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Light puzzle

  • Thread starter japam
  • Start date
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suppose an imaginary fireworker that launch a rocket at c/2 velocity, the rocket is programmed to explode at T seconds with a brilliant light
the fireman has a powerful telescope that lets he follows the rocket path
my question is :would he be able to see the explosion through the scope and when?

thankyou
 

DaveC426913

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Show your work.

The rocket takes how many seconds before it explodes?
How long does it take for the light to reach him again?
There's a small correction because the rocket is travelling relativistically.
What is the puzzle?
 

disregardthat

Science Advisor
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Well for the seconds, you have to use an equation. I'm not sure which, but i know that a little under 90% of the light speed equals 1 second from the observer = 7 seconds on the object moving at that speed.

So let's assume that at 50% (c\2) the time difference has a factor of 1.5. Which means that the observer will observe the rockets explosion (excluding the time the light have to reach his eyes) after 1.5T seconds. For example. He shoots the rocket out with a timer of 20 seconds. He will see it explode after 30 (1.5T --> 1.5*20 = 30). If you add the time the light uses to reach your eyes: The rocket travels with the speed of c\2 for 20 seconds = (300 000km*20)\2 = 3 000 000km. (3 000 000km)\300 000km\s) = 10s) ) This means the light will use 10 seconds to reach your eyes. and 30 + 10 = 40 seconds. This is only if the timer is 20, and the time difference is 1.5 (which i have no clue of, but it can't be far off...)
 

DaveC426913

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At 0.5c the relativity factor is 1.15.

Try the http://www.1728.com/reltivty.htm" [Broken]
 
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disregardthat

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nice. But you could of course use 1.15 instead of 1.5

EDIT: Okey, I read the thing about relativity factor on the relativity calculator. It said this:

At .9 times the speed of light, the factor becomes 2.294157338705618. Finally, the effects of relativity become significant. What does this factor mean though? If you were in a spaceship travelling at .9 times the speed of light:
1) the ship's mass (and you) would increase by a factor of 2.294
2) the ship (and you) would contract in the direction of travel by 2.294, meaning a 300 foot ship would shrink to 130.77 feet.
3) Perhaps the most interesting change is that 1 year to you would seem to be 2.294 years for someone back on Earth.

What does number 2 mean? I didn't understand it. How can the ship be shrinked? Wouldn't that destroy the ship?
 
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to me the question is quite silly.

if a telescope is able to follow the rocket, that means the telescope is able to follow the light emitted by the rocket. Hence, no matter what happens to the rocket, the telescope will be able to trace it as long as the event emits light.
 

DaveC426913

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1) the ship's mass (and you) would increase by a factor of 2.294
2) the ship (and you) would contract in the direction of travel by 2.294, meaning a 300 foot ship would shrink to 130.77 feet.

What does number 2 mean? I didn't understand it. How can the ship be shrinked? Wouldn't that destroy the ship?
EVERYthing is shrunk by that factor. The ship, its occupants, the space, even the distance to the star they're headed towards.

Einstein's relevation about our universe is that space itself is not fixed and immutable, but is dependent on relativity.
 
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DaveC426913

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to me the question is quite silly.

if a telescope is able to follow the rocket, that means the telescope is able to follow the light emitted by the rocket. Hence, no matter what happens to the rocket, the telescope will be able to trace it as long as the event emits light.
Perhaps before calling something silly, you should ensure you read it carefully. How long would it take for him to see the explosion?
 
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PhanthomJay

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2) the ship (and you) would contract in the direction of travel by 2.294, meaning a 300 foot ship would shrink to 130.77 feet.
What does number 2 mean? I didn't understand it. How can the ship be shrinked? Wouldn't that destroy the ship?
With the ship (and you) traveling at 0.9c relative to a stationary observer, the 300 foot ship would shrink to 130.77 feet as measured by the stationary observer, not as measured by you. Since you and the ship are travelling at the same speed, you have no speed with respect to the ship, and you would see the ship as its normal length (you won't get squished); however, to you, all space, the earth, the stationary observer, etc., would be contracted by that factor.
 

DaveC426913

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With the ship (and you) traveling at 0.9c relative to a stationary observer, the 300 foot ship would shrink to 130.77 feet as measured by the stationary observer, not as measured by you. Since you and the ship are travelling at the same speed, you have no speed with respect to the ship, and you would see the ship as its normal length (you won't get squished); however, to you, all space, the earth, the stationary observer, etc., would be contracted by that factor.
But don't misunderstand - this is not merely an illusion - a trick of the light as it were. In all meaningful ways, the contraction/dilation really does happen.
 

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