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The force of gravitational attraction between the Sun _ P
G m m
S

and the planet is a function of the mass of the Sun F = 2
r
P
(m S ), the mass of the planet (m P ), the orbital radius
(r P ), and the universal gravitation constant (G),
2
2
which is 6.67 × 10 −11 N⋅m /kg (equation 2.12). The
unit of a newton (N) is equivalent to the combined
2
units kg∙m/s .
Period (yr) The velocity of the planet in its orbital path is a V = _ P
2π r

function of the orbital radius (r P ) and the period (T ).
T
P
The centripetal force is equal to the gravitational force, 4 π r
2 3

so by setting equations 2.11 and 2.12 equal to each m = _ P
S
2
other and substituting in the velocity of the planet G T
P
in its orbital path, an equation can be derived to
determine the mass of the Sun, in kilograms, based on
the orbital radius and period. The orbital radius must
be in meters and the period in seconds.
Distance (AU)
FIGURE 15.26 Kepler’s third law describes a relationship SOLUTION
between the time required for a planet to move around the Sun and What is the mass of the Sun, in kilograms, based on the revolution
its average distance from the Sun. The relationship is that the time
period and distance from Earth?
squared is proportional to the distance cubed.
11
r = 1.50 × 1 0 m
P
T = 1 yr
Thus, a planet that takes eight times as long to complete P
−11 N⋅ m
an orbit is four times as far from the Sun as Earth. In general, G = 6.67 × 1 0 _ 2

2
Kepler’s third law means that the more distant a planet is from k g
the Sun, the longer the time required to complete one orbit. m = ?
S
Figure 15.26 shows this relationship for the planets of the solar _
2 3
4 π r

system. Kepler’s third law applies to moons, satellites, and com- m = 2
S
G T
ets in addition to the planets. In the case of moons and satellites,
the distance is to the planet that the moon or satellite is orbiting, Convert time to seconds:
not to the Sun. 7 s _
(
yr )
T = 1 yr 3.154 × 1 0
Kepler’s laws were empirically derived from the data P
7
collected by Tycho Brahe, and the reason planets followed = 3.154 × 1 0 s
these relationships would not be known or understood until ___
11
3
2
4 π (1.50 × 1 0 m)

Isaac Newton published the law of gravitation some 60 years m = 7 2
S
2
_
−11 N⋅ m

later (see chapter 2). In the meantime, Galileo constructed ( 6.67 × 1 0 kg 2 ) (3.154 × 1 0 s)
his telescope and added observational support to the heliocen- 2 11 3 3
(m )
4 π (1.50 × 1 0 )

tric theory. By the time Newton derived the law of gravitation = ___ __

7 2
−11
kg⋅m
and then improved the accuracy of Kepler’s third law, the solar (6.67 × 1 0 )(3.154 × 1 0 ) _ m 2 )

2
s
2

system was understood to consist of planets that move around ( _ (s )

2
kg
the Sun in elliptical orbits, paths that could be predicted by ap- 2 33
4 π (3.34 × 1 0 )
plying the law of gravitation. The heliocentric model of the solar = ___ kg

14
−11
system had evolved to a conceptual model that both explained (6.67 × 1 0 )(9.948 × 1 0 )
35
1.32 × 1 0
and predicted what was observed. = _

kg

6.635 × 1 0 4
30
= 1.99 × 1 0 kg
EXAMPLE 15.3
By applying the concepts of planetary motion discussed in chapter 2, a
relationship can be established between the revolution period and the EXAMPLE 15.4
distance between the planet and the Sun that allows the mass of the Sun What is the mass of the Sun, in kilograms, based on the revolution
to be determined. This relationship is established by setting the force period and distance from Jupiter?
that keeps a planet in orbit equal to the gravitational attraction of the (Answer: 1.99 × 10 kg)
30
Sun and is based on the following equations:
The centripetal force that keeps a planet in orbit _
2
m v
P P

around the Sun is a function of its mass (m P ), the F = r
P
orbital velocity (v P ), and the orbital radius (r P )
(equation 2.11).
15-23 CHAPTER 15 The Solar System 399