<p>A star with magnitude 0 would be 85 times brighter than the sun  (since Magnitude=-2.5 log(Luminiosity))</p>
<p>Referring to the <a href="https://upload.wikimedia.org/wikipedia/commons/6/6b/HRDiagram.png" rel="noreferrer">H-R diagram on Wikipedia</a> shows that there is quite a range of spectral types possible with this luminosity: from B type main sequence stars, and A type sub-giants, such as <a href="https://en.wikipedia.org/wiki/4_Scorpii" rel="noreferrer">4 Sco</a></p>
<p>There are also G and K and M type Giant stars with this luminosity. such as <a href="https://en.wikipedia.org/wiki/Rho_Ursae_Majoris" rel="noreferrer">ρ UMa</a></p>
<p>Identifying stars that are exactly this bright is made more difficult by the difficulty in measureing the distance to stars.</p>

CC BY-SA 4.0
2020-11-04T20:17:23.920Z
2020-11-04T20:22:32.227Z
2020-11-04T20:22:32.227Z
James K
<p>The absolute magnitude quantifies the luminosity of an object at a standard distance of <span class="math-container">$10\ \mathrm{pc}$</span> from Earth. For example, in the case you mentioned, Vega becomes dimmer than at its actual distance (about <span class="math-container">$7\ \mathrm{pc}$</span>).</p>
<p>To answer your question, I don't think there is an actual star with exactly 0 absolute magnitude. If there is then, following the formula of the absolute magnitude, where <span class="math-container">$d_{pc}$</span> is the actual distance of the star from Earth in parsecs:</p>
<p><span class="math-container">$$M=m-5\log_{10}\left(\frac{d_\mathrm{pc}}{10\ \mathrm{pc}}\right)$$</span></p>
<p>Then such star should satisfy the following relation between apparent magnitude and distance to have absolute magnitude zero:
<span class="math-container">$$\frac m5+1=\log_{10}(d_\mathrm{pc})$$</span></p>
<p><strong>E.g:</strong> if you have a star at the Vega's distance <span class="math-container">$d_\mathrm{pc}=7.63\ \mathrm{pc}$</span> with apparent magnitude <span class="math-container">$m\simeq-0.58$</span> (more or less as Saturn maximum apparent magnitude) then you will have <span class="math-container">$M=0$</span></p>

CC BY-SA 4.0
2020-11-04T16:44:46.770Z
2020-12-11T15:22:31.397Z
2020-12-11T15:22:31.397Z
Michele Bianco
<p>There isn't a one-to-one relationship between spectral type and absolute magnitude.</p>
<p>Instead, there is a mean relationship with a fair bit of scatter around it. The reason is that the luminosity of a star of a given effective temperature depends on its composition/metallicity and how far along in its main sequence lifetime it is.</p>
<p>Basically, late B-type main sequence stars (say B7/B8V) have an absolute magnitude of <em>about</em> zero. Alternatively there are low mass stars ascending the hydrogen shell burning giant branch (types of about K2-K5 III) that would have an absolute magnitude of zero.</p>

CC BY-SA 4.0
2020-11-04T20:08:03.643Z
2020-11-04T20:08:03.643Z
ProfRob
<p>We know that Vega is the star that serves as the zero point for the UBV color scale, and has an apparent magnitude of nearly zero (+0.02). But its absolute magnitude is +0.58, making it rather far from <span class="math-container">$M=0$</span>. So what spectral type would have an absolute magnitude of <span class="math-container">$0 \pm0.02$</span>, and are there any stars that satisfy this?</p>

CC BY-SA 4.0
2020-11-04T16:49:48.097Z
How precisely do you define an absolute magnitude of exactly 0?     The more exactly such a star is required to have an absolute magnitude of exactly 0, the fewer the stars will be that have such a magnitude.  The more exactly such a star is required to have an absoluted magnitude of exactly 0, the fewer the stars will be which have their absolute magnitudes measured and calaculated that precisesly, until eventually an increasing demand for a precise magnitude will lead to no stars having such precise magnitude known.
M. A. Golding
<star><temperature><absolute-magnitude><spectral-type>
What spectral type of star has an absolute magnitude of exactly 0?
WarpPrime
# What spectral type of star has an absolute magnitude of exactly 0?

(Asked by: username_0 on 2020-11-04)

We know that Vega is the star that serves as the zero point for the UBV color scale, and has an apparent magnitude of nearly zero (+0.02). But its absolute magnitude is +0.58, making it rather far from $M=0$. So what spectral type would have an absolute magnitude of $0 \pm0.02$, and are there any stars that satisfy this?

**Tags:** star, temperature, absolute-magnitude, spectral-type
> **Comments:**
> **username_1**: How precisely do you define an absolute magnitude of exactly 0?     The more exactly such a star is required to have an absolute magnitude of exactly 0, the fewer the stars will be that have such a magnitude.  The more exactly such a star is required to have an absoluted magnitude of exactly 0, the fewer the stars will be which have their absolute magnitudes measured and calaculated that precisesly, until eventually an increasing demand for a precise magnitude will lead to no stars having such precise magnitude known.  

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## 3 answers

### Answer by username_4 on 2020-11-04. Score: 9 | Accepted answer

There isn't a one-to-one relationship between spectral type and absolute magnitude.

Instead, there is a mean relationship with a fair bit of scatter around it. The reason is that the luminosity of a star of a given effective temperature depends on its composition/metallicity and how far along in its main sequence lifetime it is.

Basically, late B-type main sequence stars (say B7/B8V) have an absolute magnitude of *about* zero. Alternatively there are low mass stars ascending the hydrogen shell burning giant branch (types of about K2-K5 III) that would have an absolute magnitude of zero.


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### Answer by username_2 on 2020-11-04. Score: 7

A star with magnitude 0 would be 85 times brighter than the sun (since Magnitude=-2.5 log(Luminiosity))

Referring to the [H-R diagram on Wikipedia](https://upload.wikimedia.org/wikipedia/commons/6/6b/HRDiagram.png) shows that there is quite a range of spectral types possible with this luminosity: from B type main sequence stars, and A type sub-giants, such as [4 Sco](https://en.wikipedia.org/wiki/4_Scorpii)

There are also G and K and M type Giant stars with this luminosity. such as [ρ UMa](https://en.wikipedia.org/wiki/Rho_Ursae_Majoris)

Identifying stars that are exactly this bright is made more difficult by the difficulty in measureing the distance to stars.


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### Answer by username_3 on 2020-11-04. Score: 1

The absolute magnitude quantifies the luminosity of an object at a standard distance of $10\ \mathrm{pc}$ from Earth. For example, in the case you mentioned, Vega becomes dimmer than at its actual distance (about $7\ \mathrm{pc}$).

To answer your question, I don't think there is an actual star with exactly 0 absolute magnitude. If there is then, following the formula of the absolute magnitude, where $d\_{pc}$ is the actual distance of the star from Earth in parsecs:

$$M=m-5\log\_{10}\left(\frac{d\_\mathrm{pc}}{10\ \mathrm{pc}}\right)$$

Then such star should satisfy the following relation between apparent magnitude and distance to have absolute magnitude zero:
$$\frac m5+1=\log\_{10}(d\_\mathrm{pc})$$

**E.g:** if you have a star at the Vega's distance $d\_\mathrm{pc}=7.63\ \mathrm{pc}$ with apparent magnitude $m\simeq-0.58$ (more or less as Saturn maximum apparent magnitude) then you will have $M=0$


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