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Unformatted text preview: −11N ⋅ m 2 /kg 2 )(5.98 ×1024 kg)
=
(3.59 × 107 m + 6.38 ×106 m) 2 0.223 m/s 2 ____________________________________________________________________________________________ 28. REASONING AND SOLUTION
The magnitude of the net force acting on the
moon is found by the Pythagorean theorem to
be F SM Moon F
Sun F= 2
FSM + F EM 2
FEM Earth Newton's law of gravitation applied to the sunmoon (the units have been suppressed) FSM = GmSmM
2
rSM = ( 6.67 ×10−11 ) (1.99 ×1030 ) ( 7.35 ×1022 )
= 4.34 ×1020 N
11 )2
(1.50 ×10 A similar application to the earthmoon gives FEM = GmE mM
2
rEM = ( 6.67 ×10−11 ) ( 5.98 ×1024 ) ( 7.35 ×1022 )
= 1.98 ×1020 N
8 )2
( 3.85 ×10 The net force on the moon is then
F= ( 4.34 × 1020 N ) + (1.98 × 1020 N )
2 2 = 4.77 × 1020 N 178 FORCES AND NEWTON'S LAWS OF MOTION ____________________________________________________________________________________________ 29. REASONING The magnitude of the gravitational force exerted on the satellite by the
earth is given by Equation 4.3 as F = Gmsatellite mearth / r 2 , where r is the distance between
the satellite and the center of the earth. This expression also gives the magnitude of the
gravitational force exerted on the earth by the satellite. According to Newton’s second law,
the magnitude of the earth’s acceleration is equal to the magnitude of the gravitational force
exerted on it divided by its mass. Similarly, the magnitude of the satellite’s acceleration is
equal to the magnitude of the gravitational force exerted on it divided by its mass.
SOLUTION
a. The magnitude of the gravitational force exerted on the satellite when it is a distance of
two earth radii from the center of the earth is F= Gmsatellite mearth
r 2 = ( 6.67 × 10−11 N ⋅ m2 / kg2 ) ( 425 kg ) (5.98 × 1024 kg )
2
( 2 ) ( 6.38 × 106 m ) = 1.04 × 103 N b. The magnitude of the gravitational force exerted on the earth when it is a distance of two
earth radii from the center of the satellite is
F= Gmsatellite mearth
r 2 = ( 6.67 × 10−11 N ⋅ m2 / kg2 ) ( 425 kg ) (5.98 × 1024 kg )
2
( 2 ) ( 6.38 × 106 m ) = 1.04 × 103 N c. The acceleration of the satellite can be obtained from Newton’s second law.
asatellite = F
msatellite = 1.04 × 103 N
=
425 kg 2.45 m / s 2 d. The acceleration of the earth can also be obtained from Newton’s second law. aearth = F
mearth = 1.04 × 103 N
= 1.74 × 10−22 m / s 2
24
5.98 × 10 kg ____________________________________________________________________________________________ 30. REASONING The weight of a person on the earth is the gravitational force Fearth that it
exerts on the person. The magnitude of this force is given by Equation 4.3 as
Fearth = G mearth mperson
2
rearth Chapter 4 Problems 179 where rearth is the distance from the center of the earth to the person. In a similar fashion, the
weight of the person on another planet is Fplanet = G mplanet mperson
2
rplanet We will use these two expressions to obtain the weight of the traveler on the planet. SOLUTION Dividing Fplanet by Fearth we have Fplanet
Fearth G
=
G mplanet mperson
2
rplanet mearth mperson mplanet
=
m earth 2
rearth rearth rplanet 2 or
Fplanet Since we are given that mplanet
mearth mplanet
= Fearth m earth = 3 and rearth
rplanet = rearth rplanet 2 1
, the weight of the space traveler on the
2 planet is
2 1
Fplanet = ( 540.0 N )( 3) = 405.0 N
2
______________________________________________________________________________ 31. SSM REASONING According to Equation 4.4, the weights of an object of mass m on
the surfaces of planet A (mass = MA, radius = R ) and planet B (mass = MB , radius = R ) are WA = GM A m
R2 and WB = GM B m
R2 The difference between these weights is given in the problem. 180 FORCES AND NEWTON'S LAWS OF MOTION SOLUTION The difference in weights is
WA – WB = GM A m
R 2 – GM B m
R 2 = Gm
(MA – MB )
R2 Rearranging this result, we find MA – MB (W – WB ) R2 =
=A
Gm ( 3620 N ) (1.33 × 107 m )
=
(6.67 × 10−11N ⋅ m 2 /kg 2 ) ( 5450 kg )
2 1.76 × 1024 kg ____________________________________________________________________________________________ 32. REASONING AND SOLUTION The figure at the right
shows the three spheres with sphere 3 being the sphere
of unknown mass. Sphere 3 feels a force F31 due to the
presence of sphere 1, and a force F32 due to the
presence of sphere 2. The net force on sphere 3 is the
resultant of F31 and F32.
Note that since the spheres form an equilateral triangle,
each interior angle is 60°. Therefore, both F31 and F32
make a 30° angle with the vertical line as shown. 3 F F 31 32 1.20 m 30° 2 1 Furthermore, F31 and F32 have the same magnitude
given by
GMm3
F=
r2
where M is the mass of either sphere 1 or 2 and m3 is the mass of sphere 3. The components
of the two forces are shown in the following drawings: F 31
30.0° F sin θ F cos θ F cos θ F 32 30.0° F sin θ Clearly, the horizontal components of the two forces add to zero. Thus, the net force on
sphere 3 is the resultant of the vertical componen...
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 Spring '13
 CHASTAIN
 Physics, The Lottery

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