Question

The change in velocity, \(\Delta v,\) in kilometres per second, of a rocket with an exhaust velocity of \(3.1 \mathrm{km} / \mathrm{s}\) can be found using the Tsiolkovsky rocket equation \(\Delta v=\frac{3.1}{0.434}\left(\log m_{0}-\log m_{p}\right),\) where \(m_{0}\) is the initial total mass and \(m_{f}\) is the final total mass, in kilograms, after a fuel burn. Find the change in the velocity of the rocket if the mass ratio, \(\frac{m_{0}}{m_{f}},\) is 1.06 Answer to the nearest hundredth of a kilometre per second.

Short answer

The change in velocity is approximately 0.18 km/s.

Step-by-step solution

Understand the variables and constants

Given exhaust velocity is 3.1 km/s; initial total mass is denoted by \(m_0\) and final total mass is \(m_f\), the Tsiolkovsky rocket equation is \(\Delta v = \frac{3.1}{0.434} (\log m_0 - \log m_f)\). The mass ratio \(\frac{m_{0}}{m_{f}}\) is given as 1.06.

Relate the given mass ratio to logarithms

Since the mass ratio \(\frac{m_{0}}{m_{f}}\) is given as 1.06, take the logarithm of both sides: \(\log\left(\frac{m_{0}}{m_{f}}\right) = \log(1.06)\). By the property of logarithms, \(\log\left(\frac{m_{0}}{m_{f}}\right) = \log m_0 - \log m_f\).

Calculate the logarithm of the mass ratio

Use a calculator to find \(\log(1.06)\). Approximating, \(\log(1.06) \approx 0.0253\).

Substitute the values into the Tsiolkovsky rocket equation

Substitute the value of \(\log(1.06) \approx 0.0253\) into the Tsiolkovsky rocket equation: \(\Delta v = \frac{3.1}{0.434} \times 0.0253\).

Calculate the change in velocity

Perform the final calculation: \(\Delta v \approx \frac{3.1 \times 0.0253}{0.434} \approx \frac{0.078\ (rounded to 3 decimal places)}{0.434} \approx 0.18\ (rounded to 2 decimal places)\).

Key concepts

change in velocity

The change in velocity \(\Delta v\) is a crucial part of the Tsiolkovsky rocket equation. It tells us how much a rocket's speed changes due to fuel being burned and expelled as exhaust. In simpler terms, \(\Delta v\) indicates how fast the rocket becomes after burning a certain amount of fuel. For practical applications, the change in velocity is measured in kilometers per second (km/s) to match the exhaust velocity units. Understanding \(\Delta v\) helps engineers design more efficient rockets by optimizing fuel usage to achieve the desired speed. When you're working on problems involving \(\Delta v\), always ensure your units are consistent and double-check your calculations to avoid errors. Remember, small mistakes in understanding this concept can lead to significant miscalculations in real-world scenarios.

logarithms

Logarithms simplify complex multiplications and divisions by transforming them into additions and subtractions. In the Tsiolkovsky rocket equation, logarithms are used to handle the mass ratio calculations. When given a mass ratio \(\frac{m_0}{m_f}\), we can take the common logarithm (base 10) of both sides to find the difference between \(\log m_0\) and \(\log m_f\):
\(\log\frac{m_{0}}{m_{f}} = \log m_0 - \log m_f\).
In our example, with a mass ratio of 1.06, calculating \(\log(1.06)\) gives approximately 0.0253. This simplifies our work by eliminating the need for complex division directly and making it easier to substitute and calculate values in the rocket equation. Becoming comfortable with logarithms is essential for handling exponential growth and decay problems in scientific studies.

mass ratio

The mass ratio \(\frac{m_{0}}{m_{f}}\) is a vital concept in rocketry. It compares the rocket's initial mass (including fuel) \(m_0\) with its final mass \(m_f\) after the fuel is burnt. This ratio tells us how much mass has been burnt as fuel and subsequently expelled. A higher mass ratio indicates that more mass has been converted from fuel into exhaust gases, allowing for a greater change in velocity. However, it is essential to balance the mass ratio since carrying more fuel also means the rocket starts heavier. In our example, the mass ratio is given as 1.06. That means the initial mass \(m_0\) is just slightly greater than the final mass \(m_f\). Understanding mass ratios helps engineers and scientists predict how efficient a rocket will be in terms of fuel consumption and thrust-to-weight balance.

rocket propulsion

Rocket propulsion is the mechanism that allows rockets to move forward. It relies on Newton's third law of motion: for every action, there is an equal and opposite reaction. When a rocket burns fuel, it expels exhaust gases at high speeds in one direction (action), causing the rocket to move in the opposite direction (reaction). The Tsiolkovsky rocket equation integrates this principle to help calculate the change in velocity \(\Delta v\) a rocket can achieve given its exhaust velocity and fuel mass. This equation is foundational in understanding how rockets achieve liftoff and sustain their flight through space. Effective rocket propulsion requires a high exhaust velocity and an optimized mass ratio to ensure maximum efficiency and performance in reaching desired speeds and altitudes.

exhaust velocity

The exhaust velocity \(v_e\) is the speed at which exhaust gases are expelled from a rocket's nozzle. Higher exhaust velocity directly translates to greater propulsion efficiency and a larger change in velocity \(\Delta v\). For our example, the exhaust velocity is given as 3.1 km/s. This exhaust velocity is essential in the Tsiolkovsky rocket equation: \(\Delta v = \frac{v_e}{0.434} (\log m_0 - \log m_f)\). Rockets are designed to maximize exhaust velocity to achieve the highest possible speed change from a given amount of fuel. By designing nozzles and engines that can expel gases at higher velocities, engineers can create more efficient propulsion systems, allowing rockets to travel greater distances or carry more massive payloads. Knowing exhaust velocity helps in planning and optimizing the entire rocket design and its mission profile.