Physics, asked by BRAINLYBOOSTER12, 7 months ago

Distance = 2.44 billion light year
d = ( 2.44/3.2615) × 100 MPC

d = 74.8122 Mpc


Need explanation

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Answers

Answered by physicsloverhere
17

Generally, I am also confused with the mentioned numerical problem. But for a detailed explanation keep reading this answer :)

I am assuming that MPC = MEGAPARSEC

PROBABLE SOLUTION TO YOUR QUESTION :2.44 billion light years = 2440000000 light-years = 748.107 megaparsecs

Formula : For an approximate result, divide the length value by 3.262e+6 or use conversion methods in a correct way to reach the final result.

If MPC = MIlLIPARSEC, then 2440000000 light-years = 7.481 × 10^11 milliparsecs (Formula - for an approximate result, multiply the length value by 307)

Answered by Ekaro
63

Given :

Distance b/w two objects = 2.44 billion light year

To Find :

Distance b/w two objects in Mpc.

Explanation :

Remember :

:\implies\tt\:1AU=1.5\times 10^{11}\:m

:\implies\tt\:1ly=9.46\times 10^{15}\:m

:\implies\tt\:1pc=3.08\times 10^{16}\:m

  • AU denotes Astronomical unit
  • ly denotes Light year
  • pc denotes parsec

1. Astronomical unit : mean displacement of earth from sun.

2. Light year : distance travelled by light in vacuum in 1yr.

3. Parsec : distance at which the mean radius of the earth's orbit subtends an angle of one second of arc.

Relation between astronomical unit, light year and parsec :

\dashrightarrow\tt\:\dfrac{1ly}{1AU}=\dfrac{9.46\times 10^{15}}{1.5\times 10^{11}}=6.3\times 10^4

  • 1 ly = 6.3 × 10⁴ Au

\dashrightarrow\tt\:\dfrac{1pc}{1ly}=\dfrac{3.08\times 10^{16}}{9.46\times 10^{15}}=3.26

  • 1 pc = 3.26 ly

Clearly, \underline{\boxed{\bf{\blue{1pc>1ly>1Au}}}}

Let's come to the question :)

\leadsto\tt\:d=2.44\ billion\ light\ year

\leadsto\tt\:d=2.44\times (10^9)\times (9.46\times 10^{15})

\leadsto\tt\:d=23.08\times 10^{24}\:m

Conversion : m ➝ pc

\mapsto\tt\:3.08\times 10^{16}\ m=1\ pc

\mapsto\tt\:d=\dfrac{23.08\times 10^{24}}{3.08\times 10^{16}}

\mapsto\tt\:d=7.49\times 10^{8}\ pc

\mapsto\tt\:d=7.49\times 10^{8}\times 10^{-6}\ Mpc

\mapsto\:\underline{\boxed{\bf{\purple{d\approx 749\ Mpc}}}}

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