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A rocket in space where gravity is negligible has a mass (including fuel) of \(5000 \mathrm{kg}\). If it is desired to give the rocket an average acceleration of \(15.0 \mathrm{m} \mathrm{s}^{-2}\) during the first second of firing the engine and the gases leave the rocket at a speed of \(1500 \mathrm{m} \mathrm{s}^{-1}\) (relative to the rocket), how much fuel must be burned in that second?

Short Answer

Expert verified
50 kg of fuel must be burned in that second.

Step by step solution

01

- Understand the Problem

We need to find out how much fuel needs to be burned in a rocket to achieve a specific acceleration over a period of one second, with given speeds and mass.
02

- Identify Known Values

The given values are: initial mass of rocket (including fuel) \(m_0 = 5000 \text{ kg}\), acceleration \(a = 15.0 \text{ m/s}^2\), and exhaust gas velocity \(v_e = 1500 \text{ m/s}\).
03

- Use the Rocket Equation

The rocket equation is \(\frac{d(mv)}{dt} = -v_e \frac{dm}{dt}\). The thrust force due to fuel burn is provided by the mass flow rate of the fuel and exhaust velocity.
04

- Relate Force with Newton's Second Law

The force can also be described using Newton's second law as \( F = ma\). Since we need an acceleration of \(15.0 \text{ m/s}^2\), the total force required is \( F = m_0 a = 5000 \text{ kg} \times 15.0 \text{ m/s}^2 = 75000 \text{ N} \).
05

- Find the Mass Flow Rate

Using the equation \( F = v_e \frac{dm}{dt}\), solve for mass flow rate \( \frac{dm}{dt} \): \( 75000 \text{ N} = 1500 \text{ m/s} \times \frac{dm}{dt} \). Solving for \( \frac{dm}{dt} \) gives: \( \frac{dm}{dt} = \frac{75000 \text{ N}}{1500 \text{ m/s}} = 50 \text{ kg/s} \).
06

- Calculate Total Fuel Burned

As the mass flow rate is \( 50 \text{ kg/s} \) and the time interval is 1 second, the total amount of fuel burned is: \( \text{fuel burned} = \frac{dm}{dt} \times \text{time} = 50 \text{ kg/s} \times 1 \text{ s} = 50 \text{ kg} \).

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Key Concepts

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

mass flow rate
Understanding mass flow rate is crucial when dealing with rockets. In simple terms, mass flow rate is the amount of mass being expelled from the rocket per unit of time. To achieve a desired acceleration, a rocket must burn a certain amount of fuel each second. This burning process is what propels the rocket forward.
The mass flow rate can be calculated if you know the force needed (which we get from Newton's second law) and the exhaust velocity. Like in our exercise, we found the required force to achieve a certain acceleration and then solved for the mass flow rate by dividing the force by the exhaust velocity. This gave us the mass flow rate, which tells us how efficiently the rocket needs to consume fuel to achieve the desired acceleration.
Newton's second law
Newton's second law is a key principle in physics and rocket science. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration: \[F = ma\] This fundamental law helps us understand how rockets move. When we apply Newton's second law to our rocket problem, we calculate the force required to achieve a given acceleration. For our rocket, to get an acceleration of 15.0 m/s², we multiply its mass (5000 kg) by the desired acceleration, resulting in a total required force of 75,000 N. This force is what the rocket engines must provide to generate the needed acceleration. It's a straightforward but powerful equation critical to understanding rocket dynamics.
exhaust velocity
Exhaust velocity is another vital concept in rocket technology. It refers to the speed at which the gases are expelled from the rocket's engine. Essentially, higher exhaust velocity means the rocket can achieve greater thrust, given the same fuel-burning rate.
In our problem, the exhaust velocity is given as 1500 m/s. This number helps determine the efficiency of the propulsion system. Higher exhaust velocities often lead to more efficient engines, as they can produce the same amount of thrust while consuming less fuel. By using the given exhaust velocity and the calculated force required, we derive the mass flow rate, sharing how much fuel the rocket must burn. This makes exhaust velocity a key player in determining how rockets achieve the desired acceleration.
acceleration
Acceleration is the rate at which an object's velocity changes. For rockets, managing acceleration is crucial since it affects how fast the rocket can reach its intended speed.
In our exercise, we needed the rocket to accelerate at 15.0 m/s². This specific acceleration rate guided our calculations. By knowing the acceleration, we could determine the necessary force through Newton's second law. Once we had the force, we could then use it to find out how fast the rocket should burn fuel.For rockets, consistent acceleration is vital, especially during initial launch phases. This controlled acceleration ensures the rocket navigates through different stages effectively. Hence, understanding and calculating acceleration is integral for successful rocket missions.

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Most popular questions from this chapter

(a) The acceleration of free fall at the surface of a planet is \(g\) and the radius of the planet is \(R .\) Deduce that the period of a satellite in a very low orbit is given by \(T=2 \pi \sqrt{\frac{R}{s}}\). (b) Given that \(g=4.5 \mathrm{ms}^{-2}\) and \(R=\) \(3.4 \times 10^{6} \mathrm{m},\) deduce that the orbital period of the low orbit is about 91 minutes. (c) A spacecraft in orbit around this planet has a period of 140 minutes. Deduce the height of the spacecraft from the surface of the planet.

Two cars of masses \(1200 \mathrm{kg}\) and \(1400 \mathrm{kg}\) collide head-on and stick to each other. The cars are coming at each other from opposite directions with speeds of \(8.0 \mathrm{m} \mathrm{s}^{-1}\) and \(6 \mathrm{m} \mathrm{s}^{-1}\), respectively. With what velocity does the wreck move away from the scene of the accident?

Describe the energy transformations taking place when a body of mass \(5.0 \mathrm{kg}\) : (a) falls from a height of 50 m without air resistance; (b) falls from a height of 50 m with constant speed; (c) is being pushed up an incline of \(30^{\circ}\) to the horizontal with constant speed.

A satellite is in a circular orbit around the earth. The satellite turns on its engines so that a small force is exerted on the satellite in the direction of the velocity. The engines are on for a very short time and the satellite now finds itself in a new circular orbit. (a) State and explain whether the new orbit is closer to or further away from the earth. (b) Hence explain why the speed of the satellite will decrease. (c) It appears that a force, acting in the direction of the velocity, has actually reduced the speed. How do you explain this observation?

The momentum of a ball increased by \(12.0 \mathrm{N} \mathrm{s}\) as a result of a force that acted on the ball for 2.00 s. What was the average force on the ball?

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