Speed of light independent of the source ?

[Roger Dynamic Motion]

I've always heard 183,000, but neither is exact.

I have my Modern Physic Book for scientists and engineers and it said 186000 s

[Delta1212]

It's 186,000 mps, which Janus used 5 times in that post compared to the one 182,000, so I'm assuming that was just a typo.

[Janus]

It's 186,000 mps, which Janus used 5 times in that post compared to the one 182,000, so I'm assuming that was just a typo.

Which it was and I've now corrected.

[KipIngram]

Duh. Yes, of course. I don't know where I got 183 from. Maybe 186,300.

[Janus]

you mention here that ,,*The speed of light is 182,000 miles per second . Where did you get that ?

As already pointed out, that was clearly a typo considering the rest of the post.

[StringJunky]

As already pointed out, that was clearly a typo considering the rest of the post.

If people are picking at nits, you know you are doing it right.

[Silvestru]

That was not clear before, and no, it wasn't obvious. The time it takes depends on who is doing the measuring.

 

Swansont: The Astronauts

Could it be also who is doing the timing ?

 

Ahaha made my day. "The Astronauts"

[Roger Dynamic Motion]

Then there is no way you can have a 3/4 sec time difference between the arrival of the light and the ship.

Obviously not. If the measurements are being made by someone in the ship, and they turn on the light when they are 186,000 miles from the target, then by their reckoning it will take 186,000/(186,000+46500)*= 0.8 sec for the light to hit the target. They will meet up with the target in 186,000/46500 = 4 sec, or 3.2 sec after the light hits it.

 

I don't know what you are doing to get 3/4 of a sec with the numbers you have given, but whatever it is, it's wrong.

 

*The speed of light is 186,000 miles per second and 0.25 c is 46500 miles per sec. For someone in the ship the light is traveling away at 186,000 miles per sec, and the distance between themselves and the target is decreasing at a rate of 46500 miles per sec, and thus the light and target will meet when the target is 148,800 miles from the ship. The light will have traveled this distance at 186,000 miles per sec, which takes 0.8 sec. The distance between ship and target will have decreased by 37,200 miles, which at a relative velocity of 46500 miles per sec takes 0.8 sec to traverse.

 

decreasing at a rate of 46500 miles per sec

so the light is already at target after 46500 miles, while there is 3 quarter of second left for the ship to make it to target . Edited June 5, 2017 by Roger Dynamic Motion

[Janus]

so the light is already at target after 46500 miles, while there is 3 quarter of second left for the ship to make it to target .

No. The light has traveled 148,800 miles. 46500 miles per sec is the speed at which the distance between ship and target is decreasing. It takes the light(at 186,000 mps) 0.8 sec to travel that far from the ship. In that 0.8 sec the distance between target and ship decreases by 0.8 sec *46500 mps = 37200 miles from its original 186,000 miles, putting it 148,800 miles from the ship. It will take another 148,800 miles/46500 mps = 3.2 sec for the ship and target to meet.

 

3/4 of a sec isn't anywhere close to the right answer.

[zapatos]

This is like watching someone at a tennis match trying to understand how many wickets were lost during the last field goal attempt.

[studiot]

This is like watching someone at a tennis match trying to understand how many wickets were lost during the last field goal attempt.

 

Any more comments like this in one day and you are going to crack me up.

 

+1

Edited June 5, 2017 by studiot

[Roger Dynamic Motion]

No. The light has traveled 148,800 miles. 46500 miles per sec is the speed at which the distance between ship and target is decreasing. It takes the light(at 186,000 mps) 0.8 sec to travel that far from the ship. In that 0.8 sec the distance between target and ship decreases by 0.8 sec *46500 mps = 37200 miles from its original 186,000 miles, putting it 148,800 miles from the ship. It will take another 148,800 miles/46500 mps = 3.2 sec for the ship and target to meet.

 

3/4 of a sec isn't anywhere close to the right answer.

what if the''e ship travels at 186000 miles; they'll arrived a the same time ..right ?

[KipIngram]

The ship has mass and so can't travel at c.

If the ship is traveling very close to c, then the distance they measure that they have left to travel will be contracted. So they will see the light reach the target very quickly, since it's covering a small distance to them. But it will still get there ahead of them.

[Roger Dynamic Motion]

The ship has mass and so can't travel at c.

 

If the ship is traveling very close to c, then the distance they measure that they have left to travel will be contracted. So they will see the light reach the target very quickly, since it's covering a small distance to them. But it will still get there ahead of them.

yes technically.

got to go

[zapatos]

 

Any more comments like this in one day and you are going to crack me up.

 

+1

Thanks. It's been one of those days...

[Janus]

what if the''e ship travels at 186000 miles; they'll arrived a the same time ..right ?

The closer to velocity of the ship with respect to the target, the shorter the delay between the arrival of the light and ship at the target.

So at the earlier assumed 186,000 mile initial distance, it would take a relative velocity of 0.7583c to result in 3/4 delay between the arrival of light and ship as measured by the ship.

The target however would measure only a ~1/2 sec difference between light and ship arrival.