So Johnny, in Earth orbit, drives to see his girlfriend Sally at Sirius, 8.117x10^13 km away. His hot-rod accelerates at a constant rate of 9.81 meters per second squared. He will continue this until he reaches 270 km per second. He will cruise at this speed until he must then constantly decelerate at 9.81 meters per second squared to arrive at Sirius.
And no, it's not home-work help. I barely made it out of high-school algebra and people who can do calculus are wizards in my eyes. Thanks for any help or formulas on this, oh mighty math masters!
And no, it's not home-work help. I barely made it out of high-school algebra and people who can do calculus are wizards in my eyes. Thanks for any help or formulas on this, oh mighty math masters!
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Figure out the distance during acceleration.
v(f)^2 = v(i)^2 + 2ad
d = [v(f)^2 - v(i)^2]/2a
Assume that Johnny started from rest. Convert the acceleration to kilometers per second.
9.81 m/s = 0.00981 km/s
d = (270^2 - 0^2)/2(0.00981)
d = 3.7 x 10^6 km
Figure out the amount of time taken to travel this distance.
v(f) = v(i) + at
t = [v(f) - v(i)]/a
t = 270/(0.00981)
t = 2.75 x 10^4 s
Since the acceleration is equivalent to the deceleration, the amount of distance and time taken is the same.
Figure out the distance going at constant speed by find the difference between total distance and the distance covered when accelerating.
(8.117 x 10^13) - 2(3.7 x 10^6)
(8.117 x 10^13) - (7.4 x 10^6)
This difference isn't that much.
8.11699926 x 10^13 km
Figure out the amount of time taken to travel this distance.
t = d/s
t = (8.11699926 x 10^13)/(270)
t = 3.0063 x 10^11
Add this with the amount of time taken during acceleration.
t = (3.0063 x 10^11) + 2(2.75 x 10^4)
t = (3.0063 x 10^11) + (5.5 x 10^4)
t = 3.00630055 x 10^11 s
t = 9532.92 years
v(f)^2 = v(i)^2 + 2ad
d = [v(f)^2 - v(i)^2]/2a
Assume that Johnny started from rest. Convert the acceleration to kilometers per second.
9.81 m/s = 0.00981 km/s
d = (270^2 - 0^2)/2(0.00981)
d = 3.7 x 10^6 km
Figure out the amount of time taken to travel this distance.
v(f) = v(i) + at
t = [v(f) - v(i)]/a
t = 270/(0.00981)
t = 2.75 x 10^4 s
Since the acceleration is equivalent to the deceleration, the amount of distance and time taken is the same.
Figure out the distance going at constant speed by find the difference between total distance and the distance covered when accelerating.
(8.117 x 10^13) - 2(3.7 x 10^6)
(8.117 x 10^13) - (7.4 x 10^6)
This difference isn't that much.
8.11699926 x 10^13 km
Figure out the amount of time taken to travel this distance.
t = d/s
t = (8.11699926 x 10^13)/(270)
t = 3.0063 x 10^11
Add this with the amount of time taken during acceleration.
t = (3.0063 x 10^11) + 2(2.75 x 10^4)
t = (3.0063 x 10^11) + (5.5 x 10^4)
t = 3.00630055 x 10^11 s
t = 9532.92 years
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The amount of time spent accelerating and decelerating is miniscule compared to the coasting time and can be ignored without significant error.
8.117*10^13/270 = 3.006*10^11 seconds
Assuming the year is exactly 365 days long, there are 365*24*3600 = 3.1536*10^7 seconds in a year.
Johnny had better pack a lunch because it will take him (30.06/3.1536)*10^3 ≈ 9532 years to reach Sirius.
8.117*10^13/270 = 3.006*10^11 seconds
Assuming the year is exactly 365 days long, there are 365*24*3600 = 3.1536*10^7 seconds in a year.
Johnny had better pack a lunch because it will take him (30.06/3.1536)*10^3 ≈ 9532 years to reach Sirius.
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Acceleration and deceleration times are insignificant compared to total journey time.
270 km/s = 270*3600*24*365 = 8.51472*10^9 km/earth year
Time to reach Sirius = 8.117x10^13/8.51472*10^9 =
9,533 years
I hope Sally is still waiting.
270 km/s = 270*3600*24*365 = 8.51472*10^9 km/earth year
Time to reach Sirius = 8.117x10^13/8.51472*10^9 =
9,533 years
I hope Sally is still waiting.
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By my quick back-of-envelope scribblings, slightly less that 10,000 years.
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yes