An astronaut travelling at 0.70c relative to Earth uses his watch to measure the time between two radio signals arriving from a distant satellite. He measures the time as 28 min. C... An astronaut travelling at 0.70c relative to Earth uses his watch to measure the time between two radio signals arriving from a distant satellite. He measures the time as 28 min. Calculate the time difference between the signals as measured by another observer at rest on Earth. Read more

Steps to Solve

1. Identify the time measured by the astronaut The astronaut measures the time between the two radio signals as $t' = 28$ minutes.

2. Understand the time dilation formula In special relativity, the time dilation can be calculated with the formula: $$ t = \frac{t'}{\sqrt{1 - \frac{v^2}{c^2}}} $$ where

• $t$ is the time measured by an observer at rest (on Earth),
• $t'$ is the time measured by the moving observer (the astronaut),
• $v$ is the velocity of the moving observer,
• $c$ is the speed of light.

3. Substitute the given values Here, the velocity of the astronaut $v = 0.70c$ and $t' = 28$ minutes. We first calculate the denominator: $$ \sqrt{1 - \frac{(0.70c)^2}{c^2}} = \sqrt{1 - 0.49} = \sqrt{0.51} $$

4. Calculate the time measured by the Earth observer Now, substituting the values into the time dilation formula: $$ t = \frac{28 , \text{min}}{\sqrt{0.51}} $$

5. Find final numerical value Now compute the value of $t$: $$ t = \frac{28}{\sqrt{0.51}} \approx \frac{28}{0.714} \approx 39.2 , \text{min} $$

6. Calculate the time difference To find the time difference between the signals as measured on Earth, we need to subtract the astronaut's time from the Earth observer's time: $$ \Delta t = t - t' = 39.2 , \text{min} - 28 , \text{min} \approx 11.2 , \text{min} $$