Question

A star is estimated to be 4.8 x 10^15 kilometers away from the earth. When we see this star in the night sky, how old is this image? Let the speed of light c= 3.00 x 10^8 m/s.
A. 88 years
B. 500 years
C. 7800 years
D. 16 months

Answer 1

Answer:

B. 500 years

Explanation:

The light coming from the star and reaching us on the Earth travels with uniform motion (with constant velocity), so we can use the equation of uniform motion which relates distance covered, speed and time taken:

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

where

v is the speed

d is the distance covered

t is the time taken

In this problem:

$v=3.00\cdot 10^8 m/s$ is the speed at which light travels

$d=4.8\cdot 10^{15} km = 4.8\cdot 10^{18}m$ is the distance that light has to cover from the star to the Earth

Therefore, by rearranging the equation, we can find the time:

$t=\frac{d}{v}=\frac{4.8\cdot 10^{18}}{3.00\cdot 10^8}=1.6\cdot 10^{10}s$

And by converting into years, this time is

$t=\frac{1.6\cdot 10^{10}}{(365)(24)(60)(60)}=507 y$

So, approximately 500 years: this means that the image we see of the star is 500 years old.

Answer 2

Answer:

Approximately = 500 [years]

Explanation:

To solve this problem we must use the ratio of units between light year and Kilometer, that is, we must convert the length of kilometers to light years.

$1 [km]= 1.057*10^{-13} [Ly]$

$1[km]=1.057*10^{-13}[Ly]\\ 4.8*10^{15}[km]= x\\ \\x = 507.36 [LY] "Light years"$

Now we can find the time using the following equation.

v = x / t

$x = 4.08*10^{18}[m]\\ t = \frac{4.08*10^{18}}{3*10^{8}}\\ t=1.36*10^{10} [s]\\Now we have:\\1.36*10^{10}[s]*\frac{1hr}{3600s}*\frac{1day}{24hr}*\frac{1month}{30day}*\frac{1year}{12month} \\aprox = 437 [years]$