If you don't mind not stopping and accelerate all the way, the shortest distance to Neptune is around 4.5 billion km.
Now s = 1/2 at^2 gives t at a bit less than 10^6 seconds from this, which is about 11 1/2 days.
You would by this time be going at around 10^7 m/s, anout 1/30th the speed of light (which means stricly the above is not accurate, because relativity becomes noteiceable, but at this speed it only makes a difference of a couple of percent).
2007-01-28 04:09:33
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answer #1
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answered by Anonymous
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Invoke the SUVAT equation s = ut + 1/2 at^2; where u = the initial velocity at t0 = 0 and a = average acceleration = g = 9.81 m/sec^2. s = the distance between the launch point on Earth and the surface of the planet the spaceship is headed for.
Solve for time (t). To simplify this solution, assume u(t = 0) = 0, which means the space ship started with zero initial velocity. Thus, s = 1/2 at^2 = 1/2 gt^2; so that t^2 = 2s/g and t = sqrt(2s/g) = the time in seconds it would take the space ship, but if and only if it crashes head on at full velocity into the distant planet. You can look up a planet on the web and find its distance (s) from Earth.
A more realistic problem is to throw in a deceleration as the ship nears the distant planet; so the ship can actually land on it without breaking up into a million bits and pieces. Thus you would have two times to calculate: time while accelerating (a > 0) and time while decelerating (a < 0).
2007-01-28 11:42:22
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answer #2
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answered by oldprof 7
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Using the formulas of the previous answer, using the most distant planet neptune at 4.3 billion km, this would take you about 30000 seconds, which is slightly under 9 hours. Your final speed would then be about 300000 m/s 1/1000 of the speed of light.
It is true that as you approach the speed of light, speeding up will be different, as you enter the world of relativistic mechanics. This means that you can no longer refer to time as a constant (nor is space a constant). The time it takes you to get to the most distant planet will than be different for someone looking at you from earth, or for the person doing the travelling. But since you are still very far away from the speed of light, you don't have to take this into account.
2007-01-28 12:23:57
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answer #3
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answered by Anonymous
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