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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: 9780137488179Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoGiancoli Douglas 5th editionCh. 36 - The Special Theory of RelativityProblema 25
Capitolo 35, Problema 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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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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Lorentz Transformations of Velocity

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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Inertial Reference Frames
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