Exercise on single stage vs. two-stage rocket

Exercise:

1. Single stage rocket: 

Before a rocket begins to burn fuel, the rocket has a mass of  m_r,i=2.81×10⁷kg, of which the mass of the fuel is  m_f,i=2.46×10⁷kg. The fuel is burned at a constant rate with total burn time is 510 s and ejected at a speed v_e = 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 m_r,i=2.81×10⁷kg

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

t = 510 s

v_e = 3000 m/s

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

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

So, the final speed of the rocket is then: V_r,f=Δv_r=v_elnR=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 (v_e = 3000 m/s). 

In first stage, the available fuel to burn is m_f,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 m_r,i=2.81×10⁷kg

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

the discarded parts has mass m=1.4×10⁶kg

v_e = 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  m_r,1,d=m_r,i−m_f,1,i=2.81×10⁷kg- 2.03×10⁷kg = 0.78×10⁷kg and the first mass ratio is then R₁=m_r,i/m_r,1,d=3.60.

The change in speed after the first stage is complete is Δv_r,1=v_elnR₁=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 m_r,2,d=m_r,d-m_f,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 m_r,2,i=m_r,1,d-m=6.4×10⁶kg.

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

Then  R₂=m_r,2,i/m_r,2,d=3.05.

Therefore the rocket increases its speed during the second stage by an amount Δv_r,2=v_elnR₂=3342m/s.

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

v_f=Δv_r=v_elnR₁+v_elnR₂=v_eln(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/