Special relativity problem -- Clock running on a high-speed rocket from the Earth to the Moon

  • #1
arpon
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Homework Statement


An astronaut went to sun from earth by a rocket. According to the clock in the rocket, the traveling time was 5 minute. What was the velocity of the rocket?
( earth is 8 light minute away from sun)

Homework Equations


[itex]t = \frac {t_0}{\sqrt {1 - \frac{v^2}{c^2}}}[/itex]


The Attempt at a Solution


According to the reference frame of earth, the distance between sun and earth is, [itex]L_0 = 8\cdot 60 \cdot c [/itex]
Let, the time required to travel this distance be [itex]t_0[/itex] [according to earth];
So, [itex]t_0 = 5 \cdot 60 \cdot \sqrt {1 - \frac{v^2}{c^2}}[/itex]
and the velocity, [itex] v = \frac {L_0}{t_0} = \frac { 8\cdot 60 \cdot c }{ 5 \cdot 60 \cdot \sqrt {1 - \frac{v^2}{c^2}}} = \frac { 8c }{ 5 \sqrt {1 - \frac{v^2}{c^2}}} [/itex]
[itex]v^2 = \frac {64c^2}{25(1 - \frac{v^2}{c^2})}[/itex]
[itex]v^2 \cdot 25(1 - \frac{v^2}{c^2}) = 64c^2[/itex]
Simplifying, [itex]\frac{25}{c^2} (v^2)^2 - 25 (v^2) +64c^2 = 0 [/itex]
But, the solution of this equation is complex number.
 

Answers and Replies

  • #2
Orodruin
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You are doing the time dilation in the wrong way. ##t_0## is the proper time, i.e., the time for the astronaut and should be smaller than the time in the Earth reference frame.
 
  • #3
arpon
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You are doing the time dilation in the wrong way. ##t_0## is the proper time, i.e., the time for the astronaut and should be smaller than the time in the Earth reference frame.
Look, we are calculating the distance according to earth. So, shouldn't we take the time according to earth?
 
  • #4
Orodruin
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Yes, and you are not. The quantity ##t_0## is the time elapsed for the astronaut.
 
  • #5
arpon
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Yes, and you are not. The quantity ##t_0## is the time elapsed for the astronaut.
I got it. We have to calculate the time interval between two events : 1. The rocket starts from earth , 2. The rocket reaches the sun. And these two events occured at the 'same location' in the reference frame of rocket. So, the time measured from the rocket is to be taken as the proper time.
There is another way to solve the problem. According to earth, both the earth and the sun are stationary (as it is described for this problem), and according to the rocket they are moving. So, the distance between the sun and earth is contracted in the reference frame of the rocket. So, ## L = L_0 \sqrt { 1 - \frac{v^2}{c^2}}##
 
  • #6
Orodruin
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That's more like it. :)
 
  • #7
kumusta
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proper time (time moving with body) ##τ_0## = 5 min ...
apparent length (length not moving with body) L = 8 light-min ...
L = ##\upsilon##τ##~\Leftarrow~##τ = ##τ_0~[~1 - (\upsilon/c)^2~ ]^{-1/2} ## ##~ \Rightarrow## ##~L## = ## \upsilonτ## = ## \upsilonτ_0[~1 - (\upsilon/c)^2~ ]^{-1/2}## = (8 min)c ... ##\Rightarrow~## ## \upsilon##(5 min)##[~1 - (\upsilon/c)^2~ ]^{-1/2}## = (8 min)c ##~\Rightarrow~## 8##[~1 - (\upsilon/c)^2~ ]^{1/2}## = 5(## \upsilon##/c)
##\Rightarrow~64[~1 - (\upsilon/c)^2~ ]## = 64 ##-~##64##(\upsilon/c)^2## = 25##(\upsilon/c)^2## ##~\Rightarrow~## 64 = 89##(\upsilon/c)^2##
##\Rightarrow## ##(\upsilon/c)^2## = 64/89 = 0.719101 ##~\Rightarrow~## ##\upsilon## = (0.848)c
 

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