/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 A single-stage rocket is fired f... [FREE SOLUTION] | 91Ó°ÊÓ

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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)?

Short Answer

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

Step by step solution

01

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.
02

Rearranging the Equation

We need to solve for the mass ratio \( \frac{m_0}{m} \). Rearranging the rocket equation, we have: \( \ln \left( \frac{m_0}{m} \right) = \frac{v}{v_{\text{ex}}} \).
03

Substituting Known Values

Substitute the given values \( v = 8000 \ \text{m/s} \) and \( v_{\text{ex}} = 2100 \ \text{m/s} \) into the rearranged equation: \( \ln \left( \frac{m_0}{m} \right) = \frac{8000}{2100} \).
04

Calculating the Logarithm

Calculate the division: \( \frac{8000}{2100} \approx 3.8095 \). Therefore, \( \ln \left( \frac{m_0}{m} \right) = 3.8095 \).
05

Exponential of Both Sides

To find \( \frac{m_0}{m} \), exponentiate both sides of the equation: \( \frac{m_0}{m} = e^{3.8095} \).
06

Final Calculation

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

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Key Concepts

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

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.

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Most popular questions from this chapter

Energy Sharing in Elastic Collisions. A stationary object with mass \(m_{B}\) is struck head-on by an object with mass \(m_{A}\) that is moving initially at speed \(v_{0} .\) (a) If the collision is elastic, what percentage of the original energy does each object have after the collision? (b) What does your answer in part (a) give for the special cases (i) \(m_{A}=m_{B}\) and (ii) \(m_{A}=5 m_{B} ?(\mathrm{c})\) For what values, if any, of the mass ratio \(m_{A} / m_{B}\) is the original kinetic energy shared equally by the two objects after the collision?

Three odd-shaped blocks of chocolate have the following masses and center-of- mass coordinates: \((1) 0.300 \mathrm{kg},(0.200 \mathrm{m},\) \(0.300 \mathrm{m} ) ;\) (2) \(0.400 \mathrm{kg},(0.100 \mathrm{m},-0.400 \mathrm{m}) ;\) (3) 0.200 \(\mathrm{kg}\) \((-0.300 \mathrm{m}, 0.600 \mathrm{m}) .\) Find the coordinates of the center of mass of the system of three chocolate blocks.

A 1500 -kg blue convertible is traveling south, and a \(2000-\mathrm{kg}\) red SUV is traveling west. If the total momentum of the system consisting of the two cars is 7200 \(\mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\) directed at \(60.0^{\circ}\) west of south, what is the speed of each vehicle?

Automobile Accident Analysis. You are called as an expert witness to analyze the following auto accident: Car \(B,\) of mass \(1900 \mathrm{kg},\) was stopped at a red light when it was hit from behind by car \(A,\) of mass 1500 \(\mathrm{kg}\) . The cars locked bumpers during the collision and slid to a stop with brakes locked on all wheels. Measurements of the skid marks left by the tires showed them to be 7.15 \(\mathrm{m}\) long. The coefficient of kinetic friction between the tires and the road was 0.65 . (a) What was the speed of car A just before the collision? (b) If the speed limit was 35 \(\mathrm{mph}\) , was car \(A\) speeding, and if so, by how many miles per hour was it exceeding the speed limit?

Block \(A\) in Fig. E8.24 has mass \(1.00 \mathrm{kg},\) and block \(B\) has mass 3.00 \(\mathrm{kg}\) . The blocks are forced together, compressing a spring \(S\) between them; then the system is released from rest on a level, frictionless surface. The spring, which has negligible mass, is not fastened to either block and drops to the surface after it has expanded. Block \(B\) acquires a speed of 1.20 \(\mathrm{m} / \mathrm{s}\) . (a) What is the final speed of block \(A\) ? (b) How much potential energy was stored in the compressed spring?

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