Question

what is the absolute magnitude of a star that has a period of 50 days?

Explanation:

To determine the absolute magnitude of a star with a period of 50 days, we can use the Cepheid Period - Luminosity Relation (also known as the Leavitt Law), which is commonly expressed as:

$$M = -2.76(\log_{10} P - 1) - 4.16$$

where \( M \) is the absolute magnitude and \( P \) is the period in days.

Step 1: Substitute \( P = 50 \) into the formula

First, calculate \( \log_{10} 50 \). We know \( \log_{10} 50 \approx 1.69897 \) (since \( 10^{1.69897} \approx 50 \)).

Substitute \( \log_{10} 50 \) into the formula:
$$M = -2.76(1.69897 - 1) - 4.16$$

Step 2: Simplify the expression inside the parentheses

Calculate \( 1.69897 - 1 = 0.69897 \).

Now the formula becomes:
$$M = -2.76(0.69897) - 4.16$$

Step 3: Multiply \( -2.76 \) by \( 0.69897 \)

$$-2.76 \times 0.69897 \approx -1.929$$

Step 4: Subtract \( 4.16 \) from the result

$$M \approx -1.929 - 4.16 \approx -6.089$$

Step 5: Round (if needed)

The absolute magnitude is approximately \( -6.1 \) (or more precisely, \( \approx -6.09 \)).

Answer:

The absolute magnitude of the star is approximately \(\boldsymbol{-6.1}\) (or more precisely, \(\boldsymbol{-6.09}\)).