77.
(a) The relative contraction is
|∆L|
L
0
=
L
0
(1
− γ
−1
)
L
0
= 1
−
1
− β
2
≈ 1 −
1
−
1
2
β
2
=
1
2
β
2
=
1
2
630 m/s
3.00
× 10
8
m/s
2
=
2.21
× 10
−12
.
(b) Letting
|∆t − ∆t
0
| = ∆t
0
(γ
− 1) = τ = 1.00 µs, we solve for ∆t
0
:
∆t
0
=
τ
γ
− 1
=
τ
(1
− β
2
)
−1/2
− 1
≈
τ
1 +
1
2
β
2
− 1
=
2τ
β
2
=
2(1.00
× 10
−6
s)(1 d/86400 s)
[(630 m/s)/(2.998
× 10
8
m/s)]
2
=
5.25 d .