190
acted on, then it would also make sense if the determining factor in the
gravitational strength of the object exerting the force was its own mass.
Assuming there were no other factors affecting the gravitational force, then
the only other thing needed to make quantitative predictions of gravita-
tional forces would be a proportionality constant. Newton called that
proportionality constant G, and the complete form of the law of gravity he
hypothesized was
F = Gm
1
m
2
/r
2
.[ gravitational force between objects of mass
m
1
and m
2
, separated by a distance r; r is not
Newton conceived of gravity as an attraction between any two masses in the
universe. The constant G tells us the how many newtons the attractive force
is for two 1-kg masses separated by a distance of 1 m. The experimental
determination of G in ordinary units (as opposed to the special, nonmetric,
units used in astronomy) is described in section 10.5. This difficult mea-
surement was not accomplished until long after Newton’s death.
Example: The units of G
Question: What are the units of G.
Solution: Solving for G in Newton’s law of gravity gives
G=Fr
2
m
1
m
2
,
so the units of G must be N
.
m
2
/ kg
2
the value of G is 6.67x10
-11
N
.
m
2
/ kg
2
.
Example: Newton’s third law
Question: Is Newton’s law of gravity consistent with Newton’s
third law.
Solution: The third law requires two things. First, m
1
’s force on
m
2
should be the same as m
2
’s force on m
1
. This works out,
because the product m
1
m
2
gives the same result if we inter-
change the labels 1 and 2. Second, the forces should be in
opposite directions. This condition is also satisfied, because
Newton’s law of gravity refers to an attraction: each mass pulls
the other toward itself.
Example: Pluto and Charon
Question: Pluto’s moon Charon is unusually large considering
Pluto’s size, giving them the character of a double planet. Their
masses are 1.25x10
22
and 1.9x19
21
kg, and their average dis-
tance from one another is 1.96x10
4
km. What is the gravitational
force between them.
Solution: If we want to use the value of G expressed in SI
(meter-kilogram-second) units, we first have to convert the
distance to 1.96x10
7
m. The force is
6.67
×
10
—11
N
·
m
2
kg
2
1.25
×
10
22
kg
1.9
×
10
21
kg
1.96
×
10
7
m
2
= 4.1x10
18
N
1 m
1 kg
1 kg
6.67x10
-11
N
The gravitational attraction between
two 1-kg masses separated by a dis-
tance of 1 m is 6.67x10
-11
N. Do not
memorize this number!
Computer-enhanced images of Pluto
and Charon, taken by the Hubble
Space Telescope.
Chapter 10Gravity
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