Question

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.)

Answer 1

Answer:

The time is 1206 years.

Explanation:

Given that,

Distance = 27000 ly

$1\ ly =9.47\times10^{15}\ m$

Speed = 0.9990c

We need to calculate the time

Using formula of speed

$v = \dfrac{d}{t}$

$t=\dfrac{d}{v}$

Put the value into the formula

$t =\dfrac{27000\times9.47\times10^{15}}{3\times10^{8}}$

$t=8.523\times10^{11}\ sec$

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

Using formula of dilation time

$t'=t\sqrt{1-\dfrac{v^2}{c^2}}$

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

$t'=8.523\times10^{11}\times\sqrt{1-(0.9990)^2}$

$t'=3.8106\times10^{10}\ sec$

The time convert in years

We know that,

$1\ yr= 3.16\times10^{7}\ sec$

So, the time is

$t'=\dfrac{3.8106\times10^{10}}{3.16\times10^{7}}$

$t'=1206\ years$

Hence, The time is 1206 years.