6.3 Solutions
339
6.20 τ 0 = τ / γ = τ (1 − β
2 )
1/2
= 2.9 × 10
−6
× (1 − 0.6
2 )
1/2
= 2.32 × 10
−6 s
6.21 (a) Time, τ = τ 0 /(1 − β
2 )
1/2
= 2.2 × 10
−6
/(1 − 0.6
2 )
1/2
= 2.75 × 10
−6 s
Distance d = vτ = 0.6 × 3 × 10
8
× 2.75 × 10
−6
= 495 m
(b) d 0 = vτ 0 = 0.6 × 3 × 10
8
× 2.2 × 10
−6
= 396 m.
6.22 d = vt = vγ t 0 = 3 × 10
4 c
γβ = 3 × 10
4
/40 = 750
β/(1 − β
2 )
1/2
= 750
β = 0.99999956
6.23 d = vγ t 0 = βγ ct 0 = 0.99 × 3 × 10
8
× 2.5 × 10
−8
/(1 − 0.99
2 )
1/2
= 373 m.
It is therefore observed at an altitude of 1,000 − 373 = 627 m above the sea
level.
6.24 β = (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 = 2 cm
6.26 m = γ m 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 distance d from the earth then d is fixed by
the time taken by the radio signal to reach the earth is
d = ct
As observed from the earth, at t 1 = 0 the spaceship was at a distance d
approaching with speed 0.5 c. It will arrive at time
t 1 = d/βc = ct/βc = 1,125/0.5 = 2,250 s
6.28 The time t 2 recorded in the spaceship related to t 1 is shortened by γ , the
Lorentz factor.
t 2 = t 1 /γ = t(1 − β
2 )
1/2
= t(1 − 0.5
2 )
1/2
= 0.866 t
= 0.866 × 2,250 = 1,948 s
6.29 Let system S be attached to the ground and S
to the spaceship.
Let t 1 be the time when the radio signal reaches the ship. In that time the
signal traveled a distance
d 1 = ct 1
At time t 1 = 0, the ship was at a distance d.
At time t 1 it is now at a distance
d 2 = d + vt 1 = d + βct 1
Now d 1 = d 2
339
6.20 τ 0 = τ / γ = τ (1 − β
2 )
1/2
= 2.9 × 10
−6
× (1 − 0.6
2 )
1/2
= 2.32 × 10
−6 s
6.21 (a) Time, τ = τ 0 /(1 − β
2 )
1/2
= 2.2 × 10
−6
/(1 − 0.6
2 )
1/2
= 2.75 × 10
−6 s
Distance d = vτ = 0.6 × 3 × 10
8
× 2.75 × 10
−6
= 495 m
(b) d 0 = vτ 0 = 0.6 × 3 × 10
8
× 2.2 × 10
−6
= 396 m.
6.22 d = vt = vγ t 0 = 3 × 10
4 c
γβ = 3 × 10
4
/40 = 750
β/(1 − β
2 )
1/2
= 750
β = 0.99999956
6.23 d = vγ t 0 = βγ ct 0 = 0.99 × 3 × 10
8
× 2.5 × 10
−8
/(1 − 0.99
2 )
1/2
= 373 m.
It is therefore observed at an altitude of 1,000 − 373 = 627 m above the sea
level.
6.24 β = (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 = 2 cm
6.26 m = γ m 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 distance d from the earth then d is fixed by
the time taken by the radio signal to reach the earth is
d = ct
As observed from the earth, at t 1 = 0 the spaceship was at a distance d
approaching with speed 0.5 c. It will arrive at time
t 1 = d/βc = ct/βc = 1,125/0.5 = 2,250 s
6.28 The time t 2 recorded in the spaceship related to t 1 is shortened by γ , the
Lorentz factor.
t 2 = t 1 /γ = t(1 − β
2 )
1/2
= t(1 − 0.5
2 )
1/2
= 0.866 t
= 0.866 × 2,250 = 1,948 s
6.29 Let system S be attached to the ground and S
to the spaceship.
Let t 1 be the time when the radio signal reaches the ship. In that time the
signal traveled a distance
d 1 = ct 1
At time t 1 = 0, the ship was at a distance d.
At time t 1 it is now at a distance
d 2 = d + vt 1 = d + βct 1
Now d 1 = d 2