Answer:
The answer is d =10^6.43 Mpc
Explanation:
Solution
To calculate the absolute magnitude of the star, we apply or use the relation with the period
Mv = - [ 2.76 (log₁₀ (P) - 1.0)] -4.16
Here P =+ 0.5 days
Mv = - 0.569
To calculate m, we have the following relation given below:
Mₓ = - 2.5 Log₁₀ (Fₓ/Fₓ,0)
Fₓ,0 = 1 w/m² (This is assumed)
Fₓ = 2.3 * 10^-11 W/m^2 (Given)
Thus
Mₓ = - 2.5 Log₁₀ (2.3 * 10^ ⁻11
Mₓ = 26.595
Now, applying the formula we have the following given below:
d = ₁₀ ( Mₓ -Mv + 5)/5
d =₁₀ (26.595 - 0.569 +5)/5
d =10^6.43 Mpc
