A rocket of total mass 1.11 X 10⁵ kg, of which 8.70 X 10⁴ kg is fuel, is to be launched vertically. The fuel will be burned at the constant rate of 820 kg/s. Relative to the rocket, what is the minimum exhaust speed that allows liftoff at launch?

Given data:
Total mass of rocket = 1.11 × 10⁵ kg
Fuel mass = 8.70 × 10⁴ kg
Burn rate = 820 kg/s
Acceleration due to gravity = 9.81 m/s²
Step 1: Understanding the forces involved
At the moment of launch, the rocket must generate a thrust T at least equal to its weight W to lift off:

T = W
or
ṁ vₑ = Mg

where:

ṁ = fuel burn rate = 820 kg/s
vₑ = exhaust velocity (to be determined)
M = total mass at launch = 1.11 × 10⁵ kg
g = 9.81 m/s²
Step 2: Solve for exhaust speed
From the thrust equation:
820 × vₑ = (1.11 × 10⁵) × (9.81)

820 vₑ = 1.089 × 10⁶

Solving for vₑ:
vₑ = (1.089 × 10⁶) / 820
vₑ ≈ 1328.05 m/s

Final Answer:
The minimum exhaust speed required for the rocket to lift off is 1328 m/s.