Problem 113 - University Physics with Modern Physics

Question

A Multistage Rocket. 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 \(v_{\text { cx }}\) 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 \(v_{\mathrm{cx}},\) 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

Answer

The rocket's speed after fuel exhaustion in single-stage is \( v_{\text{ex}} \ln (\frac{13,000}{3,300}) \); in two-stage, the final speed is 7 km/s if \( v_{\text{ex}} \) satisfies specific conditions.

Step 1: Calculate Total Initial Mass

The total initial mass when combining both the stages into a single stage is given by:- 12,000 kg for the first stage- 1,000 kg for the second stageThus, the total combined initial mass is:\[ M_i = 12,000 + 1,000 = 13,000 \text{ kg} \]

Step 2: Calculate Total Fuel Mass

The total fuel mass in the combined single stage is:- 9,000 kg from the first stage- 700 kg from the second stageThus, the total fuel mass in the single-stage rocket is:\[ M_f = 9,000 + 700 = 9,700 \text{ kg} \]

Step 3: Calculate Final Mass After Fuel Consumption

We need to find out the final mass after the fuel is exhausted:\[ M_f = 13,000 - 9,700 = 3,300 \text{ kg} \]

Step 4: Compute Final Speed in Single-Stage Scenario

Using the rocket equation: \[ v = v_{\text{ex}} \ln \left( \frac{M_i}{M_f} \right) \]Plug in the known values:- Initial mass \( M_i = 13,000 \) kg- Final mass \( M_f = 3,300 \) kgThe speed \( v \) when the fuel is exhausted is:\[ v = v_{\text{ex}} \ln \left( \frac{13,000}{3,300} \right) \]

Step 5: Calculate Speed After First Stage Fuel Exhaustion

For two-stage rocket:The initial mass of the first stage carrying the second is \( 13,000 \) kg.After burning 9,000 kg of fuel from the first stage:- Final mass is \( 13,000 - 9,000 = 4,000 \) kg.The speed at which all the fuel in the first stage is exhausted, using the rocket equation again:\[ v = v_{\text{ex}} \ln \left( \frac{13,000}{4,000} \right) \]

Step 6: Determine Second Stage Final Speed

The initial speed of the second stage is the final speed from Step 5.Mass at the start of the second stage is 1,000 kg with 700 kg fuel, leaving 300 kg at the end.Final speed of the second stage is computed using:\[ v_{\text{final 2nd stage}} = v_{\text{initial 2nd stage}} + v_{\text{ex}} \ln \left( \frac{1,000}{300} \right) \]

Step 7: Solve for Required Exhaust Velocity

For the second stage to have a final speed of 7 km/s:\[ 7,000 = v + v_{\text{ex}} \ln \left( \frac{1,000}{300} \right) \]Solve for \( v_{\text{ex}} \) in terms of previously calculated values.Combine speeds from stages for the final result.

Key Concepts

Rocket Equation

The Rocket Equation, also known as the Tsiolkovsky rocket equation, is fundamental for understanding rocket dynamics. It relates the velocity of a rocket to the properties of the rocket and the fuel it consumes. The equation is expressed as:
\[ v = v_{\text{ex}} \ln \left( \frac{M_i}{M_f} \right) \]where:

• \( v \) is the final velocity of the rocket,
• \( v_{\text{ex}} \) is the exhaust velocity of the ejected material,
• \( M_i \) is the initial total mass (rocket plus fuel), and
• \( M_f \) is the final mass (rocket without fuel).

This equation illustrates how the velocity of the rocket depends on the mass of the consumed fuel and the speed at which the rocket expels it.
Understanding this helps predict how fast the rocket will travel when the fuel is depleted. It emphasizes the importance of the mass ratio (\( \frac{M_i}{M_f} \)) and the exhaust velocity in achieving high speeds.

Fuel Consumption Calculation

Fuel consumption is a critical aspect of rocket physics. It refers to the amount of fuel a rocket burns to gain velocity. In this problem, it first requires the total initial mass of both rocket stages combined, which was \( 13,000 \text{ kg} \).
The fuel mass is the main factor reducing the rocket's mass over time as it is consumed. For the entire rocket in this case:

• Stage One Fuel: \( 9,000 \text{ kg} \)
• Stage Two Fuel: \( 700 \text{ kg} \)
• Total Fuel Mass: \( 9,700 \text{ kg} \)

The formula \[ M_f = M_i - \text{fuel burned} \]calculates the remaining mass (\( M_f \)) after fuel consumption. Understanding fuel consumption helps determine how mass and velocity change over time, particularly crucial during the rocket's operation cycle. It fundamentally supports the use of the Rocket Equation in predicting final velocities.

Exhaust Velocity

Exhaust velocity (\( v_{\text{ex}} \)) is a principal parameter in rocket design, representing the speed at which exhaust gases are expelled from the rocket engine. It directly affects the rocket's change in velocity. A higher exhaust velocity is crucial for more efficient fuel usage.
In our scenario, this velocity remains constant and is essential in calculating the rocket's final speeds. Using the rocket equation:\[ v = v_{\text{ex}} \ln \left( \frac{M_i}{M_f} \right) \]we see that higher \( v_{\text{ex}} \) leads to a greater velocity change for a given fuel mass ratio.By comparing scenarios (single-stage and two-stage), exhaust velocity helps us understand how best to utilize fuel: reducing the amount needed for the desired speed or increasing the final speed possible with a given fuel mass. The required exhaust velocity to achieve a specific speed is an essential calculation when designing a rocket’s propulsion system.

Two-Stage Rocket Dynamics

Two-stage rockets separate into parts during flight to improve their efficiency and performance. The first stage carries the rocket a significant distance, then the second stage takes over until the fuel is depleted or reaches its destination.
For our example, the first stage carries both stages initially, consuming its 9,000 kg of fuel, leaving a remaining mass of 4,000 kg:\[ v = v_{\text{ex}} \ln \left( \frac{13,000}{4,000} \right) \]The separation of stages means less mass for the second stage to accelerate, which is beneficial for reaching higher velocities with the remaining fuel. Next, the final speed of the second stage is calculated using its initial speed plus the additional speed gained by burning its own fuel.
Designing rockets using multiple stages is a strategic way to reduce weight as each part has finished its role, adapting to the changing mass by recalculating velocities specific to each stage. This staging allows rockets to achieve speeds and altitudes that a single-stage configuration could not match with the same fuel.