Chapter 36: Problem 27
(II) A spaceship traveling at \(0.76 c\) away from Earth fires a module with a speed of \(0.82 c\) 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), as seen by an observer on Earth?
Short Answer
Step by step solution
Identify the Given Data
Understand the Problem
Compute the Velocity in the Travel Direction of Spaceship
Compute the Perpendicular Component of Velocity
Compute the Combined Speed using Pythagorean Theorem
Determine the Direction
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Special relativity
In the context of the exercise, special relativity explains why traditional Newtonian mechanics cannot apply when dealing with speeds close to \( c \). Instead, it requires us to use relativistic principles to correctly determine the relationships between velocities, such as those of the spaceship and the fired module. These principles ensure accurate descriptions of motion when velocities approach the speed of light.
Lorentz transformation
One key feature of the Lorentz transformation is time dilation, which means time can appear to move slower for an observer in motion relative to a stationary observer. Another is length contraction, where the length of an object in motion is measured as shorter by a stationary observer.
In the exercise, we use the Lorentz factor \( \gamma_{sp} = \frac{1}{\sqrt{1 - (v_{sp}/c)^2}} \) for the spaceship to correctly transform the perpendicular velocity of the module as seen by an observer on Earth. This factor accounts for the relativistic effects on the observed velocities at high speeds.
Velocity addition formula
For example, if an observer perceived one object moving at a considerable fraction of the speed of light, the velocity addition formula ensures correct calculation of another object's speed relative to the first. This formula is particularly important when velocities are significant portions of \( c \).
In the module exercise, the formula helps calculate the effective velocities in the module's perpendicular direction and combine it with the spaceship's velocity, demonstrating how special relativity modifies traditional velocity addition for high-speed travel.
Perpendicular velocities
When dealing with perpendicular velocities, we must determine the object's effective speed along each axis, factoring in both travel along the direction of the parent object (i.e., the spaceship) and perpendicular to it. Here, the Lorentz factor adjusts the perpendicular component to account for relativistic speeds.
Using the Pythagorean theorem, we then find the resultant velocity vector's magnitude. The direction can be solved through trigonometric relationships, such as tangent, to identify angles of movement. This approach integrates all aspects of relativity, ensuring an accurate representation of speed and trajectory for observers in different frames.