<p>We can estimate this using the <a href="https://en.m.wikipedia.org/wiki/Inverse-square_law" rel="nofollow noreferrer">inverse square law</a> and the <a href="https://en.wikipedia.org/wiki/Mass%E2%80%93luminosity_relation" rel="nofollow noreferrer">mass-luminosity relation</a>,
<span class="math-container">$$L=M^a$$</span>
where <span class="math-container">$L$</span> is the luminosity of the star, and <span class="math-container">$M$</span> is its mass, both measured relative to the Sun. (That is, the mass and luminosity of the Sun = <span class="math-container">$1$</span>). For a main sequence star of mass similar to the Sun <span class="math-container">$a\approx4$</span>, for larger stars <span class="math-container">$a\approx3.5$</span>. Please bear in mind that this is just an estimate, and that a star gets more luminous as it ages.</p>
<p>To keep things simple, let's assume that orbits are perfect circles. By the inverse square law, a planet with orbit radius <span class="math-container">$R$</span> orbiting a star of luminosity <span class="math-container">$L$</span> receives the same intensity of light and heat as the Earth if
<span class="math-container">$$R^2=L$$</span>
where <span class="math-container">$R$</span> is measured in <a href="https://en.wikipedia.org/wiki/Astronomical_unit" rel="nofollow noreferrer">astronomical units</a>.</p>
<p>From <a href="https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law" rel="nofollow noreferrer">Kepler's 3rd law</a>,
<span class="math-container">$$R^3=MT^2$$</span>
where <span class="math-container">$T$</span> is the orbital period in <a href="https://en.wikipedia.org/wiki/Sidereal_year" rel="nofollow noreferrer">sidereal years</a>.</p>
<p>Combining these equations we get
<span class="math-container">$$M=T^p$$</span>
where
<span class="math-container">$$p=\frac{4}{3a-2}$$</span></p>
<p>Here are the results for <span class="math-container">$T=515.375$</span> days.</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: left;">a</th>
<th style="text-align: center;">3.5</th>
<th style="text-align: center;">4</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: left;">Mass</td>
<td style="text-align: center;">1.175885</td>
<td style="text-align: center;">1.147652</td>
</tr>
<tr>
<td style="text-align: left;">Lumin</td>
<td style="text-align: center;">1.763102</td>
<td style="text-align: center;">1.734766</td>
</tr>
<tr>
<td style="text-align: left;">Radius</td>
<td style="text-align: center;">1.327818</td>
<td style="text-align: center;">1.317105</td>
</tr>
</tbody>
</table>
</div>
<p>I calculated those results using this <a href="https://sagecell.sagemath.org/?z=eJxdUmFv2jAQ_Y7Ef3jK1AkmGiAhUFAjbdK6LwNpCq32DeSSg1gEJ7OdVqjqf985brcxf7B07-6e7z1fEAT4Spb0SSqCLQjGCo2TMAarAcqG8cpIe8ZyAKFyVPpRWmiRy8Yg63bAZ19pCNSlUGSZQ1gUwngycSLIsqURVlYKnLgT2hYD33qQT6QgrYGROWkSJc7EA5SkDraAVMjF2YS-2N8rfMQSQpPLmqrk6kYxwwAZpHGgMFZXqjrJHdO1uQuCB0M8-iZGylyfcL-JPO7kMbNDNxAeey54KMRhgluWeItJK3bdqOtSHr1ZjtwXr4lQWFubxXBIKnyWR1lTLkVY6cPQRcMV-3p1F13djK7m8V9zt5rK1p53pp9aWsvGPJ7xY4Uok-qAaBTF4SgJo3G3EwSBK_2AL_9KfWCpTv_x1O2IhoWMJ_NkPruZjcKZL19fmPzmbrfTRiniKbMn03gau-rPUvFeiJ3tdnLaY9urScsqT5NxEsazhNchnfDV2Grb1LmwlH4TpaH-wku4Z0LfgWH7nIdrhieM9GK2XuAaUf_9W1Pu2aD24X__kHFofmnbW76V15rH6wXO0EUwwOoSXv6x1iUve_ZB1m7vAi_ZK1qfXjI3TLMIR_H-le0LLtm-U12Sxq6g3ZHpeHUGCFL3ql-f_m99w-qJ&amp;lang=sage" rel="nofollow noreferrer">Sage / Python script</a>.</p>
<p>A star in that mass range is a heavy G-type star, or perhaps a light F-type star.</p>

CC BY-SA 4.0
2023-05-21T07:19:35.857Z
Stars (solar type stars on the main sequence) get more luminous as they age, the temperature is fairly constant.
ProfRob
CC BY-SA 4.0
2023-05-21T08:22:43.337Z
What orbital distance is needed?
Galactic
CC BY-SA 4.0
2023-05-21T08:25:29.050Z
@Galactic Using a=4, the orbit radius is ~ 1.317 au ~= 197036000 km.
PM 2Ring
CC BY-SA 4.0
2023-05-21T08:31:32.153Z
@PM2Ring, what about using a=3.5?
Galactic
CC BY-SA 4.0
2023-05-21T08:44:47.577Z
@Galactic Like the table says, for a=3.5, radius is ~ 1.328 au. If you want the value in km, please click on the script link. You don't need to install anything to run the script.
PM 2Ring
CC BY-SA 4.0
2023-05-22T01:52:19.847Z
@PM2Ring to the rescue! For an "Earthlike planet" I wonder if it's the *fraction* of the luminosity that actually gets through the atmosphere, reaches the surface and is energetically useful for photosynthesis? In other words should there be some kind of "color penalty" that reduces $L$ to $L_{\text{useful}}$? Perhaps some exoplanetologist has done this long ago and it just needs to be looked up?
uhoh
CC BY-SA 4.0
2023-05-22T02:53:41.693Z
@uhoh Yes, there probably should be a colour correction, since the peak frequency of a more luminous star is higher. But this is only a crude estimate which ignores star age & metallicity. We only have good mass-luminosity data for a few hundred stars (mostly from nearby binary systems, see [DEBCat](https://www.astro.keele.ac.uk/jkt/debcat/)) to fit the mass-luminosity function to.
PM 2Ring
CC BY-SA 4.0
2023-05-21T06:32:29.660Z
2023-05-21T08:22:07.917Z
2023-05-21T08:22:07.917Z
PM 2Ring
<p>A planet has a year length of 515 Earth days and 9 Earth hours. It is the same size and has the same climate as Earth. What stellar conditions would be needed to produce this and what orbital distance is needed?</p>

<orbit><planet><exoplanet><habitable-zone><earth-like-planet>
What stellar conditions and orbital distance are needed to produce a year length of 515 Earth days and 9 Earth hours on an Earthlike planet?
Galactic
# What stellar conditions and orbital distance are needed to produce a year length of 515 Earth days and 9 Earth hours on an Earthlike planet?

(Asked by: username_0 on 2023-05-21)

A planet has a year length of 515 Earth days and 9 Earth hours. It is the same size and has the same climate as Earth. What stellar conditions would be needed to produce this and what orbital distance is needed?

**Tags:** orbit, planet, exoplanet, habitable-zone, earth-like-planet

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## 1 answers

### Answer by username_1 on 2023-05-21. Score: 2

We can estimate this using the [inverse square law](https://en.m.wikipedia.org/wiki/Inverse-square_law) and the [mass-luminosity relation](https://en.wikipedia.org/wiki/Mass%E2%80%93luminosity_relation),
$$L=M^a$$
where $L$ is the luminosity of the star, and $M$ is its mass, both measured relative to the Sun. (That is, the mass and luminosity of the Sun = $1$). For a main sequence star of mass similar to the Sun $a\approx4$, for larger stars $a\approx3.5$. Please bear in mind that this is just an estimate, and that a star gets more luminous as it ages.

To keep things simple, let's assume that orbits are perfect circles. By the inverse square law, a planet with orbit radius $R$ orbiting a star of luminosity $L$ receives the same intensity of light and heat as the Earth if
$$R^2=L$$
where $R$ is measured in [astronomical units](https://en.wikipedia.org/wiki/Astronomical_unit).

From [Kepler's 3rd law](https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law),
$$R^3=MT^2$$
where $T$ is the orbital period in [sidereal years](https://en.wikipedia.org/wiki/Sidereal_year).

Combining these equations we get
$$M=T^p$$
where
$$p=\frac{4}{3a-2}$$

Here are the results for $T=515.375$ days.

| a | 3.5 | 4 |
| --- | --- | --- |
| Mass | 1.175885 | 1.147652 |
| Lumin | 1.763102 | 1.734766 |
| Radius | 1.327818 | 1.317105 |

I calculated those results using this [Sage / Python script](https://sagecell.sagemath.org/?z=eJxdUmFv2jAQ_Y7Ef3jK1AkmGiAhUFAjbdK6LwNpCq32DeSSg1gEJ7OdVqjqf985brcxf7B07-6e7z1fEAT4Spb0SSqCLQjGCo2TMAarAcqG8cpIe8ZyAKFyVPpRWmiRy8Yg63bAZ19pCNSlUGSZQ1gUwngycSLIsqURVlYKnLgT2hYD33qQT6QgrYGROWkSJc7EA5SkDraAVMjF2YS-2N8rfMQSQpPLmqrk6kYxwwAZpHGgMFZXqjrJHdO1uQuCB0M8-iZGylyfcL-JPO7kMbNDNxAeey54KMRhgluWeItJK3bdqOtSHr1ZjtwXr4lQWFubxXBIKnyWR1lTLkVY6cPQRcMV-3p1F13djK7m8V9zt5rK1p53pp9aWsvGPJ7xY4Uok-qAaBTF4SgJo3G3EwSBK_2AL_9KfWCpTv_x1O2IhoWMJ_NkPruZjcKZL19fmPzmbrfTRiniKbMn03gau-rPUvFeiJ3tdnLaY9urScsqT5NxEsazhNchnfDV2Grb1LmwlH4TpaH-wku4Z0LfgWH7nIdrhieM9GK2XuAaUf_9W1Pu2aD24X__kHFofmnbW76V15rH6wXO0EUwwOoSXv6x1iUve_ZB1m7vAi_ZK1qfXjI3TLMIR_H-le0LLtm-U12Sxq6g3ZHpeHUGCFL3ql-f_m99w-qJ&lang=sage).

A star in that mass range is a heavy G-type star, or perhaps a light F-type star.

> **Comments:**
> **username_2**: Stars (solar type stars on the main sequence) get more luminous as they age, the temperature is fairly constant.  
> **username_0**: What orbital distance is needed?  
> **username_1**: @Galactic Using a=4, the orbit radius is ~ 1.317 au ~= 197036000 km.  
> **username_0**: @PM2Ring, what about using a=3.5?  
> **username_1**: @Galactic Like the table says, for a=3.5, radius is ~ 1.328 au. If you want the value in km, please click on the script link. You don't need to install anything to run the script.  
> **username_3**: @PM2Ring to the rescue! For an "Earthlike planet" I wonder if it's the *fraction* of the luminosity that actually gets through the atmosphere, reaches the surface and is energetically useful for photosynthesis? In other words should there be some kind of "color penalty" that reduces $L$ to $L_{\text{useful}}$? Perhaps some exoplanetologist has done this long ago and it just needs to be looked up?  
> **username_1**: @uhoh Yes, there probably should be a colour correction, since the peak frequency of a more luminous star is higher. But this is only a crude estimate which ignores star age & metallicity. We only have good mass-luminosity data for a few hundred stars (mostly from nearby binary systems, see [DEBCat](https://www.astro.keele.ac.uk/jkt/debcat/)) to fit the mass-luminosity function to.  

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