Transcript Example Problems for Ch. 37 Part 1

Example Problems for Ch. 37
Part 1
Problem 37.7
A spacecraft flies away from the earth with a
speed of 4.80 x 106 m/s relative to the earth
and then returns to the earth at the same
speed. The space craft returns to the earth
365 days later as measured by an atomic
clock on the earth. If there is an atomic clock
on the spaceship which was previously
synchronized to the clock on the earth, what is
the difference in elapsed times as measured
by the two clocks? Which clock shows the
smallest elapsed time?
Convert to fraction of c
This is the easiest way to deal with the
radical sign.
 4.80 x 106 m/s / 3 x 108 m/s is 0.016 so
 v=1.6%c

This is a time dilation problem

In our case, we must
determine which
clock measures
proper time
 The clock is at rest
on the moving
spaceship
 So Dt=365 days
 So now solve for Dt0
using u=1.6%c
Dt 
Dt 0
u
1 2
c
2
Substituting
So Dt=365 days
 So now solve for Dt0
using u=1.6%c

Dt 
Dt 0
u
1 2
c
2
2
u
Dt 1  2  D t 0
c
365* 1  0.016 2  364.9533  Dt0
Dt  Dt0  365  364.9533  0.046 days or 1.21 hours
Problem 37.13
A space craft of the Trade Federation flies
past the planet Coruscant at a speed of
0.600c. A scientist on Coruscant
measures the length of the moving
space craft to be 74 m. The spacecraft
later lands on Coruscant and the
scientist now measures the length of the
now stationary spacecraft. What value
does she get?
This is a length contraction problem

L0 is the proper
length
 L=74 m
 u=0.6c
2
u
L  L0 1  2
c
L
 L0
2
u
1 2
c
74
 L0
2
1  0.6
92.5m  L0
Problem 37.23
An imperial spaceship moving at high speed
relative to planet Arrakis, fires a rocket
toward the planet with a speed of 0.920c
relative to the spaceship. An observer on
Arrakis measures that the rocket is
approaching with a speed of 0.36c.
a) What is the speed of the spaceship relative
to Arrakis?
b) Is the spaceship moving towards or away
from Arrakis?
Draw It!
vx=0.36c
Arrakis
vx’=0.92c
u
Must solve for velocity, u
vx  u
vu
1  x2
c
v 'v u
vx ' x 2 x  vx  u
c
v 'v u
vx ' vx  x 2 x  u
c
 v 'v

vx ' vx  u  x 2 x  1
 c

vx ' vx
u
v
'
v
 x x 
 c 2  1


.92c  .36c
0.56c

 0.837c
 (.92c)(.36)  0.688
 1

c2


vx ' 
Since u<0, spaceship
moving away