Calculating Time Dilation for Space Travelers

[stunner5000pt]

Homework Statement

A space traveler moving at a speed of 0.70c with respect to Earth makes a trip to a distant star that is stationary relative to Earth. He measures the length of this trip to be 6.5ly. What would be the length of this same trip (in ly) as measured by a traveler moving at a speed of 0.90c with respect to Earth?

Relevant Equations

$$\Delta t = \gamma t_{0}$$

If we took the perspective of the space traveller themselves, they are stationary and the whole universe goes past them at 0.7c. THen th elapsed time of 6.5 yr looking outside is

$$\Delta t =6.5 \frac{1}{\sqrt{1-0.7^2}} = 9.11 yrs$$

THen, when the traveller looks at the person travelling at 0.9c

the time taken should be

$$9.11 \times \frac{1}{\sqrt{1-0.9^2}} = 3.97 yrs$$

but the answer is 4.01 yr. Is this just down to rounding or is there something missing in the solution above?

thanks in advance for your help

 

[Hill]

... elapsed time of 6.5 yr ...

The problem does not say anything about such time, 6.5 yr. It describes a distance, 6.5 ly.

 

[stunner5000pt]

The problem does not say anything about such time, 6.5 yr. It describes a distance, 6.5 ly.

Aha! missed that, thank you

ok so then that means the distance travelled according to the traveller is:

$$6.5c \times 24 \times 365 \times 60^2 = 6.14 \times 10^{16} m$$

when the person is in motion, how is the 6.5 ly being interpreted ? Is the noninertial observer seeing a shorter'length that the stationary length i.e. is the above number the length observed from rest?

 

[Hill]

No need to convert or to interpret ly. It is a unit of distance as good as any.
I don't see a "noninertial observer" in the problem. However, there is length contraction rather than time dilation here.

 

[Orodruin]

ok so then that means the distance travelled

The distance travelled by the Earth and the destination planet.

And as has already been said, ly is a perfectly fine length unit. Preferable to meters here really as by definition c = 1 ly/y.