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?
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15$\begingroup$ 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. $\endgroup$– David HammenFeb 9, 2022 at 0:36
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1$\begingroup$ 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. $\endgroup$– polfosol ఠ_ఠFeb 9, 2022 at 11:12
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6$\begingroup$ It'd be a pretty small planet. The total mass of the asteroid belt is about 4% that of the Moon. $\endgroup$– PM 2RingFeb 9, 2022 at 14:31
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1$\begingroup$ @polfosolఠ_ఠ chickens do come from eggs, and eggs do come from chickens. $\endgroup$– hobbsFeb 10, 2022 at 2:34
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$\begingroup$ @Moradnejad: Why have you put back the errors in the English in the question that were edited to be correct? $\endgroup$– psmearsFeb 10, 2022 at 16:06
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2 Answers
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Whether we know or believe Phaeton exited, we can estimate its orbital period from some reasonable assumptions using Kepler's 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.
With the typical semi-major axis of a main belt asteroid being about 2.7 AU, this works out to be approximately 1620 days or 4.4 years.
answered Feb 9, 2022 at 1:57
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7$\begingroup$ 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! $\endgroup$– Connor Garcia ♦Feb 9, 2022 at 2:26
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$\begingroup$ 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. $\endgroup$– DronirFeb 9, 2022 at 9:02
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$\begingroup$ Good catch. I’ll fix when I’m off mobile later today if someone doesn’t get to it first. $\endgroup$ Feb 9, 2022 at 10:33
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$\begingroup$ 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... $\endgroup$– ChrisFeb 9, 2022 at 10:51
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2$\begingroup$ Ceres has a period of 4.6 years, so looks good... $\endgroup$ Feb 9, 2022 at 15:05
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The Titus-Bode 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.
