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 .