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Sirius is about 9.0 light-years from Earth. (a) To reach the star by spaceship in 12 years (ship time), how fast must you travel? (b) How long would the trip take according to an Earth-based observer? (c) How far is the trip according to you?

Short Answer

Expert verified
(a) 0.8c, (b) 15 years, (c) 7.2 light-years.

Step by step solution

01

Understand the basics

Let's start by identifying the key concept here, which is special relativity and time dilation. We are given the distance to Sirius as 9.0 light-years and the time experienced by the traveler (ship time) as 12 years. We need to find the required speed.
02

Using the relativistic time dilation formula

According to special relativity, the time dilation formula is given by \( t = \gamma \cdot t_0 \), where \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \), \( t_0 \) is the proper time (12 years), \( t \) is the dilated time (Earth time), and \( v \) is the velocity of the spaceship. We need to find \( v \). Since the distance \( d = v \cdot t_0 \) comes out to 9 light-years, we rearrange it to find \( v \).
03

Solve for velocity

From the equation \( d = v \cdot t_0 \), substitute the values to get \( 9 = v \cdot 12 \). This leads to \( v = \frac{9}{12} = 0.75 \) light-years/year. However, this is underestimating since time dilation affects Earth time. Therefore the formula needs to be rearranged considering time dilation formula \( \gamma = \frac{t}{t_0} = \frac{9}{12} \), finding \( v \) using \( v = c \cdot \sqrt{1 - \frac{1}{\gamma^2}} \).
04

Calculate Earth observer time

Let's further use the relation \( t = \gamma \cdot t_0 \) where \( t \) is the dilated time for an Earth observer. With \( \gamma = 1.25 \) (from step 3 calculations), \( t = 1.25 \times 12 = 15 \) years, which is the time an Earth observer would measure.
05

Calculate distance according to ship

Finally, find the distance according to the spaceship. This distance is a result of length contraction, calculated by \( d' = \frac{d}{\gamma} = \frac{9}{1.25} = 7.2 \) light-years.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Time Dilation in Special Relativity
In the world of special relativity, time isn't as straightforward as we experience daily. Albert Einstein's theory introduced the concept of time dilation, which suggests that time moves slower for an object in motion relative to a stationary observer. This means if you're zooming across space at a significant fraction of the speed of light, your onboard clock would tick slower compared to an observer on Earth.
For instance, in our Sirius star exercise, the traveler aboard the spaceship perceives the journey in a timeframe of 12 years, called "proper time." Meanwhile, those observing the trip from Earth would calculate a longer duration due to time dilation. This difference stems from the formula:
  • \[ t = rac{t_0}{ ext{Time Dilation Factor }} = rac{12}{ rac{ ext{Dilated Time on Earth}}{ ext{Ship Time}}} \]
This equation highlights that time aboard the ship doesn’t slow down to keep pace with Earth—Earth watches it from a more extended clock, so to speak.
Understanding time dilation helps learners appreciate why space travelers experience shorter "trip times" even though more time passes back on Earth. It's a captivating outcome of traveling at relativistic speeds.
Length Contraction Explained
Length contraction is another fascinating outcome of Einstein's theory of special relativity. It suggests that the length measured by a moving observer is shorter than the length measured by a stationary observer. It's as if objects in motion shrink in the direction of the movement.
For the Sirius-bound traveling spaceship, this contraction phenomenon decreases the perceived journey length from 9 light-years (Earth's measure) to approximately 7.2 light-years for the traveler. This calculation utilizes the formula for length contraction:
  • \[ d' = rac{d}{ ext{Length Contraction Factor}} = rac{ ext{Original Earth Distance}}{ ext{Observer's Contraction Difference}} \]
Here, the Earth's full length measurement is compressed, reflecting the spacecraft's perception as it hurtles through space.
This contraction isn’t something you'd notice in everyday travel, as the speeds required to create noticeable effects are near lightspeed. Yet, understanding length contraction helps bridge the perception gap between stationary and moving frames.
Relativistic Velocity and Its Role
In relativity, as objects approach the speed of light, their velocities interact differently with spacetime compared to everyday speeds. Relativistic velocity calculations are central to predicting time dilation and length contraction outcomes.
Our exercise showcases using relativistic velocity to find how fast the ship must travel to appear 12 years on the ship’s clock while Sirius is 9 light-years away. We initially calculated a speed of 0.75 light-years/year using Newtonian physics, but relativistic considerations refine this further. We adjust for what is observed on Earth using:
  • \[ v = c \cdot \sqrt{1 - \frac{1}{ ext{Lorentz Factor}^2}} \]
This formula accounts for distortions in time and length, showing the necessary adjustments for accurate predictions.
You've probably gathered this isn’t the sort of math used for casual commuting—these adjustments are vital because they reflect how velocities aren’t merely added but warp space and time around them, emphasizing the complex relationship between speed and relativity.

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Most popular questions from this chapter

Two spaceships, \(A\) and \(B\), each with a speed of \(0.60 c\) relative to Earth, approach each other head on. (a) The speed of ship A relative to ship \(B\) is (1) greater than \(c,(2)\) equal to \(c,(3)\) less than \(c\). Why? (b) What is the speed of ship A relative to ship \(\mathrm{B} ?\) (c) What is the speed of ship B relative to ship A?

A spaceship containing an astronaut travels at a speed of \(0.60 c\) relative to a second inertial observer. (a) Who measures proper time intervals in the ship and the proper length of the ship: (1) the astronaut in the ship, (2) the second observer, or (3) neither? (b) How much time does a clock on board the spaceship appear to lose in a day, according to the second observer? (c) If the second observer measures a length of \(110 \mathrm{~m}\) for the ship, what is its "proper" length"? (d) What is the total energy of the astronaut according to the astronaut if her mass is \(70 \mathrm{~kg} ?\) (e) Repeat part (d) from the viewpoint of the second inertial observer.

The Sun's mass is \(1.989 \times 10^{30} \mathrm{~kg}\) and it radiates at a rate of \(3.827 \times 10^{23} \mathrm{~kW}\). (a) Over time, must the mass of the Sun (1) increase, (2) remain the same, or (3) decrease? (b) Estimate the lifetime of the Sun from this data, assuming it converts all its mass into energy. (c) The actual lifetime of the Sun is predicted to be much less than the answer to part (b), even though its energy emission rate will remain constant. What does this tell you about the \(100 \%\) conversion assumption? (d) Theoretical calculations predict the Sun's lifetime (in its current stage) to be about 5 billion years. During that time, what percentage of its mass will it lose?

The distance to Planet X from Earth is 1.00 light-year. (a) How long does it take a spaceship to reach \(X\), according to the pilot of the spaceship, if the speed of the ship is \(0.700 c\) relative to \(X ?\) (b) How long does it take the ship to make the trip according to an astronaut already stationed on Planet \(X ?\) (c) Determine the distance between Earth and Planet \(X\) according to the pilot and according to the X-based astronaut and explain why the two answers are different.

Roughly speaking, the observable mass in our universe is all attributed to stars and gas clouds in the galaxies. (a) Assuming that each galaxy contains the mass of 200 billion Suns and there are 200 billion such galaxies, what is the Schwarzschild radius of the universe? (b) Modern observations indicate that there is much more mass in the universe than can be "seen," in the form of "dark matter," neutrinos, etc. Suppose that the actual mass of the universe was 100 times larger than the visible mass. What would the Schwarzschild radius be under those conditions? (c) Current observations place the lifetime of the universe at about 13 billion years. Compare the distance light can travel in this time to your answer from part (b). Can you conclude anything about the universe itself being a black hole?

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