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Chapter 5: Problem 90

An interstellar ship has a mass of \(1.20 \times 10^{6} \mathrm{~kg}\) and is initially at rest relative to a star system. (a) What constant acceleration is needed to bring the ship up to a speed of \(0.10 c\) (where \(c\) is the speed of light, \(3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}\) ) relative to the star system in \(3.0\) days? (b) What is that acceleration in \(g\) units? (c) What force is required for the acceleration? (d) If the engines are shut down when \(0.10 c\) is reached (the speed then remains constant), how long does the ship take (start to finish) to journey \(5.0\) light-months, the distance that light travels in \(5.0\) months?

Short Answer

Expert verified
a) 1157.4 m/s虏, b) 118.0 g, c) 1.39脳10鈦� N, d) 498.6 days.

Step by step solution

01

Calculate acceleration (a)

First, we need to determine the acceleration required to reach a velocity of \(0.10c\) in 3 days. We'll use the kinematic equation \(v = u + at\), where \(v = 0.10 \, c = 0.10 \, \times \, 3.0 \, \times \, 10^8 \, \text{m/s}\), \(u = 0\), and \(t = 3 \, \text{days} \, = 259200 \, \text{seconds}\) (since 1 day = 86400 seconds). Plugging these into the equation: \[0.10 \, \times \, 3.0 \, \times \, 10^8 = 0 + a \, \times \, 259200\]Solving for \(a\) gives us \[a = \frac{0.10 \, \times \, 3.0 \, \times \, 10^8}{259200} \, \approx \, 1157.4 \, \text{m/s}^2\].
02

Convert acceleration to g units (b)

We convert the acceleration from meters per second squared to \(g\) units, where \(1 \, g = 9.81 \, \text{m/s}^2\). Thus: \[a_{g} = \frac{1157.4}{9.81} \, \approx \, 118.0 \, g\].
03

Calculate force required (c)

We use the equation \(F = ma\) to find the force, where \(m = 1.20 \, \times \, 10^6 \, \text{kg}\) and \(a = 1157.4 \, \text{m/s}^2\). Thus, \[F = 1.20 \, \times \, 10^6 \, \times \, 1157.4\]which calculates to \[F = 1.39 \, \times \, 10^9 \, \text{N}\].
04

Calculate journey time (d)

Determine the time to journey 5.0 light-months at a constant speed of \(0.10c\). First, find the distance in meters: \[5.0 \, \text{light-months} = 5.0 \, \times \, 0.10 \, \times \, 3.0 \, \times \, 10^8 \, \times \, \frac{31557600}{12} \approx 1.29 \, \times \, 10^{16} \, \text{meters}\].Time to travel this distance at \(v = 0.10c\) is \[t = \frac{1.29 \, \times \, 10^{16}}{0.10 \, \times \, 3.0 \, \times \, 10^8} \, \approx 4.3 \, \times \, 10^7 \, \text{seconds}\].Convert to days: \[\frac{4.3 \, \times \, 10^7}{86400} \, \approx \, 498.6 \, \text{days}\].

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Key Concepts

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

Kinematics
Kinematics is a branch of physics that focuses on the motion of objects without considering the forces that cause the motion. It helps us describe how an object moves through space over time. In this exercise, we used the kinematic equation \(v = u + at\) to calculate the acceleration needed to propel an interstellar ship to a certain speed. Here are some basic terms in kinematics explained:

  • Initial velocity \( (u) \): This is the speed at which an object starts. In this problem, the ship starts at rest, so \( u = 0 \).
  • Final velocity \( (v) \): This is the speed an object reaches after a certain time. For the ship, \( v = 0.10c \), where \(c\) is the speed of light.
  • Acceleration \( (a) \): It refers to how quickly an object speeds up or slows down. We calculated it using \( a = \frac{v-u}{t} \).
  • Time \( (t) \): This is the duration over which the motion occurs. Here, it was equivalent to 3 days (or 259,200 seconds).
In changing from initial to final velocity within a specified time, acceleration is needed. Understanding these concepts allows us to grasp how objects move in space and over time.
Force Calculation
Calculating the force needed to move an object involves Newton's second law of motion, which states \( F = ma \), where \( F \) is the force applied, \( m \) is the object's mass, and \( a \) is its acceleration.

In the interstellar ship's scenario:
  • The mass \( (m) \) of the ship is \( 1.20 \times 10^6 \text{ kg} \).
  • The required acceleration \( (a) \) was calculated to be approximately \( 1157.4 \text{ m/s}^2 \).
  • Multiplying these two gives the force \( (F) \) necessary to achieve such acceleration.
    The calculation shows that a force of \( 1.39 \times 10^9 \text{ N} \) is needed.
This large force indicates the sheer power required to propel such a massive spacecraft. Newton's second law is crucial in engineering and physics as it allows us to calculate necessary forces for any motion.
Light Speed Travel
Traveling at a fraction of the speed of light involves understanding both the vast distances in space and the time taken to travel those distances. Light speed \( (c) \) is a universal constant valued at \( 3.0 \times 10^8 \text{ m/s} \).

For this exercise:
  • "Light-months" specifies the distance light covers in a month. Our task was to calculate travel time for 5 light-months.
  • Even at \( 0.10c \), a velocity that seems incredibly fast, traveling such distances takes considerable time (498.6 days is the calculated travel time at this speed for 5 light-months).This illustrates how even near-light-speed travel demands significant planning and patience.
These analyses of light speed and time emphasize the vastness of space and help us appreciate the challenges and planning required for interstellar travel. Such physics problems not only test calculations but also expand our comprehension of the universe.

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Most popular questions from this chapter

A car traveling at \(53 \mathrm{~km} / \mathrm{h}\) hits a bridge abutment. A passenger in the car moves forward a distance of \(65 \mathrm{~cm}\) (with respect to the icad) while being brought to rest by an inflated air bag. What magnitude of force (assumed constant) acts on the passenger's upper torso, which has a mass of \(41 \mathrm{~kg}\) ?

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Tarzan, who weighs \(820 \mathrm{~N}\), swings from a cliff at the end of a \(20.0 \mathrm{~m}\) vine that hangs from a high tree limb and initially makes an angle of \(22.0^{\circ}\) with the vertical. Assume that an \(x\) axis extends horizontally away from the cliff edge and a \(y\) axis extends upward. Immediately after Tarzan steps off the cliff, the tension in the vine is \(760 \mathrm{~N}\). Just then, what are (a) the force on him from the vine in unit- vector notation and the net force on him (b) in unit-vector notation and as (c) a magnitude and (d) an angle relative to the positive direction of the \(x\) axis? What are the (e) magnitude and (f) angle of Tarzan's acceleration just then?

Only two horizontal forces act on a \(3.0 \mathrm{~kg}\) body that can move over a frictionless floor. One force is \(9.0 \mathrm{~N}\), acting due east, and the other is \(8.0 \mathrm{~N}\), acting \(62^{\circ}\) north of west. What is the magnitude of the body's acceleration?

The high-speed winds around a tornado can drive projectiles into trees, building walls, and even metal traffic signs. In a laboratory simulation, a standard wood toothpick was shot by pneumatic gun into an oak branch. The toothpick's mass was \(0.13\) \(\mathrm{g}\), its speed before entering the branch was \(220 \mathrm{~m} / \mathrm{s}\), and its penetration depth was \(15 \mathrm{~mm}\). If its speed was decreased at a uniform rate, what was the magnitude of the force of the branch on the toothpick?
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