Question

A spacecraft engine creates \(53.2 \mathrm{MN}\) of thrust with a propellant velocity of \(4.78 \mathrm{~km} / \mathrm{s}\). a) Find the rate \((d m / d t)\) at which the propellant is expelled. b) If the initial mass is \(2.12 \cdot 10^{6} \mathrm{~kg}\) and the final mass is \(7.04 \cdot 10^{4} \mathrm{~kg}\), find the final speed of the spacecraft (assume the initial speed is zero and any gravitational fields are small enough to be ignored). c) Find the average acceleration till burnout (the time at which the propellant is used up; assume the mass flow rate is constant until that time).

Short answer

Answer: The rate at which the propellant is expelled is \(11.13 \times 10^3\ \mathrm{kg/s}\). The final speed of the spacecraft is \(13.98 \times 10^3\ \mathrm{m/s}\). The average acceleration till burnout is \(75.94\ \mathrm{m/s^2}\).

Step-by-step solution

Find the rate at which the propellant is expelled

Recall that the thrust \(F_\mathrm{thrust}\) is related to the mass flow rate \(\frac{dm}{dt}\) and the propellant velocity \(v_\mathrm{propellant}\) by the following equation: \(F_\mathrm{thrust} = \frac{dm}{dt} v_\mathrm{propellant}\) We are given \(F_\mathrm{thrust} = 53.2\ \mathrm{MN}\) and \(v_\mathrm{propellant} = 4.78\ \mathrm{km/s}\). Convert these values to standard SI units: \(F_\mathrm{thrust} = 53.2 \times 10^6\ \mathrm{N}\) and \(v_\mathrm{propellant} = 4.78 \times 10^3\ \mathrm{m/s}\). Now, we can solve for \(\frac{dm}{dt}\): \(\frac{dm}{dt} = \frac{F_\mathrm{thrust}}{v_\mathrm{propellant}} = \frac{53.2 \times 10^6\ \mathrm{N}}{4.78 \times 10^3\ \mathrm{m/s}} = 11.13 \times 10^3\ \mathrm{kg/s}\).

Find the final speed of the spacecraft

To calculate the final speed of the spacecraft, we can apply the conservation of momentum: \(m_\mathrm{initial}v_\mathrm{initial} + \Delta m v_\mathrm{propellant} = m_\mathrm{final}v_\mathrm{final}\) Since the initial speed \(v_\mathrm{initial} = 0\), the equation simplifies to: \(\Delta m v_\mathrm{propellant} = m_\mathrm{final}v_\mathrm{final}\) Find the mass difference \(\Delta m\) and the final mass \(m_\mathrm{final}\): \(\Delta m = m_\mathrm{initial} - m_\mathrm{final} = (2.12 \times 10^6\ \mathrm{kg} - 7.04 \times 10^4\ \mathrm{kg}) = 2.0496 \times 10^6\ \mathrm{kg}\) \(m_\mathrm{final} = 7.04 \times 10^4\ \mathrm{kg}\) Now, solve for \(v_\mathrm{final}\): \(v_\mathrm{final} = \frac{\Delta m v_\mathrm{propellant}}{m_\mathrm{final}} = \frac{2.0496 \times 10^6\ \mathrm{kg} \cdot 4.78 \times 10^3\ \mathrm{m/s}}{7.04 \times 10^4\ \mathrm{kg}} = 13.98 \times 10^3\ \mathrm{m/s}\)

Find the average acceleration till burnout

Since the mass flow rate is constant until burnout, we can find the time \(t_\mathrm{burnout}\) required to use up all the propellant by dividing the mass change, \(\Delta m\), by the mass flow rate, \(\frac{dm}{dt}\): \(t_\mathrm{burnout} = \frac{\Delta m}{\frac{dm}{dt}} = \frac{2.0496 \times 10^6\ \mathrm{kg}}{11.13 \times 10^3\ \mathrm{kg/s}} = 184.17\ \mathrm{s}\) Now, we can calculate the average acceleration, \(a_\mathrm{avg}\), by dividing the final velocity, \(v_\mathrm{final}\), by the burnout time, \(t_\mathrm{burnout}\): \(a_\mathrm{avg} = \frac{v_\mathrm{final}}{t_\mathrm{burnout}} = \frac{13.98 \times 10^3\ \mathrm{m/s}}{184.17\ \mathrm{s}} = 75.94\ \mathrm{m/s^2}\) The results are: a) The rate at which the propellant is expelled is \(11.13 \times 10^3\ \mathrm{kg/s}\). b) The final speed of the spacecraft is \(13.98 \times 10^3\ \mathrm{m/s}\). c) The average acceleration till burnout is \(75.94\ \mathrm{m/s^2}\).

Key concepts

Thrust in Physics

Thrust is a critical concept when it comes to spacecraft propulsion physics. It is the force that moves a spacecraft through space and is produced by the engine's expulsion of propellant. According to Newton's third law, for every action, there is an equal and opposite reaction. In the context of spacecraft, this means that as the propellant is pushed out in one direction, the spacecraft is propelled in the opposite direction.

In the exercise provided, the spacecraft engine creates a thrust of 53.2 MN (meganewtons). By using the formula for thrust \( F_\mathrm{thrust} = \frac{dm}{dt} v_\mathrm{propellant} \), we see the direct relationship between thrust, the rate at which the propellant is expelled \( \frac{dm}{dt} \), and the velocity of the propellant \( v_\mathrm{propellant} \). Through the step-by-step solution, the mass flow rate is calculated, which conveys how much propellant mass is expelled per second to achieve the given thrust.

Conservation of Momentum

The conservation of momentum is one of the fundamental principles in physics and is particularly relevant in spacecraft propulsion. This principle states that the total momentum of a closed system remains constant if no external forces are acting on it.

During space missions, propulsion systems are designed based on this conservation law. In the absence of external forces, such as gravity in this exercise, the momentum of the spacecraft and its expelled propellant remains unchanged. The step-by-step solution demonstrates this as it approaches the final speed of the spacecraft. The initial momentum is essentially zero, as the spacecraft starts from rest. As the spacecraft ejects propellant at a known velocity, the change in the spacecraft's momentum equates to the momentum of the expelled propellant, leading to a calculation of the spacecraft's final speed using the equation \( \Delta m v_\mathrm{propellant} = m_\mathrm{final}v_\mathrm{final} \). This beautifully illustrates how the conservation of momentum governs the behavior of objects in space.

Average Acceleration

Understanding the average acceleration is essential for interpreting the motion of a spacecraft over time. Acceleration is the rate at which an object changes its velocity. In the case of a spacecraft, its acceleration may vary over time, especially as its mass changes due to the expulsion of propellant.

Average acceleration is calculated as the change in velocity divided by the time over which the change occurs. In the context of the exercise, although the spacecraft's mass decreases during the propulsion, the mass flow rate is constant, allowing for a straightforward computation of average acceleration up to the burnout moment when all the propellant is used. The formula used in the solution \( a_\mathrm{avg} = \frac{v_\mathrm{final}}{t_\mathrm{burnout}} \) provides a clear means to understand the spacecraft's average behavior over the course of its propulsion phase, culminating in an average acceleration till burnout.