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Chapter 3: Problem 174

The two orbital maneuvering engines of the space shuttle develop \(26 \mathrm{kN}\) of thrust each. If the shuttle is traveling in orbit at a speed of \(28000 \mathrm{km} / \mathrm{h},\) how long would it take to reach a speed of \(28100 \mathrm{km} / \mathrm{h}\) after the two engines are fired? The mass of the shuttle is \(90 \mathrm{Mg}\).

Short Answer

Expert verified
It takes approximately 48.1 seconds.

Step by step solution

01

Convert Units

We need to ensure all units are consistent. Convert the shuttle speed from \(\mathrm{km/h}\) to \(\mathrm{m/s}\) for both initial and final velocities.\[28000 \, \mathrm{km/h} = \frac{28000 \, \mathrm{km/h}}{3.6} \approx 7777.78 \, \mathrm{m/s}\]\[28100 \, \mathrm{km/h} = \frac{28100 \, \mathrm{km/h}}{3.6} \approx 7805.56 \, \mathrm{m/s}\]
02

Calculate the Change in Velocity

Find the change in velocity (\(\Delta v\)) by taking the difference between the final and initial speeds in \(\mathrm{m/s}\).\[\Delta v = 7805.56 \, \mathrm{m/s} - 7777.78 \, \mathrm{m/s} = 27.78 \, \mathrm{m/s}\]
03

Sum of the Thrust Forces

Calculate the total thrust force. Both engines generate \(26 \, \mathrm{kN}\) each. \[\text{Total thrust} = 2 \times 26 \, \mathrm{kN} = 52 \, \mathrm{kN} = 52000 \, \mathrm{N}\]
04

Convert Mass Units

Convert the mass from \(\mathrm{Mg}\) to \(\mathrm{kg}\) to use SI units in further calculations.\[90 \, \mathrm{Mg} = 90000 \, \mathrm{kg}\]
05

Calculate the Acceleration

Use Newton's second law to find the acceleration (\(a\)) using the formula \(F = ma\).\[a = \frac{F}{m} = \frac{52000 \, \mathrm{N}}{90000 \, \mathrm{kg}} = 0.5778 \, \mathrm{m/s^2}\]
06

Calculate the Time Required

Use the formula \(t = \frac{\Delta v}{a}\) to find the time required to reach the final speed.\[t = \frac{27.78 \, \mathrm{m/s}}{0.5778 \, \mathrm{m/s^2}} \approx 48.1 \, \mathrm{s}\]

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Key Concepts

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

Understanding Thrust
Thrust is the force that moves a spacecraft forward in its orbit or trajectory. It's generated by the engines, pushing exhaust gas in the opposite direction by Newton's third law: for every action, there is an equal and opposite reaction.
In this exercise, the two engines of the space shuttle provide a total thrust of 52,000 Newtons (where each engine contributes 26,000 Newtons). This force allows us to change the velocity of the shuttle.
  • What does thrust do? It provides the necessary force to overcome the inertia of the shuttle and change its speed.
  • What is a Newton (N)? It's the SI unit of force. One Newton is the force required to accelerate an object with a mass of 1 kg at 1 m/s虏.
Thrust is crucial for orbital maneuvers, allowing spacecraft to adjust their orbits, speed, and orientation.
Role of Acceleration
Acceleration is the rate at which an object's velocity changes with time. In the context of orbital mechanics, it can be caused by thrust from spacecraft engines.
The formula we use is based on Newton's second law of motion: \[ F = ma \], where \(F\) is the total thrust force, \(m\) is the mass, and \(a\) is the acceleration.
  • Why do we need acceleration? It helps us understand how quickly our velocity can change when a known force is applied.
  • How do we calculate acceleration? By re-arranging the formula to \( a = \frac{F}{m} \), we can determine how much the velocity of the shuttle will increase per second.
In this exercise, the acceleration due to the thrust is approximated to be 0.5778 m/s虏, meaning that the shuttle's velocity increases by this amount every second.
Calculating Velocity Change
Velocity change, denoted as \(\Delta v\), is a fundamental metric in assessing the effectiveness of thrust during an orbital maneuver. It's simply the difference between the final and initial velocities of the spacecraft.
For this problem, we compute it by finding: \[ \Delta v = v_{ ext{final}} - v_{ ext{initial}} = 7805.56 \, \text{m/s} - 7777.78 \, \text{m/s} = 27.78 \, \text{m/s} \]
  • What is velocity change? It is how much faster (or slower) the spacecraft is moving after the thrust is applied.
  • Why is it important? Knowing \(\Delta v\) helps us determine how much time is required for the engines to operate to achieve this speed change.
This change in velocity directly influences the time the shuttle needs to accelerate from its initial speed to the final speed.
Importance of Unit Conversion
Unit conversion is crucial in physics to ensure consistency and accuracy in calculations. In spatial mechanics, it's common to deal with various units of speed and mass.
Changing the speeds from kilometers per hour (km/h) to meters per second (m/s) aligns with the SI units used in the calculations.
The conversion factor between these units is 3.6, necessitating division for a consistent speed unit:
\[ 28000 \, \text{km/h} = \frac{28000}{3.6} \, \text{m/s} \]
Similarly, mass conversion from megagrams (Mg) to kilograms (kg) is straightforward because 1 Mg equals 1000 kg.
  • Why is unit conversion important? Accurate unit conversion ensures that formulas yield correct results without unit mismatches.
  • How to perform unit conversion? By applying the relevant conversion factors for each unit.
Proper unit conversion is essential to solve problems successfully in orbital mechanics.

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Most popular questions from this chapter

The cars of an amusement-park ride have a speed \(v_{1}=90 \mathrm{km} / \mathrm{h}\) at the lowest part of the track. Determine their speed \(v_{2}\) at the highest part of the track. Neglect energy loss due to friction. (Caution: Give careful thought to the change in potential energy of the system of cars.)

The robot arm is elevating and extending simultaneously. At a given instant, \(\theta=30^{\circ}, \dot{\theta}=40 \mathrm{deg} / \mathrm{s}\) \(\ddot{\theta}=120 \operatorname{deg} / \mathrm{s}^{2}, l=0.5 \mathrm{m}, \dot{l}=0.4 \mathrm{m} / \mathrm{s},\) and \(\ddot{l}=-0.3\) \(\mathrm{m} / \mathrm{s}^{2} .\) Compute the radial and transverse forces \(F_{r}\) and \(F_{\theta}\) that the arm must exert on the gripped part \(P,\) which has a mass of 1.2 kg. Compare with the case of static equilibrium in the same position.

A 30 -g tire-balance weight is attached to a vertical surface of the wheel rim by means of an adhesive backing. The tire-wheel unit is then given a final test on the tire-balance machine. If the adhesive can support a maximum shear force of \(80 \mathrm{N}\), determine the maximum rotational speed \(N\) for which the weight remains fixed to the wheel. Assume very gradual speed changes.

A boy weighing 100 lb runs and jumps on his \(20-1 b\) sled with a horizontal velocity of 15 ft/sec. If the sled and boy coast \(80 \mathrm{ft}\) on the level snow before coming to rest, compute the coefficient of kinetic friction \(\mu_{k}\) between the snow and the runners of the sled.

Car \(A\) weighing 3200 lb and traveling north at \(20 \mathrm{mi} / \mathrm{hr}\) collides with car \(B\) weighing \(3600 \mathrm{lb}\) and traveling at \(30 \mathrm{mi} / \mathrm{hr}\) as shown. If the two cars become entangled and move together as a unit after the crash, compute the magnitude \(v\) of their common velocity immediately after the impact and the angle \(\theta\) made by the velocity vector with the north direction.
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