Question

a spacecraft has a speed of 0.3c relative to earth. as measured on earth, the elapsed time between two events that happen at the same location on earth is 7 h. determine what the elapsed time between those two events would be from the spacecraft's frame.

Answer 1

The elapsed time between two events from the spacecraft's frame can be determined using the concept of time dilation in special relativity.

The spacecraft's speed of 0.3c relative to Earth causes time to appear slower for the spacecraft. Using the time dilation equation, the elapsed time between the events from the spacecraft's frame is calculated to be approximately 7.34 hours.

Step-by-step explanation:

To determine the elapsed time between two events from the spacecraft's frame, we can use the concept of time dilation in special relativity.

Time dilation states that time appears to run slower for an object moving relative to an observer at rest. In this case, the spacecraft is moving at a speed of 0.3c relative to Earth.

From the spacecraft's frame, the elapsed time between the two events would be longer than the elapsed time measured on Earth.

This can be calculated using the time dilation equation: t' = t / sqrt(1 - v^2/c^2), where t' is the elapsed time in the spacecraft's frame, t is the elapsed time on Earth, v is the spacecraft's velocity, and c is the speed of light.

Using the given information, the elapsed time between the two events from the spacecraft's frame can be calculated as follows:

t' = 7 h / sqrt(1 - (0.3c)^2/c^2)

t' = 7 h / sqrt(1 - 0.09)

t' = 7 h / sqrt(0.91)

t' ≈ 7 h / 0.953

t' ≈ 7.34 h