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RELATIVITY
33
EXERCISES
Section 33.2 Matter, Motion, and Ether
13.
INTERPRET
In this problem we are asked to take wind speed into consideration to calculate the travel time of
an airplane.
NTERPRET
DEVELOP
Since the velocities are small compared to
c
, we can use the nonrelativistic Galilean transformation of
velocities in Equation 3.7,
where
u
,
uuv
′
=+
GGG
G
is the velocity relative to the ground (
S
),
u
is that relative to the air
′
G
()
,
S
′
and v is that of
relative to
S
(in this case, the wind velocity). We used a notation consistent with that in
S
′
Equations 33.5a and 33.5b.
EVALUATE
(a)
If
(no wind), then
0
v
=
G
uu
′
=
G
G
(ground speed equals air speed), and the roundtrip travel time is
2
2(1800 km)
4.5 h
800 km/h
a
d
t
u
==
=
(b)
If
is perpendicular to
then
v
G
,
u
G
222
,
v
′
=
+
or
22
2
2
(800 km/h)
(130 km/h)
789 km/h
′
=−
=
−
=
and the roundtrip travel time is
2 /
4.56 h.
b
td
u
(c)
If
is parallel or antiparallel to
u
on alternate legs of the roundtrip, then
v
G
G
′
=
±
and the travel time is
(see Equation 33.2, but with
c
replaced by
)
u
′
1800 km
1800 km
4.62 h
800 km/h 130 km/h
800 km/h 130 km/h
c
dd
t
uvuv
=+=
+
=
′′
+−
+
−
ASSESS
We find
as mentioned in the paragraph following Equation 33.2.
,
abc
ttt
<<
14.
The difference between Equations 33.2 and 33.1 is

2
2
2
2
2
1
1
1/
cL
L
L
c
cv
v
c
v
c
⊥
⎛⎞
⎜⎟
Δ= − =
−
=
−
−−
⎝⎠
and
for the given interferometer. If
8
2 /
2(11 m)/(3 10 m/s)
73.3 ns
Lc
=×
=
/
vc
1
,
we can expand the
denominators (Appendix A) to obtain
(a)
For the Earth’s
orbital speed (30 km/s),
so that
2
2
(2 / )[1
/
(1
/2
)]
(2 / )( /2 ).
tL
c
v
c
L
c
v
c
Δ≈
+
− +
=
8
/1
0
,
−
=
8
1
2
73.3 ns(
10 )
3.67 10
s
t
−
Δ=
×
=
×
1
6
−
−
(a fraction of the period of visible
light).
(b)
If
(a few thousand periods of visible light).
(c)
At relativistic speeds, we
use the exact expression for
Δ
If
4
1
2
/
10 ,
3.67 10
s
t
−
=Δ
=
×
.
t
1
4
/,
1
/
1
/
2
/
3
,
γ
−
=
and
4
3
73.3 ns(
2/ 3)
13.1 ns.
t
−
=
(d)
If
/0
.
9
9
,
1/ 1 0.98
7.09,
=
=
and
(2 / ) (
1)
3.17 s.
c
γμ
Δ
=
Section 33.4 Space and Time in Relativity
15.
INTERPRET
This problem is about the distance between two stars, as seen by a spaceship moving at relativistic
speed.
DEVELOP
The distance between stars at rest in system
S
appears Lorentzcontracted in the spaceship’s system
,
S
′
according to Equation 33.4:
33.1
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View Full Document 33.2
Chapter 33
22
1/
x
xv
c
′
Δ=
Δ −
EVALUATE
With
and
we get
50 ly
x
0.75 ,
v
=
c
2
1
/
(50 ly) 1 (0.75)
33.1 ly
xxv
c
′
=
−
=
ASSESS
The distance appears to be shortened or “contracted” as observed by the spaceship. Note that length
contraction occurs only along the direction of motion.
16.
(a)
On Earth (system
S
, where the velocity of Pluto is much less than 0.65
c
)
9
/
(5.8 10 km)/
tx
v
Δ=Δ =
×
84
(0.65 3 10 m/s)
2.94 10 s
8.26 h.
××
=
×
=
(b)
Time dilation (Equation 33.3) means that the time interval
measured in the spacecraft (system
is
)
S
′
4
2
4
1
/
(2.94 10 s) 1 (0.65)
2.26 10 s
6.28 h.
ttv
c
′
Δ=Δ −
=
×
−
=
×
=
(Note that
(/
)1 /
/
,
vv
cx
′
Δ=Δ
−
=Δ
v
′
so Lorentz contraction could have been used to reach the same result.)
17.
INTERPRET
This is a problem about the length of the spaceship measured in its rest frame.
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This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.
 Fall '10
 Pheong

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