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Chapter 3: Problem 10

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
Acceleration needed is roughly 115.74 m/s^2, which is about 11.81 g. Required force is approximately 1.38888 x 10^8 N. Total journey time is about 1522.7 days.

Step by step solution

01

Determine the final velocity

Given the final velocity is 0.10 times the speed of light, calculate it using the provided speed of light: \[ v = 0.10 \times c = 0.10 \times 3.0 \times 10^8 \text{ m/s} = 3.0 \times 10^7 \text{ m/s} \]
02

Convert the time from days to seconds

Convert the time from 3.0 days to seconds to use in the acceleration calculation: \[ \text{Time (t)} = 3.0 \text{ days} \times 24 \text{ hours/day} \times 3600 \text{ seconds/hour} = 259200 \text{ seconds} \]
03

Calculate the acceleration

Use the formula for constant acceleration: \[ a = \frac{v}{t} \] Substitute the values for final velocity and time: \[ a = \frac{3.0 \times 10^7 \text{ m/s}}{259200 \text{ s}} \ a \approx 115.74 \text{ m/s}^2 \]
04

Convert acceleration to g units

To express this acceleration in units of standard gravity, use the conversion factor where \[ 1 \text{ g} = 9.8 \text{ m/s}^2 \]\[ a_g = \frac{115.74 \text{ m/s}^2}{9.8 \text{ m/s}^2} \ a_g \approx 11.81 \text{ g} \]
05

Calculate the required force

Use Newton's second law, \( F = ma \) to find the force:\[ F = 1.20 \times 10^6 \text{ kg} \times 115.74 \text{ m/s}^2 \]\[ F \approx 1.38888 \times 10^8 \text{ N} \]
06

Calculate the total time for the journey

Convert the journey distance from light-months to meters, where 1 light-month is \[ c \times \text{ time in seconds in one month} \]Thus,\[ 1 \text{ light-month} = 3.0 \times 10^8 \text{ m/s} \times 2.63 \times 10^6 \text{ s} (30 days) \]So, \[ 5.0 \text{ light-months} = 5.0 \times 3.0 \times 10^8 \times 2.63 \times 10^6 \ \ = 3.945 \times 10^{15} \text{ m} \]Use constant velocity formula to find the time taken: \[ t = \frac{d}{v} \]\[ t = \frac{3.945 \times 10^{15} \text{ m}}{3.0 \times 10^7 \text{ m/s}}\ t \approx 1.315 \times 10^8 \text{ s} \]Convert this to days: \[ \frac{1.315 \times 10^8 \text{ s}}{86400 \text{ s/day}} \approx 1522.7 \text{ days} \]

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Key Concepts

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

Constant Acceleration
In physics, the term 'constant acceleration' means that an object's velocity changes at a uniform rate over time. When an interstellar ship needs to reach a fraction of the speed of light, constant acceleration is applied to increase the ship鈥檚 speed steadily. The formula to calculate this is: \[ a = \frac{v}{t} \] Here, 'a' stands for acceleration, 'v' is the final velocity, and 't' is the time. For our ship problem, we know the final velocity, and we convert the given time into seconds. Using these values, we can determine the ship's acceleration easily.
Newton's Second Law
Understanding Newton's second law is crucial for calculating the force required for acceleration. This law states that the force acting upon an object is equal to the mass of that object multiplied by its acceleration: \[ F = ma \] In our given exercise, the mass of the ship is 1.20 x 10^6 kg, and we鈥檝e calculated the required acceleration. Multiplying these gives us the force needed to achieve the acceleration. This force propels the ship forward, overcoming inertia and any opposing forces in interstellar space.
Conversion of Units
Many physics problems require unit conversions to maintain consistency in calculations. For example, we converted 3 days into seconds to use in acceleration calculations. The process involves multiplying the number of days by the number of hours in a day (24), then by the number of seconds in an hour (3600): \[ \text{Time (t)} = 3 \text{ days} \times 24 \text{ hours/day} \times 3600 \text{ seconds/hour} \] Similarly, converting acceleration to 'g' units involves dividing by the standard gravity acceleration (9.8 m/s^2): \[ a_g = \frac{a}{9.8} \]
Speed of Light
The speed of light (denoted 'c') is a constant value, approximately 3.0 x 10^8 meters per second. In our problem, the interstellar ship aims to reach 0.10 times the speed of light. This means the final velocity we are considering is: \[ v = 0.10 \times c = 0.10 \times 3.0 \times 10^8 \text{ m/s} = 3.0 \times 10^7 \text{ m/s} \] The speed of light is incredibly fast, and achieving even a fraction of this speed poses significant challenges and requires robust calculations.
Distance and Time Calculation
Calculating distance and time is crucial for planning an interstellar journey. Given a distance in light-months (where one light-month is the distance light travels in one month), we convert this distance to meters. For instance, 5 light-months can be converted as: \[ \text{Distance} = 5.0 \times 3.0 \times 10^8 \times 2.63 \times 10^6 \text{ meters} \]To find the journey time at constant speed, we use: \[ t = \frac{d}{v} \] where 'd' is the distance, and 'v' is the ship's velocity. By substituting the distance and calculated velocity, we derive the total travel time, then convert it to days for practical understanding.

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Most popular questions from this chapter

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