## cc4019 Group Title The astronomical object known as Crab Nebula is thhe remnant of an exploded star. The explosion was seen by the Chinese in 1054 C.E. However, the Crab Nebula is about 3500 LY distance from the Earth In what year did the star actually explode. one year ago one year ago

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1. UnkleRhaukus Group Title
2. cc4019 Group Title

i dont understand why are we including the current year when it was viewed in 1054

3. UnkleRhaukus Group Title

hmm,

4. UnkleRhaukus Group Title

if the crab nebula is 3500 [lyr] away, then light traveling from the crab nebula will take 3500 [yr] to reach earth so to find the year that the light left the crab nebula just take away the travel time from the year it arrived at earth 1054 CE 1054-3500=

5. cc4019 Group Title

im here

6. cc4019 Group Title

2466

7. UnkleRhaukus Group Title

btw, im not sure where you got the distance to the crab nebula from because i thought it was 6500 lyr (not 3500 lyr)

8. cc4019 Group Title

my textbook says its about 3500 LY away

9. UnkleRhaukus Group Title

dont for get that the year will be before 0 CE so it will be ... BCE

10. UnkleRhaukus Group Title

(well i guess you can use that value if its in your book)

11. cc4019 Group Title

so what was the first answer you gave because that seemed to make a lot of sense if the speed of light is pretty fast

12. UnkleRhaukus Group Title

$s=\frac{d}{t}\qquad\implies t=\frac ds$ $d=3500 [\text{ly}]=3500\times c\times[\text {yr}]$$s=c$ $t=\frac ds=\frac{3500\times c\times[\text {yr}]}{c}=3500 [\text {yr}]$ $1054[\text{yr}]~C.E.-3500[\text{yr}]=-2446~C.E.=2446~B.C.E$

13. cc4019 Group Title

Ok thank you for your time and explanation.

14. UnkleRhaukus Group Title

sorry about my first post[now deleted] ( i only just woke up)

15. cc4019 Group Title

its fine you've helped me understand with the formulas you've provided