Question

The distance from earth to the center of the galaxy is about 22500 ly (1 ly = 1 light-year = 9.47 ? 10¹⁵ m), as measured by an earth-based observer. A spaceship is to make this journey at a speed of 0.9950c. According to a clock on board the spaceship, how long will it take to make the trip? Express your answer in years (1 yr = 3.16 ? 10⁷ s).

Please be as detailed as possible in your response, thank you!

Answer 1

The key thing here is this is a time dilation problem. Thelight year is a measure of distance, but supposing the earthobserver actually lives that long, he uses a clock to measure theship taking that much time for its journey in years.

We get t = D / v = 22500 L.Y. / 0.995 * 3.00 x 10⁸
                         = 9.47 x 10¹⁵ m * 22500 / (0.995 * 3.00 x10⁸) m / s
                         = 7.106 x 10¹¹ s
                         = 22533 years

The guy on the ship also uses a clock to time the journey, andbased on time dilation, will take much shorter from his referenceframe.

So we want to use: t = t₀ ? and solve fort₀ . Typically, 't' is chosen as dilated time, andthe ship time t₀ is chosen as 'normal'. ? isa time dilation factor where ? = (1 - v² /c²) ^-1/2

So:   t₀ = t * (1 - v² /c²) ^1/2
           = 22533 * (1 - 0.995² c² / c²)^1/2
               = 22533 * (1 - 0.9995²) ^1/2
               = 22533 * 0.0316188
           = 712.5 year [keep 4 sig figs]

It cuts down the time a lot, but neither the earth or spaceobserver would live that long ;-)