 _

AU actual AU actual
d model = ? _
cm model

= 0.38(25) AU actual

AU actual
= 9.5 cm model
E-29 APPENDIX E Solutions for Group A Parallel Exercises 671