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1.3 An artificial satellite is placed into a natural synchronous orbit above the equator and is
attached to the earth below by a wire. A second satellite is attached to the first by a wire of
the same length and is placed in orbit directly above the first at the same angular velocity.
assuming that the wires have zero mass, calculate the tension in the wires per unit mass of
satellite.
Answer
As usual a good sketch (Figure 1) is >50% of the solution of the problem.
Figure 1. Two satellites (m1 and m2) attached by wires with each other
and the earth. Black arrows: gravitational forces, white arrows: tension.
It is not difficult to understand that satellite 2, which is at twice the distance from the earth,
needs an extra force (the tension in the wire) to perform a synchronous circular orbit. The
necessary centripetal force Fc = mΩ 2 r increases due to the larger value of r (r=a+2h), while
the gravitational force decreases. So from Figure 1 we have that the centripetal force of
satellite 2 is the sum of the gravitational force and the tension:
Fc = Fg 2 + Ft = F5 + F6
(1)
which yields:
mΩ 2 (a + 2h ) = G
Mm
(a + 2h )2
+ Ft .
(2)
From Equation (2) we can calculate the tension (Ft), assuming m = 1 kg:
Ft = Ω 2 (a + 2h ) − G
(
= 7.292 × 10 −5
M
(a + 2h )2
) × (6370 + 2 × 35829) × 10
2
3
− 6.673 × 10 −11
5.988 × 10 24
(6370 + 2 × 35829)2 × 10 6
= 0.4149 − 0.0656 = 0.3494 ≈ 0.35 N
Note that:
1. Applying Newton’s third law (action = - reaction) learns that F1 = F3 = F4 = F6 (in
magnitude NOT in direction). The tension in both wires is equal.
2. The tensional forces on satellite 1 exactly cancel.
3. Newton’s third law also implies that there are gravitational forces from both satellites
on the earth. These have been omitted in Figure 1 for clarity.
1.4 A train is running smoothly along a curved track at the rate of 50 m s-1. A passenger
standing on a set of scales observes that his weight is 10% greater than when the train is at
rest. The track is banked so that the force acting on the passenger is normal to the floor of
the train. What is the radius of curvature of the track?
Answer
Figure 1 Left: Two forces acting on a person in circular motion on a
tilted plane: the normal force (FN) and gravity (Fg). Right: the sum of
the two forces is the centripetal force (Fc) necessary for circular
motion.
See Figure 1: the (vector)sum of Fg and FN (the normal force) is the necessary centripetal
force Fc, necessary to perform a circular motion. For the two real forces we have, if the mass
of the person on the scales is m:
Fz = mg
and
FN = 1.10 × Fz = 1.10 × mg
From Figure 1 (left) we can see that
Fc = FN2 − Fz2 = mg 1.10 2 − 1 = mg × 0.4583
We also have by definition, and using Equation (1):
Fc =
mv 2
= mg × 0.4583 ,
R
which gives:
50 2
= 9.81 × 0.4583 and R = 556 m.
R
(1)