Question

The Saturn V rocket had a mass of 2.45x10 kg, 65% of which was fuel. In the absence of gravity and starting at rest, what would be the maximum velocity attained (the "burnout velocity")? The fuel exhaust velocity was 3100m/s.​

Answer 1

Approximately
`5800\; {\rm m\cdot s^(-1)}`.

Step-by-step explanation:

Let
`m` denote the initial mass of the rocket and the fuel. The question states that the mass of the fuel is
`65\%` of the total mass, which is
`0.65\, m`. The mass of the rocket without the fuel would be
`(1 - 0.65)\, m`.

When an object of mass
`m` travels at a velocity of
`v`, the momentum
`p` of that object would be
`p = m\, v`.

Initial momentum of the rocket and the fuel, combined, would be
`0\; {\rm kg \cdot m\cdot s^(-1)}` since initial velocity was
`0\; {\rm m\cdot s^(-1)}`.

Let
`v(\text{rocket})` and
`v(\text{fuel})` denote the final velocity of the rocket and the fuel. Final momentum of the rocket would be
`((1 - 0.65)\, m)\, (v(\text{rocket}))`. Final momentum of the fuel would be
`(0.65\, m)\, (v(\text{fuel}))`.

The total final momentum of the rocket and the fuel, combined, woul dbe:

`((1 - 0.65)\, m)\, (v(\text{rocket})) + (0.65\, m)\, (v(\text{fuel}))`.

Under the assumptions, momentum would be conserved. In other words, the total momentum of the rocket and the fuel would stay the same:

`(\text{total final momentum}) = (\text{total initial momentum})`.

`((1 - 0.65)\, m)\, (v(\text{rocket})) + (0.65\, m)\, (v(\text{fuel})) = 0`.

Given that
`v(\text{fuel}) = 3100\; {\rm m\cdot s^(-1)}`:

`\begin{aligned}v(\text{rocket}) &= (0.65)/(1 - 0.65)\, v(\text{fuel}) \\ &= (0.65)/(1 - 0.65)\, (3100)\; {\rm m\cdot s^(-1)} \\ &\approx 5800\; {\rm m\cdot s^(-1)}\end{aligned}`.