The distance from earth to the center of our galaxy is about 27,000 ly (1 ly = 1 light year = 9.47 × 10^15 m), as measured by an earth-based observer. A spaceship is to make this journey at a speed of 0.9990c. According to a clock on board the spaceship, how long will it take to make the trip? Express your answer in years. (1 yr is equal to 3.16 × 10^7 s.)

Respuesta :

Answer:

The time is 1206 years.

Explanation:

Given that,

Distance = 27000 ly

[tex]1\ ly =9.47\times10^{15}\ m[/tex]

Speed = 0.9990c

We need to calculate the time

Using formula of speed

[tex]v = \dfrac{d}{t}[/tex]

[tex]t=\dfrac{d}{v}[/tex]

Put the value into the formula

[tex]t =\dfrac{27000\times9.47\times10^{15}}{3\times10^{8}}[/tex]

[tex]t=8.523\times10^{11}\ sec[/tex]

We need to calculate the time dilation the clock on the spaceship

Using formula of dilation time

[tex]t'=t\sqrt{1-\dfrac{v^2}{c^2}}[/tex]

[tex]t'=8.523\times10^{11}\times\sqrt{1-\dfrac{( 0.9990c)^2}{c^2}}[/tex]

[tex]t'=8.523\times10^{11}\times\sqrt{1-(0.9990)^2}[/tex]

[tex]t'=3.8106\times10^{10}\ sec[/tex]

The time convert in years

We know that,

[tex]1\ yr= 3.16\times10^{7}\ sec[/tex]

So, the time is

[tex]t'=\dfrac{3.8106\times10^{10}}{3.16\times10^{7}}[/tex]

[tex]t'=1206\ years[/tex]

Hence, The time is 1206 years.