Question

The discovery of dark energy was based on the variation of the observed brightness of Type la supernovas to their expected brightness. Comparing two such supernovas, A and B, what would be the expected ratio of A's brightness to that of B, if A is 1/9 th as far from Earth as B ?

Answer 1

The ratio of supernova A's brightness to that of supernova B is 1/81.

Step-by-step explanation:

The expected ratio of the brightness of Supernova A to that of Supernova B can be determined based on their relative distances from Earth.

Assuming the brightness of a Type Ia supernova is inversely proportional to the square of the distance, we can use the inverse square law to calculate the expected ratio.

Let's denote the distance of supernova A from Earth as "dA" and the distance of supernova B from Earth as "dB". According to the given information, A is 1/9th as far from Earth as B, so we can express their distances as:

dA = (1/9) * dB

Now, let's calculate the expected ratio of A's brightness to that of B:

(Brightness ratio) = (dA^2) / (dB^2)

Substituting the value of dA in terms of dB:

(Brightness ratio) = ((1/9) * dB)^2 / (dB^2)

Simplifying the equation:

(Brightness ratio) = (1/81)

Therefore, the expected ratio of supernova A's brightness to that of supernova B would be 1/81.