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\(\cdot\) An enemy spaceship is moving toward your starfighter with a speed of \(0.400 c,\) as measured in your reference frame. The enemy ship fires a missile toward you at a speed of 0.700\(c\) relative to the enemy ship. (See Figure \(27.24 . )\) (a) What is the speed of the missile relative to you? Express your answer in terms of the speed of light. (b) If you measure the enemy ship to be \(8.00 \times 10^{6} \mathrm{km}\) away from you when the missile is fired, how much time, measured in your frame, will it take the missile to reach you?

Short Answer

Expert verified
(a) 0.859c; (b) Approximately 31.1 seconds.

Step by step solution

01

List Known Variables

The speed of the enemy spaceship relative to you is given as \( u = 0.400c \). The speed of the missile relative to the enemy spaceship is \( v' = 0.700c \). We need to find the speed of the missile relative to you.
02

Use the Relativistic Velocity Addition Formula

To find the speed of the missile relative to you, use the relativistic velocity addition formula: \[v = \frac{u + v'}{1 + \frac{uv'}{c^2}}. \]Substitute \( u = 0.400c \) and \( v' = 0.700c \) into the formula.
03

Calculate the Missile Speed Relative to You

Substitute the values into the equation:\[v = \frac{0.400c + 0.700c}{1 + \frac{(0.400c)(0.700c)}{c^2}}= \frac{1.100c}{1 + 0.280}= \frac{1.100c}{1.280}.\]Calculate \( v \) to find the speed of the missile relative to you.
04

Simplify the Result

Simplify the equation from the previous step:\[v = \frac{1.100c}{1.280} = 0.859c.\]Thus, the speed of the missile relative to you is \( 0.859c \).
05

Calculate Time for Missile to Reach You

The distance to the enemy ship is \( 8.00 \times 10^6 \) km. The speed of the missile relative to you is \( 0.859c \). Use the formula for time: \[t = \frac{d}{v} = \frac{8.00 \times 10^6 \text{ km}}{0.859c}\]where \( c \approx 3.00 \times 10^5 \text{ km/s} \). Substitute \( c \) and \( v \) into the formula to calculate \( t \).
06

Calculate Time in Seconds

Calculate the time:\[t = \frac{8.00 \times 10^6 \text{ km}}{0.859 \times 3.00 \times 10^5 \text{ km/s}} = \frac{8.00 \times 10^6}{257700} \approx 31.1 \, \text{s}.\]So, it will take approximately 31.1 seconds for the missile to reach you.

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Key Concepts

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

Special Relativity
Special relativity is a theory of physics that was formulated by Albert Einstein. It revolutionizes our understanding of space, time, and energy. Unlike classical mechanics, which suggests speeds simply add up, special relativity shows us that the rules change as you approach the speed of light. At the heart of special relativity are two key postulates:
  • The laws of physics are the same in all inertial frames of reference.
  • The speed of light in a vacuum is a constant, denoted by \( c \), regardless of the observer's motion or the source of light.
When studying objects moving at significant fractions of the speed of light, like the spaceship and missile in our problem, special relativity becomes crucial because it prevents speeds from exceeding \( c \). This ensures that no object with mass can reach or exceed the speed of light.
To solve problems involving relative motion near the speed of light, like calculating the speed of a missile fired from a moving spaceship, we utilize special formulas, such as the relativistic velocity addition formula. This approach allows us to consider the effects of both time dilation and length contraction, ensuring all calculations adhere to the laws of Einstein's theory.
Missile Speed Calculation
In the context of special relativity, calculating the speed of an object, such as a missile, being launched from another moving object involves the relativistic velocity addition formula. This is different from standard addition of speeds in classical physics. Here is how it works:
  • The known parameters include the speed of the enemy spaceship \( u = 0.400c \) relative to the starfighter (your reference frame).
  • The missile's speed relative to the spaceship as \( v' = 0.700c \).
Using the relativistic velocity addition formula: \[v = \frac{u + v'}{1 + \frac{uv'}{c^2}}\] Substitute the values into the formula:
\[v = \frac{0.400c + 0.700c}{1 + \frac{(0.400c)(0.700c)}{c^2}}\] Simplifying the equation gives us: \[v = \frac{1.100c}{1.280} \] Finally, we find \( v = 0.859c \).

This means that in your reference frame, the missile moves at a speed of \( 0.859c \). The relativistic formula shows us how speeds combine differently when dealing with high velocities close to the speed of light, emphasizing the non-linear nature of velocity addition in special relativity.
Time Dilation
In problems involving high speeds similar to light, time does not behave as we are accustomed to in everyday experience. Time dilation occurs when an object is moving at high speeds compared to an observer at rest. It means that time will appear to run slower for the moving object compared to a stationary observer.

In our exercise, when measuring how long it takes for the missile to reach us, we need to consider its relativistic speed. By using the formula for time based on distance and speed, we calculate the time in our reference frame:
  • Distance to the enemy ship is \( 8.00 \times 10^6 \) km.
  • Speed of the missile relative to us is \( 0.859c \).
  • Light speed \( c \) is \( 3.00 \times 10^5 \) km/s.
The formula used is:
\[t = \frac{d}{v} = \frac{8.00 \times 10^6 \text{ km}}{0.859c} \]Substituting the constant \( c \) and velocity into the equation gives:
\[t = \frac{8.00 \times 10^6 \text{ km}}{0.859 \times 3.00 \times 10^5 \text{ km/s}} \approx 31.1 \text{ s} \] This means that, according to the observer's frame (your starfighter), it takes about 31.1 seconds for the missile to reach you.
The missile will take longer to reach you than it would seem if both were measured in a classical, non-relativistic way. This illustrates how time dilation affects measurements in different frames of motion.

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Most popular questions from this chapter

A pursuit spacecraft from the planet Tatooine is attempting to catch up with a Trade Federation cruiser. As measured by an observer on Tatooine, the cruiser is traveling away from the planet with a speed of 0.600\(c .\) The pursuit ship is traveling at a speed of 0.800\(c\) relative to Tatooine, in the same direction as the cruiser. What is the speed of the cruiser relative to the pur- suit ship?

An electron is acted upon by a force of \(5.00 \times 10^{-15} \mathrm{N}\) due to an electric field. Find the acceleration this force produces in each case: (a) The electron's speed is 1.00 \(\mathrm{km} / \mathrm{s}\) . (b) The electron's speed is \(2.50 \times 10^{8} \mathrm{m} / \mathrm{s}\) and the force is parallel to the velocity.

\(\cdot\) A cube of metal with sides of length \(a\) sits at rest in the laboratory with one edge parallel to the \(x\) axis. Therefore, in the laboratory frame, its volume is \(a^{3} .\) A rocket ship flies past the laboratory parallel to the \(x\) axis with a velocity \(v .\) To an observer in the rocket, what is the volume of the metal cube?

\(\bullet\) (a) At what speed does the momentum of a particle differ by 1.0\(\%\) from the value obtained with the nonrelativistic expression \(m v ?\) (b) Is the correct relativistic value greater or less than that obtained from the nonrelativistic expression?

A spaceship makes the long trip from earth to the nearest star system, Alpha Centauri, at a speed of 0.955\(c .\) The star is about 4.37 light years from earth, as measured in earth's frame of reference \((1\) light year is the distance light travels in a year). (a) How many years does the trip take, according to an observer on earth? (b) How many years does the trip take according to a passenger on the spaceship? (c) How many light years distant is Alpha Centauri from earth, as measured by a passenger on the speeding spacecraft? (Note that, in the ship's frame of reference, the passengers are at rest, while the space between earth and Alpha Centauri goes rushing past at 0.955\(c .\) (d) Use your answer from part (c) along with the speed of the spacecraft to calculate another answer for part (b). Do your two answers for that part agree? Should they?

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