Question

sirius is the brightest star in the night sky with an apparent magnitude of $-1.44$. only planets and the moon appear brighter. its absolute magnitude is 1.45.
what is siriuss distance from earth?
round your answer to the nearest tenth.
enter your answer in the box.
parsecs
rigel is the seventh - brightest star in the night sky with an apparent magnitude of 0.12. it is part of the constellation orion and has an absolute magnitude of $-8.1$.
what is rigels distance from earth?
round your answer to the nearest tenth. enter your answer in the box.
parsecs

Explanation:

Step1: Use the distance - modulus formula

The distance - modulus formula is \(m - M=5\log(d)-5\), where \(m\) is the apparent magnitude, \(M\) is the absolute magnitude, and \(d\) is the distance in parsecs.

For Sirius:
Given \(m=-1.44\) and \(M = 1.45\).
First, substitute the values into the formula: \(-1.44-1.45=5\log(d)-5\).
Simplify the left - hand side: \(-2.89 = 5\log(d)-5\).
Add \(5\) to both sides: \(5 - 2.89=5\log(d)\), so \(2.11 = 5\log(d)\).
Then \(\log(d)=\frac{2.11}{5}=0.422\).
Using the definition of a logarithm (\(d = 10^{\log(d)}\)), we get \(d = 10^{0.422}\approx2.64\approx2.6\) parsecs.

For Rigel:
Given \(m = 0.12\) and \(M=-8.1\).
Substitute into the formula: \(0.12-(-8.1)=5\log(d)-5\).
Simplify the left - hand side: \(0.12 + 8.1=5\log(d)-5\), so \(8.22=5\log(d)-5\).
Add \(5\) to both sides: \(8.22 + 5=5\log(d)\), so \(13.22=5\log(d)\).
Then \(\log(d)=\frac{13.22}{5}=2.644\).
Using the definition of a logarithm (\(d = 10^{\log(d)}\)), we get \(d = 10^{2.644}\approx440.6\) parsecs.

Answer:

Sirius: \(2.6\) parsecs
Rigel: \(440.6\) parsecs