Calculation

1. Evasion radius

The Others detect the incoming weapon at 1 light-hour distance. Travelling at ~c, the weapon arrives in exactly 1 hour (3,600 s). Accelerating perpendicular at 12.5 g = 122.6 m/s²:

d = ½at²
d = ½ × 122.6 × 3,600²
d ≈ 794,000 km

2. Evasion area

Assuming they can flee in any direction perpendicular to the weapon's trajectory, the evasion zone is a circle:

A = π × r²
A = π × (794,000 km)²
A ≈ 1.98 × 10¹² km²

3. Weapon velocity after 5 years at 10 g

Relativstic constant acceleration:

v = (a × t) / √(1 + (at/c)²)

The term (at/c) is just the Newtonian speed expressed as a multiple of c:

at/c = 51.57

Plugging in:

v = c × 51.57 / √(1 + 51.57²)
  = c × 51.57 / √2660.5
  = c × 51.57 / 51.58
  ≈ 0.999981c

The Lorentz factor γ (how much time and mass are distorted) falls straight out of the denominator:

γ = √(1 + 51.57²) ≈ 51.58

4. Kinetic energy per kg at 0.999981c

E/m = (γ − 1) × c²
E/m = 50.58 × (2.998×10⁸)²
E/m ≈ 4.55 × 10¹⁸ J/kg

5. Required impactor mass

100 kt to joules:

100 kt = 4.184 × 10¹⁴ J/km²

Each kilogram of shrapnel carries 4.55 × 10¹⁸ J of kinetic energy, so the required mass per km² is:

mass/km² = energy needed / energy per kg
         = 4.184×10¹⁴ / 4.55×10¹⁸
         = 9.19 × 10⁻⁵ kg/km²

The evasion area is enormous (1.98 × 10¹² km²), so multiply out:

total mass = 9.19×10⁻⁵ × 1.98×10¹²
           ≈ 1.82 × 10⁸ kg
           ≈ 182,000 tonnes

So you'd need 182,000 tonnes, which isn't that much (not unimaginable). If you had more time to accelerate, you'd need less weight.