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Chapter 9: Problem 79

A rocket that is in deep space and initially at rest relative to an inertial reference frame has a mass of \(2.55 \times 10^{5} \mathrm{~kg}\), of which \(1.81 \times 10^{5} \mathrm{~kg}\) is fuel. The rocket engine is then fired for \(250 \mathrm{~s}\) while fuel is consumed at the rate of \(480 \mathrm{~kg} / \mathrm{s}\). The speed of the exhaust products relative to the rocket is \(3.27 \mathrm{~km} / \mathrm{s}\). (a) What is the rocket's thrust? After the \(250 \mathrm{~s}\) firing, what are (b) the mass and (c) the speed of the rocket?

Short Answer

Expert verified
(a) Thrust: 1.5696 million N, (b) Remaining mass: 135,000 kg, (c) Final speed: 2080.92 m/s.

Step by step solution

01

Calculate the Thrust

The thrust of a rocket can be calculated using the formula \( F = v_e \cdot \dot{m} \), where \( v_e \) is the exhaust velocity, and \( \dot{m} \) is the mass flow rate of the fuel. Here, \( v_e = 3.27 \times 10^3 \) m/s (converted from km/s to m/s) and \( \dot{m} = 480 \) kg/s. Substituting these values:\[F = (3.27 \times 10^3) \cdot 480 = 1.5696 \times 10^6 \text{ N}\]So, the thrust of the rocket is \( 1.5696 \times 10^6 \text{ N} \).
02

Calculate the Remaining Mass

To find the mass of the rocket after 250 s, we first determine how much fuel has been consumed. Since the fuel is consumed at \( 480 \) kg/s for \( 250 \) seconds, the total fuel used is:\[\text{Fuel consumed} = 480 \times 250 = 120,000 \text{ kg}.\]Initially, the rocket had a mass of \( 2.55 \times 10^5 \) kg with \( 1.81 \times 10^5 \) kg as fuel. After burning, the fuel remaining is:\[\text{Remaining fuel} = 1.81 \times 10^5 - 1.2 \times 10^5 = 61,000 \text{ kg}.\]Therefore, the remaining mass of the rocket is:\[\text{Remaining mass} = 2.55 \times 10^5 - 1.2 \times 10^5 = 1.35 \times 10^5 \text{ kg}.\]
03

Calculate the Rocket's Final Speed

The rocket equation, derived from Newton's second law, is given by:\[v_f = v_0 + v_e \ln\left(\frac{m_i}{m_f}\right),\]where \( v_f \) is the final speed, \( v_0 \) is the initial speed which is 0 (since the rocket starts from rest), \( m_i \) is the initial mass, and \( m_f \) is the final mass after fuel consumption. Substituting the known values:\[v_f = 0 + 3270 \ln\left(\frac{2.55 \times 10^5}{1.35 \times 10^5}\right) = 3270 \ln(1.8889)\]Calculating the natural logarithm and then multiplying by the exhaust velocity gives:\[v_f = 3270 \times 0.636 = 2080.92 \text{ m/s}.\]Thus, the rocket's final speed is approximately 2080.92 m/s.

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Key Concepts

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

Thrust Calculation
Thrust is a crucial concept in rocket propulsion, as it represents the force generated by the rocket engine to propel the rocket forward. The formula to calculate thrust is:
  • \( F = v_e \cdot \dot{m} \)
  • Here, \( F \) is the thrust, \( v_e \) is the exhaust velocity, and \( \dot{m} \) is the rate at which the fuel is consumed or the mass flow rate.
In our exercise, we convert the exhaust velocity, \( v_e \), to meters per second because standard units are essential for consistent calculations. With \( v_e = 3.27 \) km/s (or \( 3270 \) m/s), and \( \dot{m} = 480 \) kg/s, substituting into the formula gives:
  • \[ F = 3270 \times 480 = 1.5696 \times 10^6 \text{ N} \]
This result indicates that the rocket engine produces a thrust of over 1.5 million Newtons, providing the force needed to achieve acceleration.
Rocket Equation
The rocket equation, also known as the Tsiolkovsky rocket equation, is a fundamental equation for understanding rocket motion. It describes how the velocity of a rocket changes as it expels mass, specifically rocket fuel:
  • \( v_f = v_0 + v_e \ln\left(\frac{m_i}{m_f}\right) \)
  • \( v_f \) is the final velocity, \( v_0 \) is the initial velocity, \( m_i \) is the initial mass, and \( m_f \) is the final mass after the fuel is burned.
In this exercise, the rocket starts from rest, so \( v_0 = 0 \). The rocket equation helps calculate the change in velocity by considering the proportion of the rocket's mass that is lost when burning fuel. The natural logarithm \( \ln\left(\frac{m_i}{m_f}\right) \) reflects the exponential nature of velocity change during fuel consumption, offering insight into performance optimization in rocket designs.
Fuel Consumption
Understanding fuel consumption is key to determining the remaining mass of the rocket and its efficiency. In our exercise, the rocket consumes fuel at a constant rate, measured as the mass flow rate:
  • Fuel consumption rate = \( 480 \) kg/s
  • Total time of firing = \( 250 \) s
Multiplying these gives us the total amount of fuel consumed:
  • \( \, \text{Total fuel consumed} = 480 \times 250 = 120,000 \text{ kg} \)
Initially, the rocket's fuel mass is \( 1.81 \times 10^5 \) kg. Subtracting the consumed fuel, we find the remaining fuel mass. Knowing the initial total mass of \( 2.55 \times 10^5 \) kg, the remaining mass after fuel consumption is crucial for further calculations like determining the final speed with the rocket equation.
Final Speed of Rocket
The final speed of the rocket is determined by the rocket equation, which takes into account the change in mass as fuel is burned. Starting from rest with a total mass of \( 2.55 \times 10^5 \) kg, and after consuming \( 120,000 \) kg of fuel, the final mass is \( 1.35 \times 10^5 \) kg.To find the final speed, we use:
  • \( v_f = v_0 + v_e \ln\left(\frac{m_i}{m_f}\right) \)
Substitute the known values:
  • \( v_f = 0 + 3270 \ln\left(\frac{2.55 \times 10^5}{1.35 \times 10^5}\right) \)
  • \( v_f = 3270 \times 0.636 \approx 2080.92 \text{ m/s} \)
This calculation gives us the final velocity of about 2080.92 meters per second. This speed reflects how efficiently the rocket engine converts fuel mass into forward momentum, illustrating the principles of conservation of momentum and specific impulse in rocket science.

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Most popular questions from this chapter

Two titanium spheres approach each other head-on with the same speed and collide elastically. After the collision, one of the spheres, whose mass is \(300 \mathrm{~g}\), remains at rest. (a) What is the mass of the other sphere? (b) What is the speed of the two-sphere center of mass if the initial speed of each sphere is \(2.00 \mathrm{~m} / \mathrm{s}\) ?

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A projectile proton with a speed of \(500 \mathrm{~m} / \mathrm{s}\) collides elastically with a target proton initially at rest. The two protons then move along perpendicular paths, with the projectile path at \(60^{\circ}\) from the original direction. After the collision, what are the speeds of (a) the target proton and (b) the projectile proton?

Particle \(A\) and particle \(B\) are held together with a compressed spring between them. When they are released, the spring pushes them apart, and they then fly off in opposite directions, free of the spring. The mass of \(A\) is \(2.00\) times the mass of \(B\), and the energy stored in the spring was \(60 \mathrm{~J}\). Assume that the spring has negligible mass and that all its stored energy is transferred to the particles. Once that transfer is complete, what are the kinetic energies of (a) particle \(A\) and (b) particle \(\bar{B}\) ?

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