Question

A rocket consists of a payload of \(4390.0 \mathrm{~kg}\) and \(1.761 \cdot 10^{5} \mathrm{~kg}\) of fuel. Assume that the rocket starts from rest in outer space, accelerates to its final velocity, and then begins its trip. What is the speed at which the propellant must be expelled to make the trip from the Earth to the Moon, a distance of \(3.82 \cdot 10^{5} \mathrm{~km},\) in \(7.00 \mathrm{~h}\) ?

Short answer

The speed at which the propellant must be expelled is approximately \(2.90\times10^{4}\mathrm{~m/s}\).

Step-by-step solution

Calculate the required final velocity of the rocket

Using the formula for velocity, \(v=d/t\), where \(v\) is the velocity, \(d\) is the distance, and \(t\) is the time taken, we can find the required final velocity of the rocket: \(v = \frac{3.82\cdot 10^5 \mathrm{~km}}{7.00\mathrm{~h}}\) To match the units, we convert kilometers to meters and hours to seconds: \(v = \frac{3.82\cdot 10^5 \mathrm{~km} \cdot 10^3\mathrm{~m/km}}{7.00\mathrm{~h} \cdot 3600\mathrm{~s/h}}\) \(v \approx 1.57\times10^4\mathrm{~m/s}\)

Apply the Rocket Equation

The Rocket Equation is given by: \(v_f = v_e \cdot \ln\left(\frac{m_{initial}}{m_{final}}\right)\) where \(v_f\) is the final velocity of the rocket, \(v_e\) is the effective exhaust velocity of the propellant, and \(m_{initial}\) and \(m_{final}\) are the initial and final total masses of the rocket, respectively. Since the payload mass is 4390 kg and the fuel mass is \(1.761\times10^5\) kg, the initial total mass \(m_{initial} = 4390 + 1.761\times10^5\) kg, and the final total mass \(m_{final} = 4390\) kg (when all fuel is used).

Rearrange the Rocket Equation to find the effective exhaust velocity

We need to calculate the \(v_e\), so we'll rearrange the Rocket Equation: \(v_e = \frac{v_f}{\ln\left(\frac{m_{initial}}{m_{final}}\right)}\) Now, substitute the calculated final velocity and the initial and final total masses into the equation: \(v_e = \frac{1.57\times10^4\mathrm{~m/s}}{\ln\left(\frac{4390 + 1.761\times10^5}{4390}\right)}\) \(v_e \approx 2.90\times10^{4}\mathrm{~m/s}\)

Interpret the result

The speed at which the propellant must be expelled for the rocket to make the trip from Earth to the Moon in 7 hours is approximately \(2.90\times10^{4}\mathrm{~m/s}\).

Key concepts

Understanding Rocket Propulsion

Rocket propulsion is a fundamental concept in aerospace engineering and physics that allows rockets to move through space. The science of rocket propulsion is based on Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. In the context of a rocket, the action is the expulsion of fuel, or propellant, out of the back of the rocket at high speed, and the reaction is the forward thrust that propels the rocket in the opposite direction.

When a rocket engine combusts its fuel, it provides the energy needed to accelerate the exhaust gases to extreme speeds, ejecting them and creating the thrust needed to propel the spacecraft. The effectiveness of this process is highly dependent on the exhaust velocity, which we'll explore in the next section. Without the principle of rocket propulsion, space exploration as we know it would be impossible.

Examining Exhaust Velocity

Exhaust velocity, often denoted as \(v_e\), plays a crucial role in rocket propulsion. It is the speed at which the propellant is ejected from the rocket's engine in the opposite direction of the desired travel. A fundamental relation in understanding how rockets accelerate is that the higher the exhaust velocity, the more efficient the rocket is in converting fuel mass into momentum.

A key aspect of exhaust velocity is that it relates directly to the specific impulse of an engine, which is a measure of how effectively a rocket uses its propellant. To make a journey from Earth to the Moon, as in the exercise, having a high exhaust velocity is crucial to achieve the necessary final velocity without requiring an impractical amount of fuel.

Conservation of Momentum in Rockets

The principle of conservation of momentum is deeply entrenched in the study of rocket motion. Momentum, the product of mass and velocity, must be conserved in a closed system according to the laws of physics. For rockets, this closed system includes the rocket and the expelled propellant gas.

During operation, when a rocket fires its engines and expels propellant out at high speeds (exhaust velocity), the rocket gains momentum in the opposite direction. As the rocket's mass decreases with the loss of fuel, a corresponding increase in the rocket's velocity occurs to ensure that the total momentum of the system remains constant. This is a perfect illustration of the conservation of momentum in action, which is the guiding force for rocket propulsion through space.

Logarithmic Functions in Rocket Equations

The rocket equation introduces a natural logarithmic function, often represented by the symbol \(\ln\), which arises from the integral nature of the conservation of momentum during the burning of the rocket's fuel. The logarithmic function appears when relating the change in velocity of the rocket to the mass ratio of the rocket before and after the propellant is expelled.

The natural logarithm of the mass ratio translates to the multiplicative effect of fuel mass on the rocket's velocity. In the provided exercise, calculating the final velocity of the rocket heavily relies on this logarithmic function, evidenced in the rearranged rocket equation to solve for exhaust velocity. This emphasizes not only the importance of logarithmic functions in analyzing rocket motion but also their prevalence in many physics equations associated with exponentially varying processes.