Question

a. To understand why rockets often have multiple stages, first consider a single-stage rocket with an empty mass of 200 kg, 800 kg of fuel, and a 2000 m/s exhaust speed. If fired in deep space, what is the rocket’s maximum speed?

b. Now divide the rocket into two stages, each with an empty mass of 100 kg, 400 kg of fuel, and a 2000 m/s exhaust speed. The first stage is released after it runs out of fuel. What is the top speed of the second stage? You’ll need to consider how the equation for v_max should be altered when a rocket is not starting from rest.

Short answer

a) v_max is equal to 3218.88 m/s

b)v_max is equal to 4240.53 m/s

Step-by-step solution

Step 1. Given information is :Part a)Empty mass of Rocket = 200 kgMass of the fuel = 800 kgExhaust Speed = 2000 m/sPart b)Empty mass of Rocket = 100 kgMass of the fuel = 400 kgExhaust Speed = 2000 m/s

We need to calculate,

Part a) Maximum speed of the rocket

Part b) Top speed of the second stage rocket

Step 2. Part a)Finding the vmax for first stage rocket

Using formula,

$$\begin{gathered} v_{max}=v_{ex}\ln \left(\frac{m_R+m_{FO}}{m_R}\right) \\ {v}_{max}=(2000m/s)\ln \left(\frac{200kg+800kg}{200kg}\right) \\ {v}_{max}=(2000m/s)\ln \left(\frac{1000kg}{200kg}\right) \\ {v}_{max}=3218.88m/s \end{gathered}$$

Step 3. Part b)Calculating vmax for multistage that is two stage rocket.

Using formula,

$$\begin{gathered} v_{max}=v_{ex}\ln \left(\frac{m_R+m_{FO}}{m_R}\right) \\ {v}_{max}=v_{ex}\ln \left(\frac{m{\text{'}}_R+m{\text{'}}_{FO}}{m{\text{'}}_R+m_{FO}}\right)+v_{ex}\ln \left(\frac{m_R+m_{FO}}{m_R}\right) \\ \left[v_{ex}\ln \left(\frac{m{\text{'}}_R+m{\text{'}}_{FO}}{m{\text{'}}_R+m_{FO}}\right)isforfirststage \\ {v}_{ex}\ln \left(\frac{m_R+m_{FO}}{m_R}\right)isforsecondstage\right] \\ {v}_{max}=v_{ex}\left[\ln \left(\frac{m{\text{'}}_R+m{\text{'}}_{FO}}{m{\text{'}}_R+m_{FO}}\right)+\ln \left(\frac{m_R+m_{FO}}{m_R}\right)\right] \\ {v}_{max}=(2000m/s)\left[\ln \left(\frac{200kg+800kg}{200kg+400kg}\right)+\ln \left(\frac{100kg+400kg}{100kg}\right)\right] \\ {v}_{max}=(2000m/s)\left[\ln \left(\frac{1000kg}{600kg}\right)+\ln \left(\frac{500kg}{100kg}\right)\right] \\ {v}_{max}=4240.53m/s \end{gathered}$$