Question

A spaceship travels at a constant speed from earth to a planet orbiting another star. When the spacecraft arrives, 14 years have elapsed on earth, and 9.7 years have elapsed on board the ship. How far away is the planet, according to observers on earth?

Answer 1

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

14 = 9.7/sqrt(1 - v^2/c^2)

now move the squareroot up by cross multiplying

14*sqrt(1-v^2/c^2) = 9.7

sqrt(1 - v^2/c^2) = 9.7/14

square both sides to get rid of root sign

(1 - v^2/c^2) = (9.7/14)^2

v^2/c^2 = 1 - (9.7/14)^2

v^2 = c^2(1- (9.7/14)^2)

v = c*sqrt(1 - (9.7/14)^2)

this means the distance from earth is

D = v * T

D = c*sqrt(1 - (9.7/14)^2)* 14years

this is probably where you make the mistake standard c values are 3*10^8 but this is m/s not m/year

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- (9.7/14)^2)* 14 years

D = 10.09 light years

D = 9460730472580.8 km

D = 9460730472580800 m

D = 9.544 x 10^16 m