Traveling to Planet X in 23 Years: An SR Challenge

[Athenian]

Homework Statement

Problem:
The planet X is far 48 light-years from Earth. Suppose that we want to travel from Earth to planet X in a time no more than 23 years, as reckoned by clocks aboard our spaceship. At what constant speed would we have to travel? How long would the trip take as reckoned by clocks on Earth?

Relevant Equations

These are the equations I used. However, there are the (what feels like hundreds of) equations out there in the SR course that I find difficult knowing which one to use. In the end, I decided the below two were my best bets in solving the assignment question.

Einstein's Equation of Time Dilation:
$$\Delta t = \frac{\Delta t_0}{(1-(\frac{v^2}{c^2}))^\frac{1}{2}}$$
&
Einstein's Equation of Length Contraction:
$$L = L_0 (1-(\frac{v^2}{c^2}))^\frac{1}{2}$$

Homework Statement: Problem:
The planet X is far 48 light-years from Earth. Suppose that we want to travel from Earth to planet X in a time no more than 23 years, as reckoned by clocks aboard our spaceship. At what constant speed would we have to travel? How long would the trip take as reckoned by clocks on Earth?
Homework Equations: These are the equations I used. However, there are the (what feels like hundreds of) equations out there in the SR course that I find difficult knowing which one to use. In the end, I decided the below two were my best bets in solving the assignment question.

Einstein's Equation of Time Dilation:
$$\Delta t = \frac{\Delta t_0}{(1-(\frac{v^2}{c^2}))^\frac{1}{2}}$$
&
Einstein's Equation of Length Contraction:
$$L = L_0 (1-(\frac{v^2}{c^2}))^\frac{1}{2}$$

To solve this, I decided to first calculate the distance between plant X and Earth (i.e. 48 light-years) in meters.

Therefore, considering that "c" is 299,792,458 m/s, I used that number to calculate the distance light would travel in 48 years. Once calculated, I continued by calculating what the distance of 23 light-years would be like using the given number of "c". Afterward, I plugged these numbers into Einstein's equation of length contraction to find the needed velocity for the spaceship to travel the distance of "48 light-years" and contract it to 23 light-years distance.

However, in the end, the velocity was - oddly - a little over 12 kilometers per second. Having a difficult time believing the number, I checked the internet and found that the Voyager 2 once traveled ~9.7 miles per second before. Thus, I am fairly certain that it is impossible for the spaceship in the question to go sailing through 48 light-years worth of space distance in 23 years through the aid of length contraction.

In short, I feel like either I went in a completely wrong direction in solving this question or I'm "close to the mark" but just not quite getting the process in solving for the answer.

If there's anybody on the forum that could kindly assist me in understanding the process to navigate through the special relativity problem, it would be much appreciated. Of course, if this question would be more appropriate if posted on the dedicated special relativity forum here on this website, please let me know and I'll be sure to post the question there instead.

Thank you!

 

[BvU]

Hello Athenian, $\qquad$ $\qquad$ !

You want to solve $$
L = L_0 (1-(\frac{v^2}{c^2}))^\frac{1}{2}
$$ for $v$. Note that $${L \over L_0} = \left (1-\left (\frac{v^2}{c^2}\right) \right)^\frac{1}{2}$$ is dimensionless. You want $v>0$ so $L$ is in the ship frame and $L_0$ is in the Earth frame.
Units don't matter, as long as they are the same: meters, furlongs, light years -- all that matters is their ratio.

in the end, the velocity was - oddly - a little over 12 kilometers per second

I wonder how you get that. Please show detailed work...

[late edit] I am making a BIG mistake here -- see @mjc123 correct me in #9 below !

 

[Abhishek11235]

Velocity is Contracted length over time.

 

[Athenian]

Hello Athenian, $\qquad$ $\qquad$ !

You want to solve $$
L = L_0 (1-(\frac{v^2}{c^2}))^\frac{1}{2}
$$ for $v$. Note that $${L \over L_0} = \left (1-\left (\frac{v^2}{c^2}\right) \right)^\frac{1}{2}$$ is dimensionless. You want $v>0$ so $L$ is in the ship frame and $L_0$ is in the Earth frame.
Units don't matter, as long as they are the same: meters, furlongs, light years -- all that matters is their ratio.

I wonder how you get that. Please show detailed work...

Hello BvU,

Thank you very much for your assistance!
I have gone ahead and recalculated my work as seen below:

Considering that units do not matter when I am calculating for the ratio, below is my calculation:

$$\frac{23 light-years}{48 light-years} = \left(1 - \frac{v^2}{299,792,458 \frac{m}{s}}\right)^\frac{1}{2} \Rightarrow v = \pm 15,197.36495 \frac{m}{s}$$

While my initial calculation of around 12 km was off, I still have trouble seeing how this is the solution to the problem (i.e. At what constant speed would the spaceship have to travel from Earth to Planet X within 23 light-years [spaceship time]?). Would the answer simply be 15,197.36495 m/s? Shouldn't the number "v" be much larger? Or am I missing a logical piece to properly answer the question?

Regardless, I sincerely appreciate your assistance. And, provided that you are available, it would be incredibly helpful if you - or anyone else reading the thread - could assist me further and help answer some of my confusion. Thank you!

 

[Athenian]

Velocity is Contracted length over time.

Thank you for your reply and assistance!

If I am thinking through this correctly, would the contracted length be taking the 23 light-years and getting the distance of that by performing the following calculation?

$$ 23 light years = 299,792,458 \frac{m}{s} \cdot 60 \cdot 60 \cdot 24\cdot 365\cdot 23 \approx 9.45425 \times 10^{15}$$

Therefore, since 23 light-years is equivalent to 9.45425 * 10^15 m, would I then take this figure and divided it by time (i.e. 23 years)?

I feel like I'm committing a logical fallacy here considering that the time (i.e. 23 years) can be canceled out the 23 in the calculation above. Please let me know where in my thought process has gone wrong. And, if possible, I would appreciate any advice to guide me in the right direction. Thank you!

 

[Abhishek11235]

Thank you for your reply and assistance!

If I am thinking through this correctly, would the contracted length be taking the 23 light-years and getting the distance of that by performing the following calculation?

$$ 23 light years = 299,792,458 \frac{m}{s} \cdot 60 \cdot 60 \cdot 24\cdot 365\cdot 23 \approx 9.45425 \times 10^{15}$$

Therefore, since 23 light-years is equivalent to 9.45425 * 10^15 m, would I then take this figure and divided it by time (i.e. 23 years)?

I feel like I'm committing a logical fallacy here considering that the time (i.e. 23 years) can be canceled out the 23 in the calculation above. Please let me know where in my thought process has gone wrong. And, if possible, I would appreciate any advice to guide me in the right direction. Thank you!

Don't care about units for now. They are human artifacts. Coming to your question,let's solve in the frame of spaceship. For him length is contracted. Therefore:

$$ v= L/\gamma t \Rightarrow (1-v^2/c^2)^{1/2} L/t$$
I urge you to do this problem in some other frame as I have done in frame in spaceship.
What answer you get using this?

 

[RPinPA]

My approach is "first do the physics, then solve the equations".

It depends on the point of view. That's the key to relativity, keeping straight whose viewpoint we are analyzing.

First let's talk about the number $\gamma = 1 / \sqrt {1 - (v/c)^2}$ that pops up in all of these equations. This measures the size of the relativistic effect. When $v$ is very small, ordinary velocities that are a tiny fraction of $c$, then $\gamma$ is 1.0 or so close to it you can't measure the difference without very sensitive instruments. When $v$ gets close to $c$, the expression $1 - (v/c)^2$ starts decreasing toward 0, and so $\gamma$ starts increasing toward $\infty$.

If $\gamma = 3$, that means lengths are contracted by a factor of 3 and times are dilated by a factor of 3, and total energy $E = \gamma mc^2$ is increased by a factor of 3 over the rest energy $E = mc^2$.

All of those statements have to do with how one observer measures things as opposed to how another observer moving at speed $v$ measures them.

If $\gamma$ is close to 1, then ordinary non-relativistic physics applies. Even if you're going 1/10 of the speed of light, $\gamma = 1.005$ only a little different from 1.0, meaning that the relativistic effects are only half a percent different fro the non-relativistic description. My example of $\gamma = 3$ requires a velocity of 94% of the speed of light.

The key lesson: $\gamma$ much different from 1 means you're going to be moving pretty close to $c$.

I am fairly certain that it is impossible for the spaceship in the question to go sailing through 48 light-years worth of space distance in 23 years through the aid of length contraction.

You're right from the point of view of earth. The ship can not move faster than light. So from the point of view of earth, it can't do the trip in less than 48 years. Since $v \lt c$, it's going to be more than 48 years. And the distance covered is going to be 48 light years. But that's not what's being asked.

From the point of view of the ship, the star is approaching at $v$. The distance to the star is therefore contracted. The Earth says the ship traveled 48 light years, which obviously you can't do in 23 years. The ship says the star was only $48/\gamma$ light years away, which could be a lot less than 48 light years. Also the ship says the star is approaching at a speed close to $c$ (I know that just by inspection, because the difference between 48 and 23 is a lot more than 1). So that trip can easily be done in in 23 years.

In fact just by inspection I can say that I want the $\gamma$ factor to be about 48/23. Since I know the speed is going to be close to $c$, then a 23 year trip (from the point of view of the ship) means the distance is about 23 light years. So the contraction factor is approximately $\gamma = 48/23$.

Use that to solve for $v/c$. Approximately.

Now I want to emphasize that this is an approximation. This $\gamma$ is close to 2, so based on the calculations I did above $v$ is going to be less than 94% of the speed of light. You'd do this to get a quick estimate, an idea of what the final answer should be, a guideline and a sanity check to the more careful calculation. But actually the Earth says the trip takes a bit more than 48 years, so the real gamma factor is $\gamma = (\text{(a bit more than 48)}) / 23$..

Here's a more careful calculation. The Earth says the ship travels at $v$ and takes time $T_0 = L_0/v$. We don't know what this is yet.

The ship sees the star approaching at the same $v$. So it says the distance to the star is $L = L_0/\gamma$. And therefore it says the time is $L/v = L_0/\gamma v$. And this is what you want to be equal to 23 years. Since we know $L_0$, then $v$ is the only unknown here and it can be solved for $v$.

You've calculated $v$ to be about 15000 km/s. The speed of light is 30000 km/s. So you're getting an answer of about $0.5c$, which certainly is in the right order of magnitude.

But please don't keep 10+ nonzero digits in your answer. You don't have that many significant figures.

 

[Athenian]

Hello RPinPA and Abhishek11235,

Thank you so much for your clarifications and clear guidance! I'll definitely be sure to calculate my answers using the methods and ideas above. However, considering that it's past 1 in the morning here, I'll be sure to give it a try and properly follow up with your replies by presenting my solutions or questions I may have on this thread tomorrow.

Once again, I sincerely appreciate your guys' help and I'll follow up with the replies soon.

Lastly, thanks for the reminder, RPinPA. I'll be sure to not include that many digits next time considering there aren't that many significant figures in my equation.

 

[mjc123]

Your major error (in post #4) is using c where you should use c². Your other error is assuming L = 23 light years, which it is not, unless the ship is traveling at c.
To simplify things, take v in light years per year (in which units c = 1). Then the distance L is 23v, and
23v/48 = (1-v²)^½

 

[jbriggs444]

Velocity is Contracted length over time.

Velocity is properly measured as plain old length divided by plain old time. You pick a frame of reference. You use that frame of reference to decide how much distance is covered and you use that frame of reference to decide how much time has elapsed. You divide the one by the other. The result is a velocity.

Relativity does not change that procedure.

 

[Athenian]

First off, thank you everybody for being so incredibly helpful in aiding me to understand how to solve this problem!

With that said, below is my attempted solution. If possible, I would sincerely appreciate it if the community could let me know if I made any errors here or if my solution is indeed correct. Thank you for your time and assistance!

Your major error (in post #4) is using c where you should use c². Your other error is assuming L = 23 light-years, which it is not, unless the ship is traveling at c.
To simplify things, take v in light years per year (in which units c = 1). Then the distance L is 23v, and
23v/48 = (1-v²)^½

Following your clear instruction mjc123, I went ahead and calculated for $v$ using the given equation above, which gave me $v = 0.90$.
Therefore, the spaceship's velocity will be 0.90 times the speed of light $c$ - where $c$ is equal to 1.

With the given magnitude of velocity in mind, I decided to go ahead and calculate for $\gamma$ using the following equation given by RPinPA:

$\gamma =\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$

Plugging in the numbers, I get $\Rightarrow$ $$\frac{1}{\left(1-\left(\frac{\frac{48}{\sqrt{2833}}}{1}\right)^2\right)^{\frac{1}{2}}} \approx 2.3$$

Since $\gamma = 2.3$, this answer directly corresponds to the below helpful quote - and "sanity check" - provided by RPinPA:

This $\gamma$ is close to 2, so based on the calculations I did above $v$ is going to be less than 94% of the speed of light.

Therefore, with the following data, I concluded that $v = 0.9c$ or $v = 270,000 \frac{km}{s}$.

And, continuing on, I performed the following operation to find the time that'll pass on Earth for the spaceship to get to Planet X $\Rightarrow T_0 = \frac{L_0}{v} = \frac{48}{0.9} \Rightarrow 53.\bar{3} \approx 53$ earth-years.

To double-check my answer, I tried to find $\gamma$ again using the following mathematical operation $\gamma = \frac{53.\bar{3}}{23} \approx 2.3$ which completely corresponds to my previous calculation of $\gamma$.

To wrap up by referring back to the question:
1. The spaceship will need to travel at a constant speed of $270,000 \frac{km}{s}$ to arrive at Planet X in no more than 23 (spaceship) years.
2. The trip from Earth to Planet X for the spaceship will take a total 53 earth-years.

With that said, is the solution correct? If not, is there anywhere I should correct to obtain the correct answer?

Once again, thank you all for your insightful assistance. In addition, thank you mjc123 for pointing out my major error in post #4. That was a careless - and tired - mistake on my part. I went ahead and corrected it in this post. Thanks for kindly pointing it out.

 

[mjc123]

I agree with your calculations.

 

[RPinPA]

Yep, I got $0.9c$ as well. You don't absolutely need to do all those other things such as calculate the time in the Earth frame, but it's always a good idea to do a sanity check and see if everything is consistent.

And I made a brainless remark earlier, saying that $c$ is 30000 km/s and therefore 15000 km/s is $0.5c$. That's off by a factor of 10 of course. However, using the approximation $\gamma=48/23$ I do get a value of $v$ close to $0.5c$. So you can see that since the exact answer is $0.9c$ the approximation is not really that good after all.

 

[Athenian]

Great! Thank you both for the confirmation and help!
I sincerely appreciate your explanations and efforts!