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would increase with the mass of the planet and decrease as
the satellite’s distance from the center of the planet increased. Fg Exercise v
m Figure 6.7 A satellite of mass m moving around the Earth at a constant speed v in a circular orbit of radius r RE h . The force Fg
acting on the satellite that causes the centripetal acceleration is the
gravitational force exerted by the Earth on the satellite. EXAMPLE 6.7 A satellite is in a circular orbit around the Earth at
an altitude of 1 000 km. The radius of the Earth is equal to
6.37 106 m, and its mass is 5.98 1024 kg. Find the speed
of the satellite, and then ﬁnd the period, which is the time it
needs to make one complete revolution. Answer 103 m/s; 6.29 7.36 103 s = 105 min. Let’s Go Loop-the-Loop! A pilot of mass m in a jet aircraft executes a loop-the-loop, as
shown in Figure 6.8a. In this maneuver, the aircraft moves in
a vertical circle of radius 2.70 km at a constant speed of
225 m/s. Determine the force exerted by the seat on the pilot
(a) at the bottom of the loop and (b) at the top of the loop.
Express your answers in terms of the weight of the pilot mg. celeration has a magnitude n bot mg, Newton’s second law
for the radial direction combined with Equation 6.1 gives
Fr
nbot Solution We expect the answer for (a) to be greater than
that for (b) because at the bottom of the loop the normal
and gravitational forces act in opposite directions, whereas at
the top of the loop these two forces act in the same direction.
It is the vector sum of these two forces that gives the force of
constant magnitude that keeps the pilot moving in a circular
path. To yield net force vectors with the same magnitude, the
normal force at the bottom (where the normal and gravitational forces are in opposite directions) must be greater than
that at the top (where the normal and gravitational forces are
in the same direction). (a) The free-body diagram for the pilot at the bottom of the loop is shown in Figure 6.8b. The
only forces acting on him are the downward force of gravity
Fg m g and the upward force n bot exerted by the seat. Because the net upward force that provides the centripetal ac- nbot
mg mg
m v2
r m v2
r
mg 1 v2
rg Substituting the values given for the speed and radius gives
nbot mg 1 (2.70 (225 m/s)2
103 m)(9.80 m/s2) 2.91mg Hence, the magnitude of the force n bot exerted by the seat
on the pilot is greater than the weight of the pilot by a factor
of 2.91. This means that the pilot experiences an apparent
weight that is greater than his true weight by a factor of 2.91.
(b) The free-body diagram for the pilot at the top of the
loop is shown in Figure 6.8c. As we noted earlier, both the
gravitational force exerted by the Earth and the force n top exerted by the seat on the pilot act downward, and so the net
downward force that provides the centripetal acceleration has 158 CHAPTER 6 Circular Motion and Other Applications of Newton’s Laws Figure 6.8 (a) An aircraft executes a loop-the-loop maneuve...

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