Problem 65 - University Physics with Modern Physics

Question

A single-stage rocket is fired from rest from a deep-space platform, where gravity is negligible. If the rocket burns its fuel in 50.0 s and the relative speed of the exhaust gas is \(v_{\text { ex }}=2100 \mathrm{m} / \mathrm{s}\) what must the mass ratio \(m_{0} / m\) be for a final speed \(v\) of 8.00 \(\mathrm{km} / \mathrm{s}\) (about equal to the orbital speed of an earth satellite)?

Step-by-Step Solution

Answer

The mass ratio \( \frac{m_0}{m} \) must be approximately 45.06.

Step 1: Understanding the Rocket Equation

The rocket equation is given by the Tsiolkovsky equation: \( v = v_{\text{ex}} \ln \left( \frac{m_0}{m} \right) \), where \( v \) is the final velocity of the rocket, \( v_{\text{ex}} \) is the relative speed of the exhaust gas, \( m_0 \) is the initial total mass of the rocket, and \( m \) is the final total mass of the rocket after burning the fuel.

Step 2: Rearranging the Equation

We need to solve for the mass ratio \( \frac{m_0}{m} \). Rearranging the rocket equation, we have: \( \ln \left( \frac{m_0}{m} \right) = \frac{v}{v_{\text{ex}}} \).

Step 3: Substituting Known Values

Substitute the given values \( v = 8000 \ \text{m/s} \) and \( v_{\text{ex}} = 2100 \ \text{m/s} \) into the rearranged equation: \( \ln \left( \frac{m_0}{m} \right) = \frac{8000}{2100} \).

Step 4: Calculating the Logarithm

Calculate the division: \( \frac{8000}{2100} \approx 3.8095 \). Therefore, \( \ln \left( \frac{m_0}{m} \right) = 3.8095 \).

Step 5: Exponential of Both Sides

To find \( \frac{m_0}{m} \), exponentiate both sides of the equation: \( \frac{m_0}{m} = e^{3.8095} \).

Step 6: Final Calculation

Calculate \( e^{3.8095} \approx 45.06 \). Thus the mass ratio \( \frac{m_0}{m} \) is approximately 45.06.

Key Concepts

Tsiolkovsky Equation

The Tsiolkovsky equation is essential in understanding how rockets accelerate through space. It's a simple yet profound formula that relates the change in velocity of a rocket to the properties of its fuel. The equation is expressed as:\[ v = v_{\text{ex}} \ln \left( \frac{m_0}{m} \right) \]Where:- \( v \) is the final velocity of the rocket.- \( v_{\text{ex}} \) is the velocity of the exhaust gas.- \( m_0 \) is the initial total mass (rocket + fuel).- \( m \) is the final total mass (rocket after fuel is burned).This equation is named after Konstantin Tsiolkovsky, a pioneer of astronautic theory. In essence, it shows how the velocity of a rocket increases as it sheds mass by expelling exhaust gases. You can think of it as the fundamental principle behind rocket motion. Understanding this equation helps explain how rockets can reach incredible speeds and travel vast distances.

Mass Ratio

The mass ratio is a crucial term in rocket science, representing the proportion of the initial mass to the final mass of a rocket. In mathematical terms, it is defined as:\[ \frac{m_0}{m} \]This ratio signifies how much of the initial mass was composed of fuel. A higher mass ratio indicates that a larger portion of the rocket's weight was fuel rather than structure. Why is this important? Because the mass ratio directly influences the final velocity (\( v \)) of the rocket, as indicated by the Tsiolkovsky equation. The greater the mass ratio, the more significant the potential velocity increase, given a constant exhaust velocity. Understanding mass ratio helps engineers design rockets that are efficient in fuel consumption while maximizing payload capacity. For students, comprehending this term reinforces the relationship between mass dynamics and rocket propulsion.

Exhaust Velocity

Exhaust velocity is the speed at which gases are expelled from the rocket's engines. In the context of the Tsiolkovsky equation, it is represented as \( v_{\text{ex}} \).The magnitude of the exhaust velocity is crucial as it effectively determines the efficiency of fuel use by the rocket. A higher exhaust velocity means that the rocket can achieve higher speeds with the same mass of fuel.Why does this matter?- Higher exhaust velocities imply that the rocket can achieve more speed with less fuel.- This efficiency can be critical when designing long-distance space missions or spacecraft for low orbits.- It affects not only how fast a rocket can go but also how much payload it can carry.In terms of engineering, achieving a high exhaust velocity requires advanced technology and innovations in rocket fuel and engine design. For students, understanding this concept provides deeper insights into what fuels rocket expeditions beyond our atmosphere.

Orbital Speed

Orbital speed refers to the velocity a rocket (or any object) must reach to achieve a stable orbit around a planet or celestial body. For rockets launching from Earth, reaching orbital speed is a critical milestone. In simple terms, if a rocket achieves this speed, it can circle Earth without falling back due to gravity. Typically, Earth's orbital speed is about 7.8 to 8.0 km/s. Here’s why orbital speed is essential: - It determines whether a satellite or spacecraft can remain in orbit. - Ensures the rocket can accomplish its mission, such as satellite deployment or docking with a space station. - Serves as a benchmark when calculating project trajectories and necessary propulsion requirements. For students and budding engineers, grasping orbital speed is key to understanding space flight mechanics, as it highlights the balance of gravitational forces and forward velocity in maintaining an orbit.