Calculating Multistage Rocket Velocities

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The discussion focuses on calculating the final speed of a multistage rocket using the rocket equation. In the first scenario, where 60% of the initial mass is fuel, the final speed is derived as a multiple of exhaust velocity. The second scenario, involving two stages of fuel burning and tank jettisoning, results in a higher final speed compared to the single-stage burn. Participants note that the calculated speeds appear low relative to exhaust velocity, which is typical for chemical rockets. The conversation concludes with an acknowledgment of the inherent inefficiency of rockets, which often consume 90-95% of their mass as fuel.
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Homework Statement


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.

Homework Equations



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

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
 
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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%.
 
Thanks, I guess it just seemed odd that they're so inefficient.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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