# Multistage Rocket

1. Sep 21, 2009

### Piamedes

1. The problem statement, all variables and given/known data
To illustrate the use of a multistage rocket consider the following:

(a) A certain rocket carries 60% of its initial mass as fuel. (That is, the mass of fuel is .6m0). What is the rocket's final speed, accelerating from rest in free space, if it burns all its fuel in one stage? Express your answer as a multiple of v exhaust.

(b) Suppose instead it burns the fuel in two stages as follows: In the first stage it burns a mass of .3m0 of fuel. It then jettisons the first-stage fuel tank, which has a mass of .1m0 and then burns the remaining .3m0 of fuel. Find the final speed in this case, assuming the same value of v exhaust throughout and compare.

2. Relevant equations

$$v - v_{0} = v_{ex} ln [\frac{m_{0}}{m}]$$

3. The attempt at a solution

For part a)

$$v - 0 = v_{ex} ln [\frac{m_{0}}{.4m_{0}}$$

$$v = v_{ex} ln [\frac{1}{\frac{2}{5}}]$$

$$v = v_{ex} ln [\frac{5}{2}]$$

For part b)

$$v_{1} - 0 = v_{ex} ln [\frac{m_{0}}{.7m_{0}}]$$

$$v_{1} = v_{ex} ln [\frac{1}{\frac{7}{10}}]$$

$$v_{1} = v_{ex} ln [\frac{10}{7}]$$

$$v_{2} - v_{1} = v_{ex} ln [\frac{.6m_{0}}{.3m_{0}}]$$

$$v_{2} - v_{1} = v_{ex} ln [2]$$

$$v_{2} = v_{ex} ln [2] + v_{1}$$

$$v_{2} = v_{ex} ln [2] + v_{ex} ln [\frac{10}{7}]$$

$$v_{2} = v_{ex} {ln [\frac{20}{7}] }$$

Is this the proper way to calculate multistage rocket velocities, or am I missing an important step? Because in both cases the final velocity seems to very small in comparison to the exhaust velocity, or is that just how rockets work?

Thanks for any and all help

2. Sep 21, 2009

### D H

Staff Emeritus
The answers you have are correct. That is just the way chemical rockets work. The rocket equation is just mean and nasty.

Most rockets the go into space burn a lot more than 60% of their mass in the form of fuel. Its more like 90 to 95%.

3. Sep 21, 2009

### Piamedes

Thanks, I guess it just seemed odd that they're so inefficient.