Substitute the given period P = 45 into the formula: M = − 2.78 ( lo g P ) − 1.35 .
Calculate lo g 45 ≈ 1.6532 .
Compute M = − 2.78 ( 1.6532 ) − 1.35 ≈ − 4.5969 − 1.35 = − 5.9469 .
Round to the nearest hundredth: M ≈ − 5.95 .
Explanation
Understanding the Problem We are given the formula that relates the absolute magnitude M of a Cepheid star to its period P (in days): M = − 2.78 ( lo g P ) − 1.35 We are asked to find the absolute magnitude of a star with a period of 45 days.
Substituting the Value of P Substitute P = 45 into the formula: M = − 2.78 ( lo g 45 ) − 1.35
Calculating the Logarithm Calculate lo g 45 . The result is approximately 1.6532. M = − 2.78 ( 1.6532 ) − 1.35
Performing the Multiplication Multiply -2.78 by 1.6532: − 2.78 × 1.6532 = − 4.596896 So, M = − 4.596896 − 1.35
Performing the Subtraction Subtract 1.35 from -4.596896: M = − 4.596896 − 1.35 = − 5.946896
Rounding the Result Round the result to the nearest hundredth: M ≈ − 5.95
Final Answer The absolute magnitude of a star that has a period of 45 days is approximately -5.95.
Examples
Understanding the relationship between a Cepheid star's period and its absolute magnitude allows astronomers to determine distances to galaxies. By measuring the period of a Cepheid variable star in a distant galaxy, they can calculate its absolute magnitude using the formula. Comparing this to the star's apparent magnitude (how bright it appears from Earth), they can estimate the distance to the galaxy. This method is crucial for mapping the universe and understanding its expansion.