Special Relativity and Binomial Expansion

  • Thread starter Xkaliber
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  • #1
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Hi all,

I am working on the last part of a problem now in which I am trying to find what velocity (as a fraction of c) must be traveled from the Earth to Andromeda (a distance of 2.00x10^6 light-years) in order for only 20 years to pass in the reference frame of the rocket. I created my equation and know what I need to do to solve it, but not sure where to employ a certain technique to solve it. My equation is:

(time in rocket frame)^2 = (time in earth frame)^2 - (distance in earth frame)^2
(20 years)^2 = (earth time)^2 - (2.00x10^6)^2

I realize that (earth time) = 2.00x10^6 / velocity
However, the book wants me to use 2 binomial approximations of the form (1+z)^n ~ 1+nz to solve the problem. I am having trouble determining where these can be applied.

Thanks
 

Answers and Replies

  • #2
Doc Al
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I'll start you off:
(earth time)^2 = (2.00x10^6)^2 + (20 years)^2
(earth time)^2 = (2.00x10^6)^2[1 + (20 years)^2/(2.00x10^6)^2]

Get the idea? Keep going.
 
  • #3
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(time in rocket frame)^2 = (time in earth frame)^2 - (distance in earth frame)^2
(20 years)^2 = (earth time)^2 - (2.00x10^6)^2
(earth time)^2 = (2.00x10^6)^2 + (20 years)^2
(earth time)^2 = (2.00x10^6)^2[1 + (20 years)^2/(2.00x10^6)^2]
(earth time) = (2.00x10^6)[1 + (20 years)^2/(2.00x10^6)^2]^1/2
(earth time) = (2.00x10^6)[1 + (20 years)^2/2(2.00x10^6)^2] (binomial approximation)
(2.00x10^6) / velocity = (2.00x10^6)[1 + (20 years)^2/2(2.00x10^6)^2]
1/velocity = 1 + 5x10^-11
(1/velocity)^-1 = [1 + (5x10^-11)]^-1
velocity = 1 - 5x10^-11 (binomial approximation)
velocity = 0.99999999995

Thanks Doc!
 

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