Unformatted text preview: 482 SPECIAL RELATIVITY SOLUTION Solving Equation 28.2 for v, we find that
2 2 L 0.500 m 8
v = c 1 − = (3.00 ×108 m/s) 1 − = 2.60 ×10 m/s
L 1.00 m 0
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10. REASONING The distance between earth and the center of the galaxy is the proper length
L0, because it is the distance measured by an observer who is at rest relative to the earth and
the center of the galaxy. A person on board the spaceship is moving with respect to them
and measures a contracted length L that is related to the proper length by Equation 28.2 as ( ) L = L0 1 − v 2 / c 2 . The contracted distance is also equal to the product of the spaceship’s
speed v the time interval measured by a person on board the spaceship. This time interval is
the proper time interval ∆t0 because the person on board the spaceship measures the
beginning and ending events (the times when the trip starts and ends) at the same location
relative to a coordinate system fixed to the spaceship. Thus, the contracted distance is also
L = v∆t0. By setting the two expressions for L equal to each other, we can find the how long
the trip will take according to a clock on board the spaceship. ( ) SOLUTION Setting L = L0 1 − v 2 / c 2 equal to L = v∆t0 and solving for the proper time
interval ∆t0 gives
∆t0 = L0
v ( 1 − v2 / c2 ) 9.47 × 1015 m 1 ly ( 23 000 ly ) ( 0.9990c )2 1 yr =
1− = 1.0 × 103 yr
2
7
8
c 0.9990 3.00 × 10 m/s 3.16 × 10 s ______________________________________________________________________________ ( 11. ) SSM REASONING AND SOLUTION The length L0 that the person measures for the
UFO when it lands is the proper length, since the UFO is at rest with respect to the person.
Therefore, from Equation 28.2 we have L0 = L
2 = 230 m
2 = 530 m v
(0.90 c)
1−
2
c
c2
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1− Chapter 28 Problems 1483 12. REASONING The Martian measures the proper time interval ∆t0, because the Martian
measures the beginning and ending events (the times when the trip starts and ends) at the
same location relative to a coordinate system fixed to the spaceship.
The given distance between Mars and Venus is the distance as measured by a person on
earth. That person is at rest relative to the two planets and, hence, measures the proper
length. The Martian, who is moving relative to the planets, does not measure the proper
length, but measures a contracted length.
According to the Martian, the time of the trip ∆t0 is equal to the contracted length that he
measures divided by the speed v of the spaceship.
SOLUTION
a. The contracted length L measured by the Martian is related to the proper length L0 by
Equation 28.2 as
L = L0 1 − v2
c 2 ( = 1.20 × 10 m
11 ) ( 0.80c )2
1−
c 2 = 7.2 × 1010 m b. The time of the trip as measured by the Martian is ∆t0 = L
7.2 × 1010 m
2
=
= 3.0 × 10 s
8
v 0.80 3.00 × 10 m/s ( ) ______________________________________________________________________________
13. REASONING Assume that traveler A moves at a speed of vA = 0.70c and traveler B moves
at a speed of vB = 0.90c, both speeds being with respect to the earth. Each traveler is
moving with respect to the earth and the distant star, so each measures a contracted length
LA or LB for the distance traveled. However, an observer on earth is at rest with respect to
the earth and the distant star (which is assumed to be stationary with respect to the earth), so
he or she would measure the proper length L0. For each traveler the contracted length is
given by the lengthcontraction equation as stated in Equation 28.2:
LA = L0 1 − 2
vA c2 and LB = L0 1 − 2
vB c2 It is important to note that the proper length L0 is the same in each application of the lengthcontraction equation. Thus, we can combine the two equations and eliminate it. Then, since
we are given values for LA, vA, and vB, we will be able to determine LB. 1484 SPECIAL RELATIVITY SOLUTION Dividing the expression for LB by the expression for LA and eliminating L0,
we obtain LB
=
LA 1−
LB = LA
1− L0 1 −
L0 1 − 2
vB c2 =
2
vA
c2 2
vB c 2 = 6.5 lightyears
(
)
2
vA
c2 1−
1− 2
vB c2
2
vA c2 (
)
1 − 0.90c
2
c
= 4.0 lightyears
( 0.70c )2
1−
c2
2 14. REASONING
a. The two events are the creation of the particle and its subsequent disintegration. Relative
to a stationary reference frame fixed to the laboratory, these two events occur at different
locations, because the particle is moving relative to this reference frame. The proper
−3
distance L0 is the distance (1.05 × 10 m) given in the statement of the problem, because
this distance is measured by an observer in the laboratory who is at rest with respect to these
locations.
b. The distance measured by a hypothetical person traveling with the particle is a contracted
distance, because it is measured by a person who is moving relative to the two locations.
The contracted distance L is related to the proper distance L0 by the lengthcontraction
formula, L = L0 1 − v 2 / c 2 (Equation 28.2).
c. The proper lifetime ∆t0 of the particle is the lifetime as reg...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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