Answer to Question #218971 in Astronomy | Astrophysics for annoonymus0082

Question

You have used Cepheid variable to measure the distance to a nearby spiral galaxy. The spiral galaxy is 5.0 Mpc (megaparsecs) away. Three years ago, a Type 1a supernova was observed in this galaxy, and had a peak flux of 6.2 × 10^-14W m^-2.

You have just discovered a new Type 1a supernova in a very distant galaxy. This new supernova stayed bright for the same time as the one in the nearby spiral galaxy. The flux of the new supernova was 1.3 × 10^-18W m^-2.

What is the distance to the new supernova (in Mpc)?:

Answer 1

If we don't take the cosmologic expansion into account, we assume the flux to be proportional to $r^{-2}$ , so $\dfrac{I_1}{I_2} = \dfrac{r_2^2}{r_1^2}, \; r_2 = 218r_1 = 1.1\;\mathrm{Gpc}.$

But if we take the expansion into account, we should write $I = \dfrac{L}{4\pi r_2^2(1+z)^2}$ . For such distances the redshift is approximately 0.2, if we calculate it in form of $z = Hr_2/c$ , therefore, $(1+z)^2 \approx 1.5,$ $(1+z)\approx 1.24$ and we should correct the estimate of $r_2$ from the by $1/1.24$ and get the value of 880 Mpc.