Canopus, a bright star in our sky, has an apparent magnitude of -0.62.The star is located 96pc from Earth.?

Question

Canopus, a bright star in our sky, has an apparent magnitude of -0.62. The star is located 96pc from Earth.

Would the absolute magnitude of Canopus be greater or less than the apparent magnitude? Please Explain.


Now, I know that the lower the magnitude the closer it is to Earth. So, if you use the equation m-M=5log(d)-5 would that mean that the absolute magnitude of Canopus would be greater than the apparent magnitude? I'm so lost...please help!

Answer 1

If a star is farther than 10 parsecs (or 32.616334 lightyears) from Earth, then its absolute magnitude is TO THE LEFT of its apparent magnitude on the number line.

If a star is closer than 10 parsecs from Earth, then its absolute magnitude is TO THE RIGHT of its apparent magnitude on the number line.

If a star is exactly 10 parsecs from Earth, then its absolute magnitude is equal to its apparent magnitude.

D = 10 parsecs = 32.616334 lightyears
m - M = 5 log(d/D)

where D is the standard distance of 10 pc.
where d is the distance from Earth to the star.
where M is the absolute magnitude of the star.
where m is the apparent magnitude of the star.

Ls = 3.826E26 watts
M = 4.75 - 2.5 log(L/Ls)

where Ls is the luminosity of the sun.
The sun's absolute magnitude is +4.75.

I combine the equations,
L/Ls = { 10^(1.9 - m/2.5) } (d/D)^2

I input your data for Canopus,
m = -0.62
d = 96 pc

L/Ls = 12958.1

Canopus is 12958.1 times more luminous than the sun.

M = 4.75 - 2.5 log(L/Ls)

Canopus has an absolute magnitude of -5.53.

Answer 2

The absolute magnitude would be brighter, which is to say a lower number, than the apparent magnitude, because lower magnitude numbers signify greater brightness. This is true for any star more than 10 parsecs away, because absolute magnitude is defined as the apparent magnitude it would have if it were exactly 10 parsecs away.

Answer 3

You can use this equation to convert apparent to absolute magnitude -- Mv = V + 5 log (10/d)
'Mv' is absolute magnitude
'V' is apparent magnitude
'd' is distance in parsecs

Mv = -0.62 + 5log (.1042) =
Mv = -0.62 + 5 * -0.98227 =
Mv = -0.62 + -4.911356 =
Mv = -5.531 <== absolute magnitude greater than apparent

Check my math..!

Answer 4

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