Spacecraft Question: How Fast to Travel for 2 Days on Earth?

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To determine the speed a spaceship must travel for one day onboard to equal two days on Earth, the concept of time dilation is applied using the Lorentz factor, gamma (γ). The equation γ = Earth time / spaceship time indicates that γ equals 2 in this scenario. By substituting this value into the relativistic equation for gamma, the calculation for velocity (v) can be simplified. However, care must be taken to avoid yielding a negative value when solving for v, as this indicates a mathematical error. The discussion emphasizes the importance of using the correct relativistic equations to ensure accurate results.
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How fast must a spaceship be traveling relative for one day on the spacecraft to equal 2 days on Earth? I'm pretty sure that t(0) = 172,800 seconds, and t = 86,000 seconds. Do I just need to use ∆t = t – t(0), simplify that to t = γt(0), and then since γ = 1+ (1/2)(v^2)/(c^2); take the resulting equation and just resolve for v?
 
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If I remember correctly, you will want to use the relativistic equation for gamma and not the approximation for speeds much lower than the speed of light. So,
gamma = 1/(sqrt(1-(v^2/c^2))) Otherwise, using the aproximation will yeild a speed faster than the speed of light which is not possible.
 
I'm ending up with v = sqrt{(((t(0)^2)(c^2))/(∆t^2))-c^2} Can someone please check my math?
 
Close, it may be easier to notice that gamma = (earth time) / (spaceship time)

So, gamma = 2 in your case. Should be easy from here . . .
 
How does gamma equal Earth time/ spaceship time or 2? Is this just though simplification?
 
So I can just set gamma equal to 2, and solve for v!
 
Yes, that should do the trick.
 
Thanks a bunch Olaf.
 
If gamma =2, then 2 = 1/(1+(v^2)/(c^2)), thus leaving you with a negative value when you simplify for v.
 
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