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How long does it take to get to the nearest star? 
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#1
Jun1211, 01:00 PM

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1. The problem statement, all variables and given/known data
People hoping to travel to other worlds are faced with huge challenges. One of the biggest is the time required for a journey. The nearest star is 4.1 x 10^16 m away. Suppose you had a spacecraft that could accelerate at 1.5g for half a year, then continue at a constant speed. (This is far beyond what can be achieved with any known technology.) How long would it take you to reach the nearest star to earth? 2. Relevant equations Kinematic Equations, primarily: Delta(x) = v(i)t + 1/2(a(x))t^2 3. The attempt at a solution Never got that far after a long time of attempting to work it out. Try not to provide me with the answer but just some hint or way to go about it if thats possible. Thanks so much. 


#2
Jun1211, 01:04 PM

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Try solving the problem in two steps:
1.How far do you travel while accelerating? 2. How much distance is left to cover? How long does this take going at that constant speed? 


#3
Jun1211, 01:08 PM

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P: 11,675

Hi frostedpoptar. Welcome to Physics Forums.
You should try to write out what you think might be the relevant equations, and show an example of a calculation that you attempted. Your problem statement seems to be missing some information (numerical value for distance, units for acceleration) that could make a big difference in how the problem is to be solved. 


#4
Jun1211, 01:48 PM

P: 15

How long does it take to get to the nearest star?
G01, I was attempting to do that yesterday and it didn't work out. I erased it from my whiteboard but I will attempt to do it again and post my results.
gneill, You're completely write, I left some things out in the main post, I just edited it. Thanks for noticing. I'll attempt to work it out again and reply as soon as I do. 


#5
Jun1211, 01:49 PM

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#6
Jun1211, 01:58 PM

P: 15

I just tried to work it out partially.
I used v(f) = v(i) + a*t As v(f) = 14.7(1314000) so I got v(f) as 19,315,800 m/s 


#7
Jun1211, 02:00 PM

P: 15

I also plugged that into
v(f)^2 = v(i)^2 + 2a(delta x) and got delta x as 1.26904806 x 10^13 meters. 


#8
Jun1211, 02:04 PM

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#9
Jun1211, 02:07 PM

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Δx = v_{i}t + (1/2)at^{2} Since you already know a, you already computed t, and you know that v_{i} = 0. Either way, when you substitute in quantities like v_{f} or t that were computed in intermediate steps, be sure to carry over a lot of digits in order to avoid accumulation of error. You can round off to the correct number of sig figs when you get your final answer at the end. 


#10
Jun1211, 02:09 PM

P: 15

Gotcha.
I plugged in the new value (15,768,000 s) and got these values for v(f) and delta x: v(f) = 231,789,600 m/s delta x = 1.827429206 x 10^15 


#11
Jun1211, 02:13 PM

P: 15

I got 5.3589 years as the time it takes from the moment the spacecraft stops accelerating to the moment it reaches the star..



#12
Jun1211, 02:13 PM

P: 15

I got it! its 5.9 when you factor in the .5 year of acceleration.
Thanks so much guys. 


#13
Jun1211, 06:16 PM

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Thanks
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What about stopping?



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