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It will get trickier for interstellar distances. In Newtonian mechanics v = at, so it'd choose rather less than the usual calendar year to succeed in c at one g acceleration. But relativity will not let that, we can easily only get near c.
How to go about calculating the size and Power needs for a space-primarily based linear accelerator or railgun to accelerate little probes? five
Through the stationary observer's viewpoint the traveler's charge of acceleration would gradual because they approached the velocity of light. The traveler would see no change concerning their pace and the pace of sunshine. Alternatively they would working experience time at an progressively slower rate which would effectively lead to the space for the destination to become shorter.
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Obviously, you don't need to position straight up, and TidalWave's assumption that Anything you intended is 9.8m/s/s might be right - but Be aware that Nevertheless, his response provides you with a minimum, eg assuming you could turn off gravity and the ambiance (and the assumptions he mentions at the top). $endgroup$
James JenkinsJames Jenkins 27.2k2323 gold badges108108 silver badges242242 bronze badges $endgroup$ three four $begingroup$ "1g of thrust" pointed straight up will balance gravity, and cause you floating. "1g" (as I go through it), will be the acceleration caused by the Earth's gravity; if that's how you actually define it, then your acceleration decreases as you have further more (and 'really feel significantly less pull') from Earth.
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seventy six c plus the passengers perceiving three hundred days. I am not sure which is proper, I am not at ease with Specific relativity. I do not Feel possibly an out of doors notice nor the accelerating travellers would see what appears being a Newtonian universe. $endgroup$
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$begingroup$ When you've got the Strength for a constant 1G thrust, just how long wouldn't it consider for getting for the planets inside our photo voltaic program? Just how long for the 5 nearest photo voltaic units?
But that point is in the viewpoint of stationary observers with the departure place. The vacation's period from the traveler's viewpoint could be considerably less mainly because of the time dilation effect predicted by Einstein's Theory of Relativity. The greater the distance, the greater the velocity from the stationary observer's viewpoint.
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1 $begingroup$ @Mark not really correct. In spite of everything, for those who notice points falling in continual gravity, you still cannot see them go beyond velocity of light, which means that Newtonian physics does not keep even as observed through the ship. $endgroup$
HopDavidHopDavid 16k4444 silver badges8080 bronze badges $endgroup$ 9 20 $begingroup$ Your "increase somewhat more than a calendar year for each lightyear length" is accurate for an outside observer, but for somebody aboard the ship, the Newtonian model is correct for all distances (as calculated before starting acceleration): Lorentz contraction will shrink the universe all through journey to give the appearance of Newtonian physics. $endgroup$
Because of the time dilation influence, 1G acceleration needs to be ample to travel wherever inside our galaxy in lower than a life time within the viewpoint in the traveler, but not the stationary observer. For more info on the time dilation impact read through Stephen Hawking's "Temporary Background of Time"