Exercise on single stage vs. two-stage rocket

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Exercise:

1. Single stage rocket: 

Before a rocket begins to burn fuel, the rocket has a mass of  mr,i=2.81×107kg, of which the mass of the fuel is  mf,i=2.46×107kg. The fuel is burned at a constant rate with total burn time is 510 s and ejected at a speed ve = 3000 m/s relative to the rocket. If the rocket starts from rest in empty space, what is the final speed of the rocket after all the fuel has been burned?

What we know:

initial rocket mass mr,i=2.81×107kg

initial mass of the fuel mf,i=2.46×107kg

t = 510 s

ve = 3000 m/s

So, we can calculate the dry mass of the rocket mr,d=mr,i-mf,i=0.35×107kg.

The mass ratio is then R = mr,i /mr,d=8.03.

So, the final speed of the rocket is then: Vr,f=Δvr=velnR=6250m/s.

2. Two-stage rocket: 

Now, we have similar example, the same rocket as in the previous example, but the fuel will be burnt in two stages ejecting the fuel in each of the stages at the same relative speed (ve = 3000 m/s). 

In first stage, the available fuel to burn is mf,1,i=2.03×107kg and it will burn time 150 s. Then the empty tank and the engine will be jettisoned away. These parts have a mass of  m=1.4×106kg. 

The remaining fuel is burnt during the second stage in time 360 s. All together the burning time is 510 s (two seconds shorter than in the first case). 

The question is the same as in the first example: What is the final speed of the rocket after all the fuel has been burned?

What we know:

the initial rocket mass has mr,i=2.81×107kg

the initial mass of the fuel is mf,1,i=2.03×107kg

the discarded parts has mass m=1.4×106kg

ve = 3000 m/s

the first burn time is 150 s

the second burn time is 360 s.

Stage 1: The calculation is then, the mass of the rocket after all the fuel in the first stage is burned is  mr,1,d=mr,i−mf,1,i=2.81×107kg- 2.03×107kg = 0.78×107kg and the first mass ratio is then R1=mr,i/mr,1,d=3.60.

The change in speed after the first stage is complete is Δvr,1=velnR1=3843m/s. 

Into the stage 2: After the empty fuel tank and accessories from the first stage one are disconnected from the rest of the rocket, we can calculate the dry mass of the rocket mr,2,d=mr,d-mf,i-m=0.35×107kg-1.4×106kg=2.1×106kg, which is the remaining mass of the rocket. 

The mass of the rocket plus the unburned fuel at the beginning of the second stage is mr,2,i=mr,1,d-m=6.4×106kg.

The remaining fuel has mass  mf,2,i=mr,2,i-mr,2,d=4.3×106kg. 

Then  R2=mr,2,i/mr,2,d=3.05.

Therefore the rocket increases its speed during the second stage by an amount Δvr,2=velnR2=3342m/s.

The final speed of the rocket is the sum of the change in speeds due to each stage,

vf=Δvr=velnR1+velnR2=veln(R1R2)=7185m/s,

which is higher number than if the fuel were burned in one stage.

After solving their exercises, i need to use their plot. But my numbers are very close to their. source: https://phys.libretexts.org/ 


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