Problem 26 - University Physics with Modern Physics

Question

A spacecraft travels along a straight line from Earth to the Moon, a distance of \(3.84 \cdot 10^{8} \mathrm{~m}\). Its speed measured on Earth is \(0.50 c\). a) How long does the trip take, according to a clock on Earth? b) How long does the trip take, according to a clock on the spacecraft? c) Determine the distance between Earth and the Moon if it were measured by a person on the spacecraft.

Step-by-Step Solution

Answer

Answer: a) The trip takes approximately 1.7067 seconds according to a clock on Earth. b) The trip takes approximately 1.477 seconds according to a clock on the spacecraft. c) The distance between Earth and the Moon, if measured by a person on the spacecraft, is approximately \(3.32 \cdot 10^{8}\ \mathrm{m}\).

Step 1: Find the time taken according to a clock on Earth

To find the time taken according to a clock on Earth, we can use the formula: time = distance / speed. We are given the distance between Earth and the Moon as \(3.84 \cdot 10^{8} \mathrm{~m}\) and the speed of the spacecraft as \(0.50 c\), where c is the speed of light (\(3.00 \cdot 10^{8} \mathrm{\ m/s}\)). Plug in the values, and we get: time = \((3.84 \cdot 10^{8} \mathrm{~m}) / (0.50 \cdot 3.00 \cdot 10^{8} \mathrm{\ m/s})\) time = \(2.56 \cdot 10^{8} / 1.5 \cdot 10^{8}\) time = 1.7067 s So, the trip takes approximately 1.7067 seconds, according to a clock on Earth.

Step 2: Calculate the relativistic gamma factor

Before we move on to find the time taken according to a clock on the spacecraft, we need to calculate the relativistic gamma factor. The gamma factor, represented by \(\gamma\), is given by the formula: \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\) Plug in the values, and we get \(\gamma = \frac{1}{\sqrt{1 - \frac{(0.50\cdot 3.00 \cdot 10^{8}\mathrm{\ m/s})^2}{(3.00 \cdot 10^{8}\mathrm{\ m/s})^2}}}\) After simplifying, we get \(\gamma = \frac{1}{\sqrt{1 - 0.25}} = \frac{1}{\sqrt{0.75}} = \frac{1}{0.866} = 1.155\)

Step 3: Find the time taken according to a clock on the spacecraft

Now that we have the relativistic gamma factor, we can find the time taken according to a clock on the spacecraft using the time dilation formula: time on spacecraft = time on Earth / \(\gamma\) Plug in the values, and we get the time on the spacecraft = \(1.7067\ \mathrm{s} / 1.155 = 1.477\ \mathrm{s}\) So, the trip takes approximately 1.477 seconds, according to a clock on the spacecraft.

Step 4: Determine the distance between Earth and the Moon if measured by a person on the spacecraft

To find the distance measured by a person on the spacecraft, we will use the length contraction formula: distance on spacecraft = distance on Earth / \(\gamma\) Plug in the values, and we get the distance on the spacecraft = \((3.84 \cdot 10^{8} \mathrm{~m}) / 1.155 = 3.32 \cdot 10^{8}\ \mathrm{m}\) So, the distance between Earth and the Moon, if measured by a person on the spacecraft, is approximately \(3.32 \cdot 10^{8}\ \mathrm{m}\).

Key Concepts

Time Dilation

When discussing relativity, time dilation is one of the most fascinating effects described by Einstein's theory of special relativity. It refers to the way time is perceived differently by observers moving relative to each other.
In simple terms, when an object is moving at significant speeds close to the speed of light, time appears to slow down for that object compared to an observer who is at rest.
This means that if you were travelling in a speedy spacecraft, time for you aboard would tick slower compared to someone observing you from Earth.

• In our example, the spacecraft is traveling at 0.50 times the speed of light (c).
• The time on Earth was calculated to be approximately 1.7067 seconds for the trip to the Moon.
• Using the time dilation formula: \[\text{time on spacecraft} = \frac{\text{time on Earth}}{\gamma}\]we calculated that time on the spacecraft would be approximately 1.477 seconds.

Time dilation plays a crucial role in understanding how time is not absolute but relative to the observer's state of motion. This effect becomes more pronounced as one approaches the speed of light.

Length Contraction

Length contraction is another intriguing concept in relativity, referring to the shortening of objects in the direction they are moving as perceived by an observer at rest. It only becomes significant at speeds close to the speed of light.
The faster an object moves, the more contracted its length appears to an observer for whom the object is moving.

• In the exercise, the spaceship is traveling at half the speed of light, 0.50c.
• The initial measured distance from Earth to the Moon is given as \(3.84 \cdot 10^{8} \text{ m} \).
• Using the length contraction formula:\[\text{distance on spacecraft} = \frac{\text{distance on Earth}}{\gamma}\]we find that the measured distance by someone on the spacecraft is approximately \(3.32 \cdot 10^{8} \text{ m} \).

This illustrates how space is not absolute; motion affects how distances are perceived, and the solid objects appear to be shorter or squished in the direction of travel at high speeds.

Gamma Factor

The gamma factor, denoted as \(\gamma\), is a key term in understanding relativistic effects like time dilation and length contraction. It ensures we can quantitatively describe how time slows down and how objects contract in length at relativistic speeds.
The formula for the gamma factor is given by:\[\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\]where \(v\) is the velocity of the moving object and \(c\) is the speed of light.

• For the problem at hand, the spacecraft's velocity is 0.50c.
• Substituting the values, we obtained a gamma factor of approximately 1.155.

This factor indicates precisely how much slower time ticks and how much shorter objects appear at high velocities. The greater the speed, the larger the gamma factor, leading to more pronounced relativistic effects.
It's the cornerstone for making predictions about the behavior of objects as they approach the speed of light, ensuring we can comprehend and calculate the altered perceptions of time and length.