Who sees distance contract in a 10.4 light-year trip at 0.95c

[noone123]

Can someone please clear this up for me?
Lets say the distance of Point A from Earth is 10.4 light years. Some spacecraft is traveling 0.95c.

So. The person in the spacecraft will see Earth contract but Earth will see the spacecraft contract in length too. The thing I'm confused on is who sees the distance of the flight contract(Earth to point A)? Is it the person inside the spacecraft or the Earthlings?

Also for time dilation, Earth will see the spacecraft take around 11 years. Is this right? Then if this is right, the spacecraft will experience less time.
So in the formula, tv is the time seen on Earth which is around 11 years and t0 is time seen in spacecraft will be much less?

Thx. If you can include calculations for me that would be awsome, but an explanation is more than fine.

 

[ghwellsjr]

Can someone please clear this up for me?
Lets say the distance of Point A from Earth is 10.4 light years. Some spacecraft is traveling 0.95c.

So. The person in the spacecraft will see Earth contract but Earth will see the spacecraft contract in length too. The thing I'm confused on is who sees the distance of the flight contract(Earth to point A)? Is it the person inside the spacecraft or the Earthlings?

Also for time dilation, Earth will see the spacecraft take around 11 years. Is this right? Then if this is right, the spacecraft will experience less time.
So in the formula, tv is the time seen on Earth which is around 11 years and t0 is time seen in spacecraft will be much less?

Thx. If you can include calculations for me that would be awsome, but an explanation is more than fine.

You have everything correct.

You need to calculate the reciprocal of gamma which is the square root of 1 minus the square of the speed which is 0.3122 and multiply that by the time which results in about 3.4 years.