Physics 43 HW 3
Serway Chapter 39 & Knight Chapter 37
Serway 7th Edition Chapter 39 Problems: 15, 18, 52, 57, 60, 65 15. Review problem. An alien civilization occupies a brown dwarf, nearly stationary relative to the Sun, several lightyears away. The extraterrestrials have come to love original broadcasts of I Love Lucy, on our television channel 2, at carrier frequency 57.0 MHz. Their line of sight to us is in the plane of the Earth’s orbit. Find the difference between the highest and lowest frequencies they receive due to the Earth’s orbital motion around the Sun. P39.15 The orbital speed of the Earth is as described by
v=
GmS
=
r
( 6.67 × 10
−11
N ⋅m
2
∑ F = m a:
)(
kg2 1.99 × 1030 kg
1.496 × 10 m
11
G m Sm E
r2
m E v2
r
=
) = 2.98 × 10
4
m s.
The maximum frequency received by the extraterrestrials is
fobs = fsource
1+ v c
= 57.0 × 106 H z
1− v c
(
)
(
1− ( 2.98 × 10
) ( 3.00 × 10
s) ( 3.00 × 10
1+ 2.98 × 104 m s
4
m
) = 57.005 66 × 10
s)
8
m s
8
m
6
H z.
The minimum frequency received is
fobs = fsource
(
) 1+ 2.98 × 10
(
1− v c
= 57.0 × 106 H z
1+ v c
(
) ( 3.00 × 10
s) ( 3.00 × 10
1− 2.98 × 104 m s
4
m
) = 56.994 34 × 10
s)
8
m s
8
m
6
H z.
The difference, which lets them figure out the speed of our planet, is
( 57.005 66 − 56.994 34) × 106 H z =
1.13 × 104 H z .
18. A physicist drives through a stop light. When he is pulled over, he tells the police officer that the Doppler shift made the red light of wavelength 650 nm appear green to him, with a wavelength of 520 nm. The police officer writes out a traffic citation for speeding. How fast was the physicist traveling, according to his own testimony? P39.18
For the light as observed
1+ v c
1+ v c c
c
fobs =
=
fsource =
1− v c
1 − v c λsource
λobs
1 + v c λsource 650 nm
=
=
1− v c
520 nm
λobs
v
v
= 1.562 − 1.562
c
c
v = 0.220c = 6.59 × 107 m s
1+
1+ v c
= 1.252 = 1.562
1− v c
v 0.562
=
= 0.220
c 2.562
52. Ted and Mary are playing a game of catch in frame S’, which is moving at 0.600c with respect to frame S, while Jim, at rest in frame S, watches the action. Ted throws the ball to Mary at 0.800c (according to Ted) and their separation (measured in S’) is 1.80 × 1012 m. (a) According to Mary, how fast is the ball moving? (b) According to Mary, how long does it take the ball to reach her? (c) According to Jim, how far apart are Ted and Mary, and how fast is the ball moving? (d) According to Jim, how long does it take the ball to reach Mary? P39.52
(a)
Since Mary is in the same reference
frame, S ′, as Ted, she measures the ball to have the same speed Ted observes, namely
ux′ = 0.800c .
Lp
(b)
Δt ′ =
(c)
L = Lp 1 −
ux′
1.80 × 1012 m
=
(
)
0.800 3.00 × 10 m s
8
v2
= 1.80 × 1012 m
c2
(
)
= 7.50 × 103 s
1−
( 0.600c)2
c2
= 1.44 × 1012 m
Since v = 0.600c and u′x = −0.800c , the velocity Jim measures for the ball is
ux =
(d)
u′x + v
( −0.800c) + ( 0.600c)
=
= −0.385c
2
1 + ( −0.800)( 0.600)
1 + u ′x v c
Jim measures the ball and Mary to be initially separated by 1.44 × 1012 m. Mary’s motion at
0.600c and the ball’s motion at 0.385c nibble into this distance from both ends. The gap closes at
the rate 0.600c + 0.385c = 0.985c , so the ball and catcher meet after a time
Δt =
1.44 × 1012 m
(
)
0.985 3.00 × 10 m s
8
= 4.88 × 103 s
60. Imagine that the entire Sun collapses to a sphere of radius Rg such that the work required to remove a small mass m from the surface would be equal to its rest energy mc2. This radius is called the gravitational radius for the Sun. Find Rg. (It is believed that the ultimate fate of very massive stars is to collapse beyond their gravitational radii into black holes.) P39.60 If the energy required to remove a mass m from the surface is equal to its rest energy mc2 ,
then
GM sm
= mc2
Rg
and Rg =
GM s
c2
6.67 × 10−11 N ⋅ m 2 kg 2 )( 1.99 × 1030 kg )
(
=
2
( 3.00 × 108 m s)
Rg = 1.47 × 103 m = 1.47 km
57. An alien spaceship traveling 0.600c toward the Earth launches a landing craft with an advance guard of purchasing agents and physics teachers. The lander travels in the same direction with a speed of 0.800c relative to the mother ship. As observed on the Earth, the spaceship is 0.200 ly from the Earth when the lander is launched. (a) What speed do the Earth observers measure for the approaching lander? (b) What is the distance to the Earth at the time of lander launch, as observed by the aliens? (c) How long does it take the lander to reach the Earth as observed by the aliens on the mother ship? (d) If the lander has a mass of 4.00 × 105 kg, what is its kinetic energy as observed in the Earth reference frame? P39.57
(a)
Take the spaceship as the primed frame, moving toward the right at v = +0.600c .
Then u′x = +0.800c , and
Lp
ux =
ux′ + v
0.800c + 0.600c
=
= 0.946c
1 + ( ux′ v ) c2 1 + ( 0.800)( 0.600)
L = ( 0.200 ly ) 1 − ( 0.600) = 0.160 ly
2
(b)
L=
(c)
The aliens observe the 0.160-ly distance closing because the probe nibbles into it from one end
at 0.800c and the Earth reduces it at the other end at 0.600c.
(d)
γ
:
0.160 ly
= 0.114 yr
0.800c + 0.600c
Thus,
time =
⎛
⎞
1
K =⎜
− 1⎟ mc2 :
⎜ 1 − u 2 c2
⎟
⎝
⎠
⎛
⎞
1
K =⎜
− 1⎟ 4.00 × 105 kg 3.00 × 108 m s
⎜ 1 − 0.946 2
⎟
(
)
⎝
⎠
(
K = 7.50 × 1022 J
)(
)
2
65. Suppose our Sun is about to explode. In an effort to escape, we depart in a spacecraft at v = 0.800c and head toward the star Tau Ceti, 12.0 ly away. When we reach the midpoint of our journey from the Earth, we see our Sun explode and, unfortunately, at the same instant we see Tau Ceti explode as well. (a) In the spacecraft’s frame of reference, should we conclude that the two explosions occurred simultaneously? If not, which occurred first? (b) What If? In a frame of reference in which the Sun and Tau Ceti are at rest, did they explode simultaneously? If not, which exploded first? P39.65
We choose to write down the answer to part (b) first.
(b)
Consider a hermit who lives on an asteroid halfway between the Sun and Tau Ceti, stationary
with respect to both. Just as our spaceship is passing him, he also sees the blast waves from
both explosions. Judging both stars to be stationary, this observer concludes
that the two stars blew up simultaneously .
(a)
We in the spaceship moving past the hermit do not calculate the explosions to be simultaneous.
We measure the distance we have traveled from the Sun as
2
2
⎛v⎞
L = Lp 1 − ⎜ ⎟ = ( 6.00 ly ) 1 − ( 0.800) = 3.60 ly
⎝ c⎠
We see the Sun flying away from us at 0.800c while the light from the Sun approaches at 1.00c.
Thus, the gap between the Sun and its blast wave has opened at 1.80c, and the time we
calculate to have elapsed since the Sun exploded is
3.60 ly
= 2.00 yr
1.80c
We see Tau Ceti as moving toward us at 0.800c, while its light approaches at 1.00c, only 0.200c
faster. We measure the gap between that star and its blast wave as 3.60 ly and growing at
0.200c. We calculate that it must have been opening for
3.60 ly
= 18.0 yr
0.200c
and conclude that Tau Ceti exploded 16.0 years before the Sun .
Knight 2nd Edition Chapter 37 Exercises & Problems: 12, 13, 15, 54, 61, 72 37.12. You are standing at x = 9.0km.Lightning bolt 1 strikes at x = 0 km and lightning bolt 2 strikes at 12.0km. Both
flashes reach your eye at the same time. Your assistant is standing at x = 3.0km. Does your assistant see the flashes at
the same time? If not, which does she see first and what is the time difference between the two?
Model: You and your assistant are in the same reference frame. Light from the two lightning bolts travels toward you and your
assistant at 300 m/μs. You and your assistant have synchronized clocks. “flashes reach your eye” means that you see the lightning
strikes as simultaneous but since they are at different distances, you know that they struck at different times.
Visualize:
Solve:
Bolt 1 is 9.0 km away, so it takes 30 μs for the light to reach you ( 9000 m ÷ 300 m/μ s ) . Bolt 2 is 3.0 km away from you, so
it takes 10 μ s to reach you. Since both flashes reach your eye at the same time, event 1 happened 20 μs before event 2. If event 1
happened at time t1 = 0 then event 2 happened at time t2 = 20 μs. For your assistant, it takes light from bolt 1 10 μ s to reach her and
light from bolt 2 30 μs to reach her. She sees the flash from bolt 1 at t = 10 μs and the flash from bolt 2 at t = 50 μs. That is, your
assistant sees flash 2 40 μs after she sees flash 1.
37.13. You are standing at x = 9.0km.Lightning bolt 1 strikes at x = 0 km and lightning bolt 2 strikes at 12.0km. You see
the flash from bolt 2 at t= 10 μs and the flash from bolt 1 at t = 50μs. Your assistant is standing at x = 3.0km. Does your
assistant see the flashes as simultaneous? If not, which does she see first and what is the time difference between the
two?
Model: You and your assistant are in the same reference frame. Light from the two lightning bolts travels toward you and your
assistant at 300 m/μs. You and your assistant have synchronized clocks.
Visualize: same as above
Solve: Bolt 1 hits 9.0 km away, so the light takes 30 µs to reach you (9000 m ÷ 300 m/μs). You see this flash at t =
50 μs, so the lightning hit at t1 = 20 μs. Light from bolt 2, which hits 3.0 km away, takes 10 μs to reach you. You see it at 10 μs, so
the lightning hit at t2 = 0 μs. The strikes are not simultaneous. Bolt 2 hits first, 20 μs before bolt 1. Your assistant is in your inertial
reference frame, so your assistant agrees that bolt 2 hits first, 20 μs before bolt 1.
Assess: A simple calculation would show that your assistant sees the flashes at the same time. When the flashes are seen is not the
same as when the events happened.
37.15. Model: Your personal rocket craft is an inertial frame moving at 0.9c relative to stars A and B.
Solve: In your frame, star A is moving away from you and star B is moving toward you. When you are exactly halfway between
them, both the stars explode simultaneously. The flashes from the two stars travel toward you with speed c. Because (i) you are at
rest in your frame, (ii) the explosions are equally distant, and (iii) the light speed is c, independent of the fact that the stars are
moving in your frame, the light will arrive simultaneously.
37.54. Model: Let the earth be frame S and the rocket be frame S´. S´ moves with speed v relative to S.
Solve: (a) The round-trip distance is 860 ly. If the rocket takes time ∆t to make the round trip, as measured on earth, its speed (as
a fraction of c) is
v 860 ly 860 yr
=
=
Δt
c
c Δt
where we used c = 1 ly/yr (1 light year per year). The astronaut’s elapsed time ∆t´ is the proper time, so ∆τ = 20 yr. The time
dilation equation is
Δτ
20 yr
Δt =
=
⇒ 1 − (860 yr/ Δt ) 2 = ( 20 yr/ Δt ) 2
2
1 − ( v/c )
1 − (860 yr/Δt ) 2
Solving for ∆t gives ∆t = 860.2325 y, and thus
v
860 y
=
= 0.99973 ⇒ v = 0.99973c
c 860.2325 y
(b) The rocket starts with rest energy Ei = mc2 and accelerates to have energy Ef = γpmc2. Thus the energy needed to accelerate the
rocket is
∆E = Ef – E1 = (γp – 1)mc2
This is just the kinetic energy K gained by the rocket. We know the rocket’s speed, so
⎛
⎞
1
ΔE = ⎜
− 1⎟ (20,000 kg)(3.0 × 108 m/s) 2 = 7.6 × 1022 J
2
⎜ 1 − (0.99973)
⎟
⎝
⎠
(c) The total energy used by the United States in 2000 was ≈1.0 × 1020 J. To accelerate the rocket would require roughly 760 times
the total energy used by the United States.
37.61. Model: Let S be the ground’s reference frame and S′ the muon’s reference frame. S′ travels with a speed of v relative to
S.
Solve: (a) The half-life of a muon at rest is 1.5 μs. That is, the half-life in the muon’s rest frame S′ is 1.5 μs. So, Δt′ = Δτ =
1.5 μs. The half-life of 7.5 μs, when muons have been accelerated to very high speed, means that Δt = 7.5 μs. Thus
Δτ
Δt = 7.5 μs =
1 − (v c)
2
=
1.5 μs
1 − v2 c2
⇒ 1 − v 2 c 2 = 0.20 ⇒ v = 0.98c
(b) The muon’s total energy is
E = γ p mc 2 =
2
⎛ 1 ⎞
−31
−11
8
mc 2 = ⎜
⎟ ( 207 ) ( 9.11× 10 kg )( 3.0 × 10 m/s ) = 8.5 × 10 J
0.20
⎝
⎠
1− v c
1
2
2
37.72. Model: Mass and energy are equivalent and given by Equation 37.43.
(a) The sun radiates energy for 3.154 × 107 s per year. The amount of energy radiated per year is
(3.8 × 1026 J/s)(3.154 × 107 s) = 1.198 × 1034 J/y
Since E0 = mc2, the amount of mass lost is
Solve:
m=
E0
1.198 × 1034 J
=
= 1.33 × 1017 kg ≈ 1.3 × 1017 kg
2
2
8
c
( 3.0 × 10 m/s )
(b) Since the mass of the sun is 2.0 × 1030 kg, the sun loses 6.7 × 10
(c) The lifetime of the sun can be estimated to be
T=
−12
% of its mass every year.
2.0 × 1030 kg
= 1.5 × 1013 years
1.33 × 1017 kg/y
The sun will not really last this long in its current state because fusion only takes place in the core and it will become a red giant
when the core hydrogen is all fused.