6.3 Solutions 339
6.20τ 0 =τ/γ=τ(1−β^2 )^1 /^2
= 2. 9 × 10 −^6 ×(1− 0. 62 )^1 /^2
= 2. 32 × 10 −^6 s
6.21 (a) Time,τ=τ 0 /(1−β^2 )^1 /^2 = 2. 2 × 10 −^6 /(1− 0. 62 )^1 /^2
= 2. 75 × 10 −^6 s
Distanced=vτ= 0. 6 × 3 × 108 × 2. 75 × 10 −^6 =495 m
(b)d 0 =vτ 0 = 0. 6 × 3 × 108 × 2. 2 × 10 −^6 =396 m.
6.22d=vt=vγt 0 = 3 × 104 c
γβ= 3 × 104 / 40 = 750
β/(1−β^2 )^1 /^2 = 750
β= 0. 99999956
6.23d=vγt 0 =βγct 0 = 0. 99 × 3 × 108 × 2. 5 × 10 −^8 /(1− 0. 992 )^1 /^2
=373 m.
It is therefore observed at an altitude of 1, 000 − 373 =627 m above the sea
level.
6.24β=(β 1 +β 2 )/(1+β 1 β 2 )=(0. 9 + 0 .9)/(1+ 0. 9 × 0 .9)= 0. 994475
6.25 E=T+m 0 c^2 =^100 +^0.^51 =^100 .51 MeV
γ=E/m 0 c^2 = 100. 51 / 0. 51 = 197
L=L 0 /γ= 4 / 197 = 0 .02 m=2cm
6.26m=γm 0
γ=m/m 0 = 101 / 100 = 1. 01
β=(γ^2 −1)^1 /^2 /γ= 0. 14
6.27 If the space station is located at a distancedfrom the earth thendis fixed by
the time taken by the radio signal to reach the earth is
d=ct
As observed from the earth, att 1 = 0 the spaceship was at a distance d
approaching with speed 0.5c. It will arrive at time
t 1 =d/βc=ct/βc= 1 , 125 / 0. 5 = 2 ,250 s
6.28 The timet 2 recorded in the spaceship related tot 1 is shortened byγ,the
Lorentz factor.
t 2 =t 1 /γ=t(1−β^2 )^1 /^2 =t(1− 0. 52 )^1 /^2 = 0. 866 t
= 0. 866 × 2 , 250 = 1 ,948 s
6.29 Let systemSbe attached to the ground andS′to the spaceship.
Lett 1 be the time when the radio signal reaches the ship. In that time the
signal traveled a distance
d 1 =ct 1
At timet 1 =0, the ship was at a distanced.
At timet 1 it is now at a distance
d 2 =d+vt 1 =d+βct 1
Nowd 1 =d 2