Question

A spaceship travels at a constant speed from earth to a planet orbiting another star. When the spacecraft arrives, 13 years have elapsed on earth, and 8.6 years have elapsed on board the ship. How far away is the planet, according to observers on earth?

Answer 1

the equations you need

D' =v * T'
D = v * T

this means that the times and distances seen from a moving frame of reference is different then the ones seen by the non moving one but the equations still hold for the values inside the frame

the transformation for time
T = T'/sqrt(1 - v^2/c^2)
we know the time the ship takes in earth time thats T frame
and the time that it takes in the moving ship thats T' frame

13 = 8.6/sqrt(1 - v^2/c^2)
now move the squareroot up by cross multiplying
13*sqrt(1-v^2/c^2) = 8.6
sqrt(1 - v^2/c^2) = 8.6/13
square both sides to get rid of root sign
(1 - v^2/c^2) = (8.6/13)^2
v^2/c^2 = 1 - (8.6/13)^2
i switched v^2/c^2 and (8.6/13)^2

solve for v
v^2 = c^2(1- (8.6/13)^2)
v = c*sqrt(1 - (8.6/13)^2)

this means the distance from earth is
D = v * T
D = c*sqrt(1 - (8.6/13)^2)* 13years
this is probably where you make the mistake standard c values are 3*10^8 but this is m/s not m/year
so we either have to change years to seconds which seems stupid or the speed of light to years

if we use 1 light year thats distance/year
3*10^8 m/s * s/y = m/y which is a light year
so C can be 1

D in light years = sqrt(1- (8.6/13)^2)* 13 years
= 9.75 light years

since a light year is 9,460,730,472,580.8 km
that is 9,460,730,472,580,800 m

this is about 9.461*10^15 m/ly * 9.75 ly = 92244750000000000 = 9.22 * 10^16 m