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Ch. 36 - The Special Theory of Relativity
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
All textbooksGiancoli Douglas 5th editionCh. 36 - The Special Theory of RelativityProblem 25
Chapter 35, Problem 25

A spaceship traveling at 0.76c away from Earth fires a module with a speed of 0.85c at right angles to its own direction of travel (as seen by the spaceship). What is the speed of the module, and its direction of travel (relative to the spaceship’s direction), seen by an observer on Earth?

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Understand the problem: The spaceship is moving at a velocity of 0.76c relative to Earth, and it fires a module at a velocity of 0.85c perpendicular to its direction of travel (relative to the spaceship). We need to determine the speed and direction of the module as observed from Earth. This involves relativistic velocity addition in two dimensions.
Set up the coordinate system: Let the spaceship's motion be along the x-axis with velocity \( v_s = 0.76c \) relative to Earth. The module's velocity relative to the spaceship is \( v_m = 0.85c \) along the y-axis. The goal is to find the module's velocity components \( v_{x,\text{Earth}} \) and \( v_{y,\text{Earth}} \) relative to Earth, and then calculate the resultant speed and direction.
Apply the relativistic velocity addition formula for the x-component: The x-component of the module's velocity relative to Earth is given by \( v_{x,\text{Earth}} = \frac{v_s + v_{x,\text{module}}}{1 + \frac{v_s v_{x,\text{module}}}{c^2}} \). Since the module's velocity relative to the spaceship is entirely in the y-direction, \( v_{x,\text{module}} = 0 \), so \( v_{x,\text{Earth}} = v_s = 0.76c \).
Apply the relativistic velocity addition formula for the y-component: The y-component of the module's velocity relative to Earth is given by \( v_{y,\text{Earth}} = \frac{v_{y,\text{module}}}{\gamma (1 + \frac{v_s v_{x,\text{module}}}{c^2})} \), where \( \gamma = \frac{1}{\sqrt{1 - \frac{v_s^2}{c^2}}} \). Substituting \( v_{y,\text{module}} = 0.85c \) and \( v_{x,\text{module}} = 0 \), simplify to find \( v_{y,\text{Earth}} \).
Calculate the resultant speed and direction: The speed of the module relative to Earth is \( v_{\text{Earth}} = \sqrt{v_{x,\text{Earth}}^2 + v_{y,\text{Earth}}^2} \). The direction \( \theta \) relative to the x-axis is given by \( \theta = \arctan\left(\frac{v_{y,\text{Earth}}}{v_{x,\text{Earth}}}\right) \). Use these formulas to express the final speed and direction of the module as observed from Earth.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Relativity of Velocity

In special relativity, the velocity of an object is not simply additive due to the finite speed of light. When calculating the velocity of an object moving at high speeds, one must use the relativistic velocity addition formula, which accounts for the effects of time dilation and length contraction. This is crucial for determining how fast the module travels relative to an observer on Earth.
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Intro to Relative Motion (Relative Velocity)

Lorentz Transformation

The Lorentz transformation equations relate the space and time coordinates of events as observed in different inertial frames. They are essential for converting measurements from one frame to another, particularly when dealing with speeds close to the speed of light. Understanding these transformations helps in accurately calculating the observed speed and direction of the module from Earth's perspective.
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Reference Frames

A reference frame is a perspective from which measurements are made, including position, velocity, and time. In this problem, there are two reference frames: one for the spaceship and one for Earth. Recognizing how different observers perceive motion differently is key to solving the problem, as the speeds and directions of objects can vary significantly depending on the observer's frame of reference.
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Related Practice
Textbook Question

(I) An observer on Earth sees an alien vessel approach at a speed of 0.70c. The fictional starship Enterprise comes to the rescue (Fig. 36–17), overtaking the aliens while moving directly toward Earth at a speed of 0.90c relative to Earth. What is the relative speed of one vessel as seen by the other?

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Textbook Question

An observer in reference frame S notes that two events are separated in space by 180 m and in time by 0.80μs. How fast must reference frame S' be moving relative to S in order for an observer in S' to detect the two events as occurring at the same location in space?

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Textbook Question

A stick of length ℓ₀, at rest in reference frame S, makes an angle θ with the x axis. In reference frame S', which moves to the right with velocity v→\(\overrightarrow{v}\) = vî with respect to S, determine (a) the length l of the stick, and (b) the angle θ it makes with the x' axis.

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Textbook Question

In the old West, a marshal riding on a train traveling 35.0 m/s sees a duel between two men standing on the Earth 55.0 m apart parallel to the train. The marshal’s instruments indicate that in his reference frame the two men fired simultaneously.

(a) Which of the two men, the first one the train passes (A) or the second one (B) should be arrested for firing the first shot? That is, in the gunfighter’s frame of reference, who fired first?

(b) How much earlier did he fire?

(c) Who was struck first?

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Textbook Question

Two spaceships leave Earth in opposite directions, each with a speed of 0.50c with respect to Earth.

(a) What is the velocity of spaceship 1 relative to spaceship 2?

(b) What is the velocity of spaceship 2 relative to spaceship 1?

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Textbook Question

(I) Repeat Problem 19 using the Lorentz transformation and a relative speed v = 1.60 x 10⁸ m/s, but choose the time t to be (a) 3.5μs and (b) 10.0 μs .

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