Question How far away is a RR Lyrae variable star if it has a period of 0.5 days and a brightness of 2.3 x 10^-11 W/m^2?

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 Log in to Reply

Answer:The answer is d =10^6.43 Mpc

Explanation:SolutionTo 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