Problem 73 - University Physics with Modern Physics

Question

Two stationary space stations are separated by a distance of \(100 .\) light- years, as measured by someone on one of the space stations. A spaceship traveling at \(0.950 c\) relative to the space stations passes by one of the space stations heading directly toward the other one. How long will it take to reach the other space station, as measured by someone on the spaceship? How much time will have passed for a traveler on the spaceship as it travels from one space station to the other, as measured by someone on one of the space stations? Round the answers to the nearest year.

Step-by-Step Solution

Answer

Answer: It will take approximately 105 years for the spaceship to reach the other space station, as measured by someone on the spaceship. For a traveler on the spaceship, 337 years will have passed as measured by someone on one of the space stations during the journey.

Step 1: Identify the given values

We're given: - Distance between the space stations (D): 100 light-years - Speed of the spaceship (v): 0.950 c

Step 2: Calculate the time it takes to reach the other space station, as measured by someone on the spaceship

We can use the formula for time as: \(time = \frac{distance}{speed}\) which can be rewritten as: \(t_{spaceship} = \frac{D}{v}\) Then, \(t_{spaceship} = \frac{100}{0.950 c}\) \(t_{spaceship} = \frac{100}{0.950} \,\text{years}\) \(t_{spaceship} \approx 105 \,\text{years}\)

Step 3: Calculate time dilation factor

We can find the time dilation factor using the formula: \(\gamma = \frac{1}{\sqrt{1- \frac{v^2}{c^2}}}\) where \(c\) is the speed of light. \(\gamma = \frac{1}{\sqrt{1- \frac{(0.950c)^2}{c^2}}}\) \(\gamma = \frac{1}{\sqrt{1- 0.9025}}\) \(\gamma \approx 3.203\)

Step 4: Calculate the time that passes for a traveler on the spaceship, as measured by someone on the space station

We can use the time dilation factor to find the time that passes for a traveler on the spaceship (\(t_{station}\)) as: \(t_{station} = \gamma \times t_{spaceship}\) \(t_{station} = 3.203 \times 105\,\text{years}\) \(t_{station} \approx 337\,\text{years}\) So, it will take approximately 105 years for the spaceship to reach the other space station, as measured by someone on the spaceship, and 337 years will have passed for a traveler on the spaceship as it travels from one space station to the other, as measured by someone on one of the space stations.

Key Concepts

Time Dilation

Time dilation is a fascinating concept from Einstein's theory of relativity. It describes the effect of time moving at different rates depending on the relative speed of observers. In simpler terms, if you are moving very fast, close to the speed of light, time for you (in your spaceship) seems to move slower compared to someone who is stationary. This is one of the strange but fascinating outcomes of moving at such high speeds.

• For our spaceship problem, someone on the spaceship will experience 105 years passing.
• Meanwhile, someone at the stationary space station observes the journey taking 337 years!

This difference is quite significant and shows how time dilation can greatly affect time perception at high velocities.

Reference Frames

A reference frame is like an invisible platform or viewpoint from which an observer measures and describes physical events. Each observer can have their own reference frame, meaning they may see the same events differently depending on their motion. In our exercise:

• The people on the spaceship have a moving reference frame.
• The space station has a stationary reference frame.

Different structures because of their distinct reference frames lead to the question: how much time actually passes? Understanding different reference frames helps to clarify that time and space can be measured differently relative to where you stand. It's a fundamental concept to see why two observers may not agree on the duration of a physical event.

Lorentz Factor

The Lorentz factor is a crucial part of relativity that helps calculate how much time dilation occurs. It can be thought of as a correction factor needed when speeds approach that of light to adjust measurements of time and space.The formula for the Lorentz factor, denoted as \(\gamma = \frac{1}{\sqrt{1- \frac{v^2}{c^2}}} \)shows how it relates the velocity of the moving object (\(v\)) to the speed of light (\(c\)).

• In our exercise, the spaceship travels at 0.950c, and we found the Lorentz factor to be approximately 3.203.
• This means the onboard measurements of time and space need to be adjusted by this factor when viewed from the stationary space station.

The Lorentz factor hence acts as a bridge, enabling us to connect and translate observations between moving and stationary observers.