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 E 16 m away. Suppose you had a spacecraft that could accelerate 1 g for half a year, then continued at constant speed. How long would it take you to reach the nearest star to Earth? I really need help solving this problem. I am stumped. I have been working on it for hours!! :/
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Given the fact that classically the spacecraft would ultimately have a velocity
(v = g t) of approximately 154,000 km/s , which is about half the speed of light, I will nevertheless assume that the problem asks for a classical solution, not a relativistically correct one...
Then the solution is:
after 0.5 y ( 1.57 * 10^6 seconds) the velocity will be
v = g t = 9.81 m/s^2 * 1.57*10^6 s = 1.54 * 10^8 m/s
In that half year, already a distance will be covered of:
s = 1/2 g t^2
= 1/2 * 9.81 m/s^2 * (1.57*10^6 s s)^2
= 1.21 * 10^13 m
The remaining distance then is
4.1 * 10^16 m - 1.21 * 10^13 m = 4.1 * 10^16 m
(at the given level of accuracy, the distance covered in the acceleration phase can be ignored!)
The additional time it takes (in the uniform speed phase) is:
t = s / v
= 4.1 * 10^16 m / ( 1.54 * 10^8 m/s)
= 2.6 * 10^8 s
=2.6 * 10^8 s / ( 3.15 * 10^6 s/year)
= 8.2 years
So 8.7 years for the total journey .
(v = g t) of approximately 154,000 km/s , which is about half the speed of light, I will nevertheless assume that the problem asks for a classical solution, not a relativistically correct one...
Then the solution is:
after 0.5 y ( 1.57 * 10^6 seconds) the velocity will be
v = g t = 9.81 m/s^2 * 1.57*10^6 s = 1.54 * 10^8 m/s
In that half year, already a distance will be covered of:
s = 1/2 g t^2
= 1/2 * 9.81 m/s^2 * (1.57*10^6 s s)^2
= 1.21 * 10^13 m
The remaining distance then is
4.1 * 10^16 m - 1.21 * 10^13 m = 4.1 * 10^16 m
(at the given level of accuracy, the distance covered in the acceleration phase can be ignored!)
The additional time it takes (in the uniform speed phase) is:
t = s / v
= 4.1 * 10^16 m / ( 1.54 * 10^8 m/s)
= 2.6 * 10^8 s
=2.6 * 10^8 s / ( 3.15 * 10^6 s/year)
= 8.2 years
So 8.7 years for the total journey .
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EDIT***********
Dr Zorro and I both made a mistake. Below is the corrected solution:
1g = 9.8 m/s^2
half a year = 1/2(365*24*60*60) = t = 15,768,000s = 1.58 E7 s
The spacecraft speed after half a year = v = at = 9.8(1.58 E7) = 154,526,400 m/s = 1.55 E8 m/s
Now, think of the remainder of the problem in this way: The spacecraft starts out at 0 m/s, and after 1/2 year the space craft is going 1.55 E8 m/s, then it continues at that speed for the rest of the journey.
u =0
v = 1.55 E8 m/s,
Use s = 1/2(u+v)*t to find the distance the spacecraft travels in 1/2 year:
s = 1/2(0 + 1.55 E8)(1.58 E7) = 1.22 E 15 m
Now find how much further it has to go:
s = 4.1 E 16 m - 1.22 E 15 m = 3.98 E16m
Now calculate how long it takes to travel 3.98 E16m at a speed of 1.55 E8 m/s:
t = s/v = 3.98 E16m/1.55 E8 = 256774193 s = 2.49 E8 s
So, the spacecraft took T = 1.58 E7 s + 2.49 E8 s = 2.57 E8 s = 8.1 years
This is mainly a lot of number crunching; go through the calculation yourself to check my result.
Hope this helped!
Dr Zorro and I both made a mistake. Below is the corrected solution:
1g = 9.8 m/s^2
half a year = 1/2(365*24*60*60) = t = 15,768,000s = 1.58 E7 s
The spacecraft speed after half a year = v = at = 9.8(1.58 E7) = 154,526,400 m/s = 1.55 E8 m/s
Now, think of the remainder of the problem in this way: The spacecraft starts out at 0 m/s, and after 1/2 year the space craft is going 1.55 E8 m/s, then it continues at that speed for the rest of the journey.
u =0
v = 1.55 E8 m/s,
Use s = 1/2(u+v)*t to find the distance the spacecraft travels in 1/2 year:
s = 1/2(0 + 1.55 E8)(1.58 E7) = 1.22 E 15 m
Now find how much further it has to go:
s = 4.1 E 16 m - 1.22 E 15 m = 3.98 E16m
Now calculate how long it takes to travel 3.98 E16m at a speed of 1.55 E8 m/s:
t = s/v = 3.98 E16m/1.55 E8 = 256774193 s = 2.49 E8 s
So, the spacecraft took T = 1.58 E7 s + 2.49 E8 s = 2.57 E8 s = 8.1 years
This is mainly a lot of number crunching; go through the calculation yourself to check my result.
Hope this helped!
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what is 1g?