A spaceship is traveling at 0.8c. An observer on Earth measures the speed of light from the ship’s headlight to be
1.6c
0.8c
0.2c
1.0c
The correct answer and explanation is:
The correct answer is 1.0c.
In special relativity, the speed of light in a vacuum is always constant, regardless of the relative motion between the observer and the light source. This principle is one of the foundational postulates of Einstein’s theory of special relativity.
Even though the spaceship is traveling at 0.8c (80% the speed of light) relative to Earth, the observer on Earth will still measure the speed of light emitted by the spaceship’s headlight as 1.0c. This is because the speed of light does not add or subtract based on the motion of the source, which contrasts with how velocities in everyday situations combine.
To explain this in more detail: Suppose two objects are moving relative to each other, and one of them emits light. In classical mechanics, if the spaceship were moving towards an observer at 0.8c, you might think the speed of the light emitted from the headlight would be the sum of the spaceship’s speed and the speed of light (i.e., 0.8c + c = 1.8c). On the other hand, if the spaceship were moving away from the observer, you might think the speed of light would be reduced (c – 0.8c = 0.2c).
However, special relativity changes this. The speed of light remains constant at 1.0c for any observer, regardless of the motion of the source or the observer. This is a result of how space and time are interrelated in special relativity: as the spaceship approaches the speed of light, time and space adjust in such a way that the speed of light remains constant.
Thus, even though the spaceship is moving at a significant fraction of the speed of light, the observer on Earth will still measure the speed of light from the spaceship’s headlight as 1.0c, in accordance with the invariance of the speed of light.