At the speed of light it would take 1,000,000,000 years.
In days that would be 365,250,000,000
The speed of light is roughly 670,616,629.2 mph (1,079,252,848.8 kph) so if your traveler is traveling at a slower speed you can divide this down. This was just a reference point because if mass is accelerated this fast mass and time become altered and the rules change.
So roughly if your traveling at say around 670 mph it would take 1,000,000,000,000,000 years
2007-07-31 15:38:17
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answer #1
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answered by Kev 3
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You must enter your speed into the equation to calculate an answer.
A light year is the distance travelled by a beam of light in a year, so if you are going the speed of light it will take 1,000,000,000 years.
2007-08-05 10:29:13
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answer #2
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answered by eferrell01 7
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What is your speed?
At the speed of light, it would take 1,000,000,000 years.
At half the speed of light it would take twice as long.
At one tenth the speed of light it would take 10 times as long.
At one one-hundredth the speed of light it would take 100 times 1,000,000,000 years.
S0 anyway, you get the picture. You have to know the speed of the traveler before you can determine how long it takes to travel a given distance.
2007-07-31 15:39:44
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answer #3
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answered by VampireDog 6
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Assuming you could travel at the speed of light, which you can't it would take you one billion years to travel one billion light years.
2007-07-31 16:01:59
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answer #4
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answered by Anonymous
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First we have to know which calender you are using Julian or Gregorian
Julian multiply by 365.25
Gregorian multiply by 365.2425
2007-07-31 15:39:32
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answer #5
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answered by born_tomd 3
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More than my lifetime could ever fit.
2007-08-02 15:42:04
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answer #6
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answered by fitman 6
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KEV is SMART! I think he's probably right. Sounds good to me...
2007-08-06 09:43:20
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answer #7
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answered by SarahBear 2
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probably 7
2007-07-31 19:42:47
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answer #8
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answered by Dr. Illegal Morphine 2
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Agree with BORN what ca lander???
2007-08-01 17:08:34
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answer #9
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answered by vfisher27 2
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well all your life.. why?
2007-07-31 19:59:46
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answer #10
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answered by Anonymous
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