Very Simple Math Problem!!!!!

Taking time dilation into account, how long would it take to travel 1100 light years going .999999 C? how much time would have passed on the craft and how long would have passed on earth? If there is a super genius or extraterrestrial biological entities among this forum please come forward.

 

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Traveling through a wormhole you could make that trip in less than 3 minutes.

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At what speed?

 

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It depends on how fast you are going.

And there is a huge difference between .999991 and .999999 % of the speed of light.

At .999991% c, then 1 day on the spacecraft is 86,031 days back at earth

But

At .999999% c 1 day on the spacecraft is 258,094 days back at earth

It also depends on how long it takes you to accelerate to that speed and then slow down from that speed. If there are people on board this craft it's unlikely that you want to accelerate at faster than 1g for long periods of time.

Arthur

 

so how long would it take to go 1100LY with time dilation accounted for? how many days on the craft and how many years on earth?

 

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http://www.1728.com/reltivty.htm

 

http://www.cthreepo.com/lab/math1.shtml

Arthur

 

ive been up all night trying to solve this and giving me things i already know isnt helping its just showing nobody else can solve this problem. like ive said before, how long would it take in days or years to go 1100LY on the craft going the speed of light(c) and how many earth years would have passed.

 

It's relatively easy to solve, but no one can do so since you won't state HOW FAST are you travelling on the trip.

BUT I gave you a simple way to compute it.

At .999999% c 1 day on the spacecraft is 258,094 days back at earth

So in this case, if you assume this is your speed, then simply divide 1100 years by .999999 to get the time it would take (in years) at that speed to get to the star (assuming instantaneous acceleration to that speed) and then multiply that result by (258,094 / 365) to get how many years would have passed back on Earth.

Arthur

 

so solve it if im going .999999 c. how long would it take on the craft and how long would have passed on earth. why would i wanna go any other speed but C just gemme the answer if you know.

 

You can't go c, that's a limit you can approach, but never get to.

You've been given how to work it. If you're doing homework, then I would assume your teacher wants you to show work anyway, so take what's here and work it out.

 

im not doing homework. i just wanna know how long it would take going .999999 C to go 1100LY which is the distance it is to leave the milky way galaxy and how long it would be inside the craft and how long of time would have passed on earth. just a theoretical answer would be nice.

 

FFS!
Post 10 gives you the numbers.
All you have to do is some division and multiplication.

 

Solved 3.01 years on the craft and 777818.7 years on earth. i believe that is what the answer is according to arthur's numbers

 

Sorry,
I think I led you astray......

Let's look at it in parts.

First from the point of view of the observer on Earth.

The observer sees the spaceship go by on the way to the distant star.

It's speed is calculated between two fixed points to be .999999 c.

The distance to the star is 1100 light years and so the Earth Bound observer correctly predicts that in 1100/.999999 years the spacecraft will arive at the star.

1100/.999999 = 1100.0011 years, which makes sense since light itself would have only been a bit quicker.

Also this makes sense because there are no relativistic effects to be considered for the Earth bound observer. His clock is unaffected by how fast the spaceship is travelling.

Which leaves the question, how long does a clock carried by the person on the spaceship measure this same trip?

And here is where there are relativistic affects.

As you speed up, your clock runs slower.

Or alternatively, as you speed up, the length is contracted.

Either calculation though gives the same Relativistic answer.

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html

If you plug in .999999 c you get a Relativistic value (gamma) of 707.10685 which if you divide into 1100 light years you see that the time dialation results in a trip time for the people on board of 1.55 years.

Or if you plug it .999999c into the length contraction formula you get a value of .0014142 which multiplied by 1100 gives you a distance of 1.55 light years, which is the length contracted distance you travel, so at your speed that's how many years it takes.

Arthur

 

so obviously the time that has passed on earth is 1100 years its a bit confusing

 

Learn to read:

How can it take less time to a stationary observer if the ship is moving slower than light?

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i got the 707 number the first time around in my math i was just counting that as equal to one day on the craft so i know where i made my mistakes. i got the 1100 number and thought that was the amount of days on the craft and divided that by 365 to get my 3.01 years its all good it makes sense now thank you arthur. but the only thing that doesnt seem to click with my brain is why you told me to mulitiply the 1100.0011 by 258,094 and then dividing that by 365 would give the 777818 years on earth.

 

that is what you said arthur on post 9