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Suppose the first stage of a two stage rocket has total mass \(12,000 \mathrm{kg}\) , of which 9000 \(\mathrm{kg}\) is fuel. The total mass of the second stage is \(1000 \mathrm{kg},\) of which 700 \(\mathrm{kg}\) is fuel. Assume that the relative speed \(\boldsymbol{U}_{\mathbf{e x}}\) of ejected material is constant, and ignore any effect of gravity. (The effect of gravity is small during the firing period if the rate of fuel consumption is large) (a) Suppose the entire fuel supply carried by the two-stage rocket is utilized in a single-stage rocket with the same total mass of \(13,000 \mathrm{kg}\) . In terms of \(v_{e x}\) what is the speed of the rocket, starting from rest, when its fuel is exhausted? (b) For the two-stage rocket, what is the speed when the fuel of the first stage is exhausted if the first stage carries the second stage with it to this point? This speed then becomes the initial speed of the second stage. At this point, the second stage separates from the first stage.(c) What is the final speed of the second stage? (d) What value of \(v_{\mathrm{ex}}\) is required to give the second stage of the rocket a speed of 7.00 \(\mathrm{km} / \mathrm{s}\) ?

Step by step solution

01

Calculate Total Exhausted Fuel

The total fuel mass is the sum of the fuel in both stages: \(9000 \text{ kg} + 700 \text{ kg} = 9700 \text{ kg}\).

02

Single-stage Rocket Speed Calculation (a)

The initial total mass is \(13,000 \text{ kg}\) and final mass once the fuel is burned is \(13,000 \text{ kg} - 9700 \text{ kg} = 3300 \text{ kg}\). Using the rocket equation \(v = v_{ex} \ln\left(\frac{m_0}{m_1}\right)\), where \(m_0\) is the initial mass and \(m_1\) is the final mass, the speed \(v\) is \( v = v_{ex} \ln\left(\frac{13000}{3300}\right)\).

03

Two-stage Rocket First Stage Speed (b)

The first stage has an initial mass of \(12,000 \text{ kg}\) and burns \(9000 \text{ kg}\) of fuel. The final mass is \(12,000 \text{ kg} - 9000 \text{ kg} = 3000 \text{ kg}\) plus the \(1000 \text{ kg}\) of the second stage, giving \(4000 \text{ kg}\). Using the rocket equation again, \(v_1 = v_{ex} \ln\left(\frac{12000}{4000}\right)\).

04

Determine Initial Speed of Second Stage

At separation, the speed of the second stage becomes the speed \(v_1\) calculated in Step 3: \(v_1 = v_{ex} \ln\left(\frac{12000}{4000}\right)\).

05

Final Speed of Second Stage (c)

The second stage exhausts \(700\ \text{kg}\) of fuel, bringing its mass from \(1000 \text{ kg}\) to \(300\ \text{kg}\). Thus, using the rocket equation from its initial speed \(v_1\): \(v_2 = v_1 + v_{ex} \ln\left(\frac{1000}{300}\right)\).

06

Calculate Required \(v_{ex}\) for Specific Final Speed (d)

Set \(v_2\) equal to 7000 m/s (\(7.00 \text{ km/s}\)) and solve: \(7000 = v_{ex} \left(\ln\left(\frac{12000}{4000}\right) + \ln\left(\frac{1000}{300}\right)\right)\). Simplify to find \(v_{ex}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Two-stage rocket

A two-stage rocket operates by utilizing separate segments, or "stages," each with its own set of engines and fuel. When the first stage's fuel depletes, it detaches, allowing the second stage to continue the journey with reduced weight, which enhances efficiency and velocity. This concept improves fuel efficiency and achieves higher velocities compared to single-stage rockets. The first stage encompasses the entire payload and the second stage, and once jettisoned, reduces the mass dramatically.
In our exercise, the total starting mass of the two-stage rocket is 13,000 kg, with 9,000 kg of fuel allocated for the first stage and 700 kg for the second stage. After the first stage is expended, only the remaining structure and second stage proceed, minimizing the inertia that must be propelled in the following phase.

Rocket equation

The rocket equation, also known as the Tsiolkovsky rocket equation, is fundamental in calculating velocity changes in rockets. This equation showcases the efficiency at which a rocket expends its fuel to achieve acceleration. The equation is expressed as:\[ v = v_{ex} \ln\left(\frac{m_0}{m_1}\right) \]where:

• \( v \) is the final velocity of the rocket
  
• \( v_{ex} \) is the exhaust velocity
  
• \( m_0 \) is the initial total mass
  
• \( m_1 \) is the final total mass after fuel is burned
  

In essence, it links mass ratios and exhaust velocity to the rocket's speed. Within this framework, when the mass of a rocket decreases due to burned fuel, the velocity increases, assuming a constant exhaust velocity. Let's apply this to our two-stage scenario:
For instance, in calculating the speed of the rocket after the first stage is burned, the initial mass includes the entire fuel and payload—ending with the expended first stage but retaining the second stage as new payload.

Exhaust velocity

The exhaust velocity \( v_{ex} \) refers to the speed at which gases exit the rocket's propulsion system. It is a critical factor affecting the overall thrust efficiency. The higher the exhaust velocity, the greater the escape speed, and hence, more thrust is produced. This component is crucial when considering the kinetic energy imparted from the fuel to the rocket.
Higher exhaust velocities demand advanced engineering in designing engines and fuel types, while higher values can significantly improve the rocket's ability to overcome gravitational forces and atmospheric drag. In our exercise, the task revolved around finding an exhaust velocity that enables the rocket stages to achieve specific velocities, such as the final speed of 7.00 km/s for the second stage.

Fuel consumption rate

Fuel consumption rate in rocket propulsion outlines the rate at which a rocket engine burns its fuel load. During launches, this rate is typically extremely high, as a significant portion of the rocket's mass is the propellant itself. A rapid consumption rate leads to a significant decrease in mass, thereby enhancing acceleration according to the rocket equation.
The faster the fuel is consumed, the quicker the mass drops, elevating the rocket's velocity. In our case study, ignoring gravity allows analysis focused purely on fuel expulsion effects, given the brief fuel-burning duration. The main challenge and strategy lie in balancing the fuel rate to match the stages and to achieve desired velocity increases at each stage. Each stage is carefully planned with specific consumption rates to maximize efficiency.