Question

A rocket at rest in space, where there is virtually no gravity, has a mass of \(2.55 \times 10^{5} \mathrm{~kg}\), of which \(1.81 \times 10^{5} \mathrm{~kg}\) is fuel. The engine consumes fuel at the rate of \(480 \mathrm{~kg} / \mathrm{s}\), and the exhaust speed is \(3.27 \mathrm{~km} / \mathrm{s}\). The engine is fired for \(250 \mathrm{~s}\). (a) Find the thrust of the rocket engine. \((b)\) What is the mass of the rocket after the engine burn? \((c)\) What is the final speed attained?

Short answer

The thrust of the rocket engine is approximately \(1.57 \times 10^{6} N\). The mass of the rocket after the engine burn is approximately \(1.35 \times 10^{5} kg\). The final speed of the rocket is approximately 2.9 km/s.

Step-by-step solution

Calculate the Thrust of the Rocket Engine

We can calculate the thrust of the rocket by considering the momentum of outflowing exhaust gases. The speed of the exhaust gases is given as 3.27 km/s (or 3270 m/s to convert to SI units). The thrust \(T\) (which is a force) is the rate of change of momentum and can be found using the formula \(T = dm/dt \times v\), where \(dm/dt\) is the rate of fuel consumption (480 kg/s), and \(v\) is the speed of exhaust gases. Plugging in the given values, we get, \(T = 480 \times 3270 = 1.57 \times 10^{6} N\).

Determine the Mass of the Rocket After Engine Burn

The mass of the rocket after the engine burn can be found by subtracting the mass of the fuel consumed by the rocket from its initial mass. The rate of fuel consumption is 480 kg/s and the engine is fired for 250 seconds. So the total fuel consumed is \(480 \times 250 = 12 \times 10^{4} kg\). The initial mass of the rocket is \(2.55 \times 10^{5} kg\). So the final mass of the rocket after the engine burn is \((2.55 - 1.2) \times 10^{5} kg = 1.35 \times 10^{5} kg\).

Calculate the Final Speed of the Rocket

Using the law of conservation of momentum, the final speed of the rocket can be calculated. We know that the initial momentum (when the rocket was at rest) is 0. If we consider the momentum of the rocket after consuming some amount of fuel, the total momentum would be the momentum of the rocket (mass times velocity) and the momentum of the outflowing exhaust gases (fuel consumed times exhaust speed). By the conservation of momentum, these should sum up to 0. This gives us the equation: \(v \times m_{final} - v_{exhaust} \times m_{fuel} = 0\). From this, we can express the speed of the rocket as \(v = v_{exhaust} \times (m_{fuel}/ m_{final})\). Substituting the known values, we get \(v = 3270 \times (1.2 \times 10^{5} / 1.35 \times 10^{5}) = 2.9 \times 10^{3} m/s\) or 2.9 km/s.

Key concepts

Thrust Calculation

The thrust of a rocket is a crucial concept in rocket propulsion, as it determines the force that moves the rocket forward. Thrust is generated by the reactive force of expelling exhaust gases at high speed in the opposite direction. According to Newton's Third Law, every action has an equal and opposite reaction. In simple terms, when the rocket expels fuel gases backward, a forward thrust is created, propelling the rocket forward.

Thrust can be calculated using the formula:
\[T = \frac{dm}{dt} \times v\]where \( \frac{dm}{dt} \) is the rate of fuel consumption, and \( v \) is the exhaust velocity. In our problem, the rate of fuel consumption is 480 kg/s, and the exhaust velocity is 3270 m/s. By substituting these values into the equation, we find:

The thrust, \( T = 480 \times 3270 = 1.57 \times 10^{6} \) N.

This enormous thrust is what allows rockets to overcome gravitational pull and enter space.

Conservation of Momentum

Conservation of momentum is a fundamental principle in rocket physics. It states that the total momentum of a closed system remains constant if no external forces act on it. For rockets, this system includes both the rocket and its exhaust gases.

When the rocket is at rest, its initial momentum is zero. As the rocket burns fuel and expels exhaust gases, the momentum of these gases is matched by an equal and opposite change in the momentum of the rocket. This principle is expressed in the equation:
\[v \times m_{\text{final}} - v_{\text{exhaust}} \times m_{\text{fuel}} = 0\]This means the momentum of the exhausted fuel (fuel mass times exhaust velocity) is equal and opposite to the momentum gained by the rocket. In our solution, the rocket increases its velocity to maintain the total momentum at zero. This results in a final velocity of 2.9 km/s. Understanding this concept helps in visualizing how rockets move in space, where they have no other means of propulsion.

Fuel Consumption Rate

The fuel consumption rate is a measure of how quickly a rocket consumes its fuel, which is critical in determining both thrust and how long a rocket can stay powered. It's defined as the amount of fuel (in kilograms) used per second by the rocket's engines.

In our described problem, the fuel consumption rate is given as 480 kg/s. This means that for every second the rocket's engine is fired, 480 kg of fuel is burned. Knowing this rate is essential for predicting how long the fuel will last and ultimately how the mission can proceed.

To calculate the total amount of fuel consumed during the burn time, multiply the fuel consumption rate by the burn duration:
\[\text{Total Fuel Consumed} = \frac{dm}{dt} \times t = 480 \times 250 = 12 \times 10^{4} \text{ kg}\]Understanding fuel consumption rate allows engineers to design efficient engines and properly manage the fuel resources onboard.

Exhaust Velocity

The exhaust velocity is a measure of how fast the expelled gases leave the rocket's engine. This is a critical factor in determining the rocket's thrust and efficiency. The higher the exhaust velocity, the more momentum the exhaust gases carry, and consequently, the more thrust is generated.

In our problem, the exhaust velocity is 3270 m/s. This value contributes to the calculation of the thrust force, which is crucial for propelling the rocket forward. The speed at which exhaust gases are ejected determines how effectively the rocket can push against space to propel itself.

Exhaust velocity is influenced by the design of the rocket motor and the type of propellant used. Efficient rockets strive to achieve higher exhaust velocities to maximize their thrust. This concept highlights the importance of engineering precision in rocket design and the role of high-velocity nozzle exhaust in achieving desired missions.