Special Relativity rocket’s speed

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SUMMARY

The discussion focuses on calculating the speed of a rocket traveling between two planets one light-year apart, with the goal of ensuring that only one year passes for the captain aboard the rocket. The relevant equations include the Lorentz factor (ϒ) and the relationship between distance, time, and velocity, specifically v = d/t' and d' = d/ϒ. The correct approach involves manipulating these equations to avoid complex roots, ultimately leading to the conclusion that the rocket's speed must be v = sqrt(1/2) * c, where c is the speed of light.

PREREQUISITES
  • Understanding of special relativity concepts, including time dilation and the Lorentz factor (ϒ).
  • Familiarity with algebraic manipulation of equations.
  • Knowledge of the relationship between distance, time, and velocity in physics.
  • Basic understanding of the speed of light (c) as a constant.
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of special relativity and their applications in real-world scenarios.

br0shizzle1
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Homework Statement


A rocket flies between two planets that are one light-year apart. What
should the rocket’s speed be so that the time elapsed on the captain’s watch
is one year?

Homework Equations


I have v = d'/t'
d'=d/ϒ
t'=1 year
d=1 light year

The Attempt at a Solution


Will the equations I use vϒ=d/t' but I get complex roots as an answer. What am I doing wrong?
 
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Hello and welcome to PF!

I believe you have it set up correctly. You'll need to show your algebra steps in order for us to see where you are making a mistake.
 
TSny said:
Hello and welcome to PF!

I believe you have it set up correctly. You'll need to show your algebra steps in order for us to see where you are making a mistake.
v(sqrt(1-(v/c)^2)=d/t
v^2(1-(v/c)^2)=d^2/t^2
v^2-v^4/c^2=d^2/t^2
v^2-v^4/c^2-d^2/t^2=0
I used wolfram to solve for roots and it gave back complex numbers.
 
br0shizzle1 said:
v(sqrt(1-(v/c)^2)=d/t
Did you use the correct expression for ϒ?
 
TSny said:
Did you use the correct expression for ϒ?
oops, v^2/(1-(v/c)^2)=d^2/t^2
Should I multiply the LHS by the denominators conjugate?
 
br0shizzle1 said:
oops, v^2/(1-(v/c)^2)=d^2/t^2

EDIT: OK

Should I multiply the LHS by the denominators conjugate?

No, try multiplying both sides by the denominator on the left side.
 
TSny said:
EDIT: OK
No, try multiplying both sides by the denominator on the left side.
Ah! thank you, everything works now.
edit: is there any way I can give your points or something of that regard?
 
Good work! Don't worry about any points.
 
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br0shizzle1 said:
Ah! thank you, everything works now.
edit: is there any way I can give your points or something of that regard?
Hey! Would you mind sharing what you got as a result? I'm having a little trouble with the equation myself.
 
  • #10
Harjot said:
Hey! Would you mind sharing what you got as a result? I'm having a little trouble with the equation myself.
PF custom is to help find answers. So show your work to let us help you where you get stuck.

@br0shizzle1: TSny is too modest -- befitting someone of his (/her?) status in PF. But there is a "Like" link at the lower right in every posting (except your own :) ).
 
Last edited:
  • #11
BvU said:
PF custom is to help find answers. So show your work to let us help you where you get stuck.

@br0shizzle1: TSny is too modest -- befitting someone of his (/her?) status in PF. But there is a "Like" link at the lower right in every posting (except your own :) ).
Fair enough. I also had v(gamma)=d/t' --> (v^2)=((d^2)/(t'^2))*(1-(v/c)^2) --> here I considered d^2/t'^2 to be equal to c^2 because d=1 light year and t we set to 1 year so my equation became --> v^2 = c^2(1-(v^2/c^2)) --> (v^2)/(c^2) = 1-(v^2/c^2) --> ((v^2)/(c^2))+((v^2)/(c^2)) = 1 --> 2((v^2)/(c^2)) = 1 --> ((v^2)/(c^2)) = 1/2 ----> v/c=sqrt(1/2) and then v=sqrt(1/2)*c.

Appreciate the response :)
 
  • #12
And where do you think you have trouble with the equation ?

Oh, and: Hello Harjo, welcome to PF! :)

A little unusual to make your debut tagging onto an existing thread, but your question is clear and it's a start...
 
  • #13
Oh I was just wondering if I had done it right. If my assumption was fair to make.

I realize now I should have put the work down first and then asked for clarification. woopsie daisy
 
  • #14
You're doing fine. And, just so you know: I was surprised by the answer, too ! Shows you're never too old to learn
 
  • #15
Awesome! True enough :w

Thanks a bunch!
 

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