The apparent brightness of a Cepheid variable star in a far-away galaxy is measured to be 640,000 times fainter compared to an equivalent star 2,000 light years away from the Sun in our Galaxy. Assuming that there is no light absorption between galaxies, what is the distance to the host galaxy of the Cepheid?
Use the inverse square law for brightness.
To be 640,000 times weaker, the star must be sqrt640,000 = 800 times farther way, if the two stars have the same abolute magnitude.
2000 ly x 800 = ____ light years
To determine the distance to the host galaxy of the Cepheid variable, we can use the method known as "inverse square law of brightness." This law states that the apparent brightness of an object decreases with the square of its distance.
In this scenario, the apparent brightness of the Cepheid variable in the distant galaxy is measured to be 640,000 times fainter compared to an equivalent star 2,000 light years away from the Sun in our Galaxy.
Let's denote the distance to the host galaxy of the Cepheid as "d" light years.
Using the inverse square law of brightness, we can set up the following equation:
(Apparent brightness of star in our galaxy) / (Apparent brightness of Cepheid in the host galaxy) = (Distance to star in our galaxy)^2 / (Distance to host galaxy of Cepheid)^2
Plugging in the values we know:
(1 / 640,000) = (2,000 light years)^2 / d^2
To solve for "d," we can cross-multiply and then take square roots:
1 * d^2 = 640,000 * (2,000 light years)^2
d^2 = 640,000 * (2,000 light years)^2
d^2 ≈ 2,560,000,000 * (2,000 light years)^2
Now, we can take the square root of both sides to solve for "d":
d ≈ √(2,560,000,000 * (2,000 light years)^2)
Calculating:
d ≈ √(2,560,000,000 * 4,000,000 light years^2)
d ≈ √(10,240,000,000,000,000 light years^2)
d ≈ 3,200,000,000 light years
Therefore, the distance to the host galaxy of the Cepheid is approximately 3,200,000,000 light years (3.2 billion light years).