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A two-stage rocket is fired vertically up and is above the atmosphere when the first stage burns out and the second stage separates and ignites. The second stage carries \(1200 \mathrm{kg}\) of fuel and has an empty mass of \(200 \mathrm{kg}\). Upon ignition the second stage burns fuel at the rate of \(5.2 \mathrm{kg} / \mathrm{s}\) and has a constant exhaust velocity of \(3000 \mathrm{m} / \mathrm{s}\) relative to its nozzle. Determine the acceleration of the second stage 60 seconds after ignition and find the maximum acceleration and the time \(t\) after ignition at which it occurs. Neglect the variation of \(g\) and take it to be \(8.70 \mathrm{m} / \mathrm{s}^{2}\) for the range of altitude averaging about \(400 \mathrm{km}\).

Short Answer

Expert verified
Acceleration at 60s: 14.34 m/s²; Maximum acceleration: 68.3 m/s² at t ≈ 230s.

Step by step solution

01

Calculate Remaining Mass 60 Seconds After Ignition

The second stage of the rocket starts with a mass of 1400 kg (sum of 1200 kg fuel and 200 kg empty mass). After 60 seconds, the fuel burned is \(5.2 \mathrm{kg/s} \times 60 \mathrm{s} = 312 \mathrm{kg}\). Therefore, the remaining mass of the rocket is:\[M = 1400 \mathrm{kg} - 312 \mathrm{kg} = 1088 \mathrm{kg}\]
02

Use Rocket Equation to Find Velocity Change After 60 Seconds

The rocket equation (Tsiolkovsky's equation) is given as:\[v_e = v \ln\left(\frac{M_i}{M_f}\right)\]where \(v_e\) is the exhaust velocity, \(M_i\) is the initial total mass (1400 kg), and \(M_f\) is the final mass after burning 312 kg of fuel (1088 kg). Substituting the values, we have:\[v = 3000 \ln\left(\frac{1400}{1088}\right)\]
03

Calculate Thrust of the Rocket

The thrust \(F\) can be calculated using the following formula:\[F = v_e \cdot \frac{dM}{dt}\]where \(v_e = 3000\, \mathrm{m/s}\) and \(\frac{dM}{dt} = 5.2\, \mathrm{kg/s}\). Thus:\[F = 3000 \times 5.2 = 15600\, \mathrm{N}\]
04

Compute Acceleration 60 Seconds After Ignition

The net acceleration \(a\) at 60 seconds can be found using Newton's Second Law, \(F = ma\): \[a = \frac{F}{M} - g\]Substituting in the values gives:\[a = \frac{15600}{1088} - 8.7 = 14.34\, \mathrm{m/s^2}\]
05

Determine Maximum Acceleration and Time Occurrence

The acceleration is highest when the mass is lowest (meaning the fuel is almost burnt out). Use Newton's Second Law:\[a(t) = \frac{F}{200 + 1200 - 5.2t} - 8.7\]Set \(\frac{da}{dt} = 0\) and solve for \(t\). After solving the equation, one finds the time \(t\) when maximum acceleration occurs is approximately when the fuel is nearly depleted:\[t \approx \frac{1200}{5.2} \approx 230\, \mathrm{s}\]Substitute \(t = 230\) back to determine acceleration:\[a_{max} = \frac{15600}{200 + 1200 - 1200} - 8.7 = 68.3\, \mathrm{m/s^2}\]

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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 is a type of rocket that has two distinct propulsion phases. In the first stage, the rocket expends a certain amount of fuel to propel itself into a higher altitude, usually getting out of the densest parts of Earth's atmosphere. After the first stage has used up its fuel, it detaches, allowing the second stage to ignite. This separation allows the two-stage rocket to be more efficient, as the rocket becomes lighter once the first stage is jettisoned.

In the exercise, the focus is on the second stage which has an empty mass of 200 kg and carries 1200 kg of fuel. The second stage becomes the new 'rocket' once the first stage is off, and is crucial for reaching the final velocity or altitude. The two-stage design is vital for reducing the weight the rocket has to carry, thereby increasing the overall efficiency and effectiveness of the launch.
Tsiolkovsky's Rocket Equation
Tsiolkovsky's rocket equation is central to understanding how a rocket changes its velocity over time. It is given by the formula:

\[ v = v_e \ln\left(\frac{M_i}{M_f}\right) \]

where:
  • \(v\) is the final velocity change,
  • \(v_e\) is the exhaust velocity,
  • \(M_i\) is the initial total mass,
  • \(M_f\) is the final mass after burning the fuel.
This equation is important because it shows that the velocity change of a rocket is logarithmically related to the mass ratio, \(\frac{M_i}{M_f}\).

In simpler terms, as more fuel is used up, the mass of the fuel becomes less compared to the mass of the rocket itself, meaning the rocket speeds up considerably as it burns fuel. This helps in calculating the velocity changes at different points of the rocket’s journey.
Acceleration Calculation
In rocketry, calculating acceleration is key to knowing how the rocket will perform during flight. Using Newton’s second law, you can determine the acceleration by understanding the forces acting on the rocket.

The net force acting on the rocket is the thrust minus gravitational force. The thrust is calculated as:

\[ F = v_e \times \frac{dM}{dt} \]

Assuming constant exhaust velocity and fuel burn rate, this offers a consistent thrust force.
  • Once thrust is known, acceleration \(a\) can be calculated with:
  • \[ a = \frac{F}{M} - g \]
where \(M\) is the mass at any point and \(g\) is gravitational acceleration.

This means the acceleration is maximum when the fuel is nearly burnt out, as the rocket becomes lighter and the constant thrust now acts on a smaller mass, increasing acceleration much more.
Newton's Second Law
Newton's Second Law is fundamental in understanding how forces affect the motion of objects. It is often summarized as \( F = ma \), where:
  • \(F\) is the total force in Newtons,
  • \(m\) is the mass in kilograms,
  • \(a\) is the acceleration in \(\mathrm{m/s^2}\).
This means that the acceleration of an object is dependent on the net force acting upon it, and inversely proportional to its mass.

In the context of our rocket, Newton's second law is applied to find the net acceleration at any given time. The law shows that with a lighter rocket (due to burned fuel), the same amount of force results in greater acceleration. Adjusting for gravity, this provides the actual net acceleration used to predict the rocket's flight path.

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Most popular questions from this chapter

The upper end of the open-link chain of length \(L\) and mass \(\rho\) per unit length is released from rest with the lower end just touching the platform of the scale. Determine the expression for the force \(F\) read on the scale as a function of the distance \(x\) through which the upper end has fallen. (Comment: The chain acquires a free-fall velocity of \(\sqrt{2 g x}\) because the links on the scale exert no force on those above, which are still falling freely. Work the problem in two ways: first, by evaluating the time rate of change of momentum for the entire chain and second, by considering the force \(F\) to be composed of the weight of the links at rest on the scale plus the force necessary to divert an equivalent stream of fluid.)

A small rocket-propelled vehicle weighs \(125 \mathrm{lb}\), in cluding 20 lb of fuel. Fuel is burned at the constant rate of 2 lb/sec with an exhaust velocity relative to the nozzle of \(400 \mathrm{ft} / \mathrm{sec}\). Upon ignition the vehicle is released from rest on the \(10^{\circ}\) incline. Calculate the maximum velocity \(v\) reached by the vehicle. Neglect all friction.

The rocket shown is designed to test the operation of a new guidance system. When it has reached a certain altitude beyond the effective influence of the earth's atmosphere, its mass has decreased to \(2.80 \mathrm{Mg},\) and its trajectory is \(30^{\circ}\) from the vertical. Rocket fuel is being consumed at the rate of \(120 \mathrm{kg} / \mathrm{s}\) with an exhaust velocity of \(640 \mathrm{m} / \mathrm{s}\) relative to the nozzle. Gravitational acceleration is \(9.34 \mathrm{m} / \mathrm{s}^{2}\) at its altitude. Calculate the \(n\) - and \(t\) -components of the acceleration of the rocket.

The magnetometer boom for a spacecraft consists of a large number of triangular-shaped units which spring into their deployed configuration upon release from the canister in which they were folded and packed prior to release. Write an expression for the force \(F\) which the base of the canister must exert on the boom during its deployment in terms of the increasing length \(x\) and its time derivatives. The mass of the boom per unit of deployed length is \(\rho .\) Treat the supporting base on the spacecraft as a fixed platform and assume that the deployment takes place outside of any gravitational field. Neglect the dimension \(b\) compared with \(x\).

A jet of fresh water under pressure issues from the \(3 / 4\) -in.-diameter fixed nozzle with a velocity \(v=120 \mathrm{ft} / \mathrm{sec}\) and is diverted into the two equal streams. Neglect any energy loss in the streams and compute the force \(F\) required to hold the vane in place.

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