<p>The <a href="https://en.m.wikipedia.org/wiki/Titius%E2%80%93Bode_law" rel="nofollow noreferrer">Titus-Bode law</a> predicts<sup>1</sup> a semimajor axis of 2.7 AU which gives it a period of 4.4 years.</p>
<p><span class="math-container">$$a = \frac{4 + 3 \times 2^n}{10} \ \ \text{AU}$$</span></p>
<p>where</p>
<p><span class="math-container">$$n = -\infty, 0, 1, 2, 3...$$</span></p>
<p>Earth with <span class="math-container">$n=1$</span> is at 1 AU, an <span class="math-container">$n=3$</span> planet between Mars and Jupiter would be at 2.7 AU.</p>
<p>For a given star a small planet's period is proportional to <span class="math-container">$a^{3/2}$</span> and 2.7<span class="math-container">$^{3/2}$</span> is about 4.4</p>
<hr />
<p><sup>1</sup>there's no science behind the &quot;law&quot; so I hesitate to call it a <em>bona fide</em> envelope-back spherical cow estimate. It's probably no better than taking the geometrical average of the period of Mars and Jupiter and getting 4.7 years.</p>

CC BY-SA 4.0
2022-02-09T03:41:48.697Z
2022-02-09T11:40:54.557Z
2022-02-09T11:40:54.557Z
uhoh
<p>Whether we know or believe <a href="https://en.wikipedia.org/w/index.php?title=Phaeton_(hypothetical_planet)&amp;oldid=1053985712" rel="noreferrer">Phaeton</a> exited, we can estimate its orbital period from some reasonable assumptions using <a href="https://en.wikipedia.org/w/index.php?title=Kepler%27s_laws_of_planetary_motion&amp;oldid=1062045365#Third_law" rel="noreferrer">Kepler's third law</a>.</p>
<p>Kepler's third law states:</p>
<blockquote>
<p>The ratio of the square of an object's orbital period with the cube of
the semi-major axis of its orbit is the same for all objects orbiting
the same primary.</p>
</blockquote>
<p>Assuming a circular radius and a mass significantly less than that of the sun, a good approximation to the orbital period is given by:
<span class="math-container">$$\frac{a^3}{T^2} \approx 7.496 \cdot 10^{-6}\left(\frac{AU^3}{days^2}\right)$$</span>
<span class="math-container">$$T^2 \approx 1.334 \cdot 10^5 \left(\frac{days^2}{AU^3}\right) \cdot a^3$$</span>
<span class="math-container">$$T \approx \sqrt{1.334 \cdot 10^5 \left(\frac{days^2}{AU^3}\right) \cdot a^3}$$</span>
where <span class="math-container">$T$</span> is the orbital period in days and <span class="math-container">$a$</span> is the radius of the circular orbit in <a href="https://en.wikipedia.org/w/index.php?title=Astronomical_unit&amp;oldid=1060186457" rel="noreferrer">astronomical units</a>.</p>
<p>With the typical semi-major axis of a <a href="https://en.wikipedia.org/w/index.php?title=Asteroid_belt&amp;oldid=1068744292" rel="noreferrer">main belt asteroid</a> being about 2.7 AU, this works out to be approximately 1620 days or 4.4 years.</p>

CC BY-SA 4.0
2022-02-09T02:26:35.273Z
As you point out, we can't measure a hypothesized orbital period, but rather estimate it.  Kepler's 3rd is the perfect tool for this question! Excellent!
Connor Garcia
CC BY-SA 4.0
2022-02-09T09:02:02.053Z
You forgot to flip the AU^3/days^2 term when dividing, so the the units aren't right on your second and third line, though the numerical values are correct.
Dronir
CC BY-SA 4.0
2022-02-09T10:33:10.837Z
Good catch. I’ll fix when I’m off mobile later today if someone doesn’t get to it first.
GrapefruitIsAwesome
CC BY-SA 4.0
2022-02-09T10:51:42.467Z
are the AU and days just units? Dronir's comment and your reply seem to suggest that they are except you've said "AU is the radius[..]" which suggests it has a value. Should that have been "a is the radius" since a seems otherwise undefined. Also my maths seems to suggest that comes out as 1620...
Chris
CC BY-SA 4.0
2022-02-09T11:17:25.610Z
a is the radius and AU the unit. It needs fixed up. Should be fixed now
GrapefruitIsAwesome
CC BY-SA 4.0
2022-02-09T15:05:49.980Z
Ceres has a period of 4.6 years, so looks good...
Jon Custer
CC BY-SA 4.0
2022-02-09T01:57:22.610Z
2022-02-09T11:20:10.953Z
2022-02-09T11:20:10.953Z
GrapefruitIsAwesome
<p>Phaeton was a planet hypothesized to have existed between the orbits of Mars and Jupiter, the destruction of which supposedly led to the formation of the asteroid belt.</p>
<p>Can we measure its orbital period based on what we know of the asteroid belt?</p>

CC BY-SA 4.0
2022-02-09T00:36:28.563Z
Re *Phaeton is believed to have existed between the orbits of Mars and Jupiter* No, it's not. That is a very old hypothesis based on the now discarded numerological Titus-Bode Law. The idea that Phaeton ever existed has been widely discarded.
David Hammen
CC BY-SA 4.0
2022-02-09T11:12:51.367Z
I thought planets are gradually formed as a combination of smaller asteroids. Now for the first time I hear that the asteroid belt might be formed from destruction of a planet, it feels like a chicken and egg problem.
polfosol ఠ_ఠ
CC BY-SA 4.0
2022-02-09T14:31:03.550Z
It'd be a pretty small planet. [The total mass of the asteroid belt is about 4% that of the Moon.](https://en.wikipedia.org/wiki/Asteroid_belt)
PM 2Ring
CC BY-SA 4.0
2022-02-10T02:34:08.907Z
@polfosolఠ_ఠ chickens *do* come from eggs, and eggs *do* come from chickens.
hobbs
CC BY-SA 4.0
2022-02-10T16:06:39.497Z
@Moradnejad: Why have you put back the errors in the English in the question that were edited to be correct?
psmears
CC BY-SA 4.0
2022-02-11T18:30:31.903Z
@psmears The introductory sentence of the question is completely correct in its initial form and there was no reason to rephrase that.
Moradnejad
CC BY-SA 4.0
2022-02-11T23:05:56.357Z
@Moradnejad: The first sentence, as it currently stands, certainly is grammatically correct, but it means that _Jupiter_ was destroyed to form the asteroid belt. Is that really what you mean?
psmears
CC BY-SA 4.0
2022-02-12T10:06:13.170Z
@psmears That is interesting and there could be ambiguity. But based on the context, no one would assume that. However, I can only note that this is the sentence structure from Wikipedia:Phaeton (hypothetical planet).
Moradnejad
CC BY-SA 4.0
2022-02-12T22:36:23.010Z
@Moradnejad: The sentence in Wikipedia isn't great, and is a little ambiguous (remember there are lots of people with poor English editing Wikipedia!). But its structure is not the same as yours. In the wikipedia version it has the words _hypothetical planet_, and "which" refers back to that. But that is not present in your sentence, so the "which" can only (in good English) refer to Jupiter. If you want to refer back to the subject (Phaeton) you need to use "it" here rather than "which". (Yes, people _can_ figure out what you probably meant, but it just makes it harder for them...)
psmears
<asteroid-belt><sidereal-period>
Can we estimate the orbital period of the planet Phaeton (destroyed planet between Mars and Jupiter)?
Moradnejad
# Can we estimate the orbital period of the planet Phaeton (destroyed planet between Mars and Jupiter)?

(Asked by: username_0 on 2022-02-08)

Phaeton was a planet hypothesized to have existed between the orbits of Mars and Jupiter, the destruction of which supposedly led to the formation of the asteroid belt.

Can we measure its orbital period based on what we know of the asteroid belt?

**Tags:** asteroid-belt, sidereal-period

> **Comments:**
> **username_1**: Re *Phaeton is believed to have existed between the orbits of Mars and Jupiter* No, it's not. That is a very old hypothesis based on the now discarded numerological Titus-Bode Law. The idea that Phaeton ever existed has been widely discarded.  
> **username_2**: I thought planets are gradually formed as a combination of smaller asteroids. Now for the first time I hear that the asteroid belt might be formed from destruction of a planet, it feels like a chicken and egg problem.  
> **username_3**: It'd be a pretty small planet. [The total mass of the asteroid belt is about 4% that of the Moon.](https://en.wikipedia.org/wiki/Asteroid_belt)  
> **username_4**: @username_2 chickens *do* come from eggs, and eggs *do* come from chickens.  
> **username_5**: @username_0: Why have you put back the errors in the English in the question that were edited to be correct?  
> **username_0**: @username_5 The introductory sentence of the question is completely correct in its initial form and there was no reason to rephrase that.  
> **username_5**: @username_0: The first sentence, as it currently stands, certainly is grammatically correct, but it means that _Jupiter_ was destroyed to form the asteroid belt. Is that really what you mean?  
> **username_0**: @username_5 That is interesting and there could be ambiguity. But based on the context, no one would assume that. However, I can only note that this is the sentence structure from Wikipedia:Phaeton (hypothetical planet).  
> **username_5**: @username_0: The sentence in Wikipedia isn't great, and is a little ambiguous (remember there are lots of people with poor English editing Wikipedia!). But its structure is not the same as yours. In the wikipedia version it has the words _hypothetical planet_, and "which" refers back to that. But that is not present in your sentence, so the "which" can only (in good English) refer to Jupiter. If you want to refer back to the subject (Phaeton) you need to use "it" here rather than "which". (Yes, people _can_ figure out what you probably meant, but it just makes it harder for them...)  

---

## 2 answers

### Answer by username_7 on 2022-02-09. Score: 15 | Accepted answer

Whether we know or believe [Phaeton](https://en.wikipedia.org/w/index.php?title=Phaeton_(hypothetical_planet)&oldid=1053985712) exited, we can estimate its orbital period from some reasonable assumptions using [Kepler's third law](https://en.wikipedia.org/w/index.php?title=Kepler%27s_laws_of_planetary_motion&oldid=1062045365#Third_law).

Kepler's third law states:

> The ratio of the square of an object's orbital period with the cube of
> the semi-major axis of its orbit is the same for all objects orbiting
> the same primary.

Assuming a circular radius and a mass significantly less than that of the sun, a good approximation to the orbital period is given by:
$$\frac{a^3}{T^2} \approx 7.496 \cdot 10^{-6}\left(\frac{AU^3}{days^2}\right)$$
$$T^2 \approx 1.334 \cdot 10^5 \left(\frac{days^2}{AU^3}\right) \cdot a^3$$
$$T \approx \sqrt{1.334 \cdot 10^5 \left(\frac{days^2}{AU^3}\right) \cdot a^3}$$
where $T$ is the orbital period in days and $a$ is the radius of the circular orbit in [astronomical units](https://en.wikipedia.org/w/index.php?title=Astronomical_unit&oldid=1060186457).

With the typical semi-major axis of a [main belt asteroid](https://en.wikipedia.org/w/index.php?title=Asteroid_belt&oldid=1068744292) being about 2.7 AU, this works out to be approximately 1620 days or 4.4 years.

> **Comments:**
> **username_8**: As you point out, we can't measure a hypothesized orbital period, but rather estimate it.  Kepler's 3rd is the perfect tool for this question! Excellent!  
> **username_9**: You forgot to flip the AU^3/days^2 term when dividing, so the the units aren't right on your second and third line, though the numerical values are correct.  
> **username_7**: Good catch. I’ll fix when I’m off mobile later today if someone doesn’t get to it first.  
> **username_10**: are the AU and days just units? username_9's comment and your reply seem to suggest that they are except you've said "AU is the radius[..]" which suggests it has a value. Should that have been "a is the radius" since a seems otherwise undefined. Also my maths seems to suggest that comes out as 1620...  
> **username_7**: a is the radius and AU the unit. It needs fixed up. Should be fixed now  
> **username_11**: Ceres has a period of 4.6 years, so looks good...  

---

### Answer by username_6 on 2022-02-09. Score: 5

The [Titus-Bode law](https://en.m.wikipedia.org/wiki/Titius%E2%80%93Bode_law) predicts1 a semimajor axis of 2.7 AU which gives it a period of 4.4 years.

$$a = \frac{4 + 3 \times 2^n}{10} \ \ \text{AU}$$

where

$$n = -\infty, 0, 1, 2, 3...$$

Earth with $n=1$ is at 1 AU, an $n=3$ planet between Mars and Jupiter would be at 2.7 AU.

For a given star a small planet's period is proportional to $a^{3/2}$ and 2.7$^{3/2}$ is about 4.4

---

1there's no science behind the "law" so I hesitate to call it a *bona fide* envelope-back spherical cow estimate. It's probably no better than taking the geometrical average of the period of Mars and Jupiter and getting 4.7 years.


---

