Any estimate of the common period of more than two planets (i.e., after how much time do they approximately align in heliocentric longitude again?) depends very strongly on how much deviation from perfect alignment is acceptable.
If the period of planet $i$ is $P_i$, and if the acceptable deviation in time is $b$ (in the same units as $P_i$), then the combined period $P$ of all $n$ planets is approximately $$P approx frac{prod_i P_i}{b^{n-1}}$$ so reducing the acceptable deviation by a factor of 10 means increasing the common period by a factor of $10^{n-1}$, which for 8 planets is a factor of 10,000,000. So, it is meaningless to quote a common period if you don't also specify how much deviation was acceptable. When the acceptable deviation declines to 0 (to achieve "perfect alignment"), then the common period increases to infinity. This corresponds to several commenters' statements that there is no common period because the periods are not commensurate.
For the planets' periods listed by harogaston, $prod_i P_i approx 1.35times10^6$ when the $P_i$ are measured in Julian years of 365.25 days each, so the common period in years is approximately $$P approx frac{1.35times10^6}{b^7}$$ if $b$ is measured in years as well. If the periods are approximated to the nearest day, then $b approx 0.00274$ years and $P approx 1.2times10^{24}$ years. If the periods are approximated to the nearest 0.01 day, then $b approx 2.74times10^{-5}$ and $P approx 1.2times10^{38}$ years.
The derivation of the above formula is as follows:
Approximate the planets' periods by multiples of a base unit $b$: $P_i approx p_i b$ where $p_i$ is a whole number. Then the common period is at most equal to the product of all $p_i$. That product is still measured in units of $b$; we must multiply by $b$ to go back to the original units. So, the common period is approximately $$P approx b prod_i p_i approx b prod_i frac{P_i}{b} = b frac{prod_i P_i}{b^n} = frac{prod_i P_i}{b^{n-1}}$$
The above derivation doesn't take into account that the $p_i$ might have common factors so that the alignment occurs sooner than $prod_i p_i$ suggests. However, whether or not any two $p_i$ have common factors depends strongly on the chosen base period $b$, so it is effectively a random variable and does not affect the global dependence of $P$ on $b$.
If you express the acceptable deviation in terms of angle rather than time, then I expect you'll get answers that depend on the size of the acceptable deviation as strongly as for the above formula.
See http://aa.quae.nl/en/reken/periode.html for a graph of $P$ as a function of $b$ for all planets including Pluto.
EDIT:
Here is an estimate with acceptable deviation in terms of angle. We
want all planets to be within a range of longitude of width $δ$
centered on the longitude of the first planet; the longitude of the
first planet is free. We assume that all planets move in the same
direction in coplanar circular orbits around the Sun.
Because the planets' periods are not commensurate, all combinations of
longitudes of the planets occur with the same probability. The
probability $q_i$ that at some specific moment of time the longitude
of planet $i > 1$ is within the segment of width $δ$ centered on the
longitude of planet 1 is equal to $$q_i = frac{δ}{360°}$$
The probability $q$ that planets 2 through $n$ are all within that
same segment of longitude centered on planet 1 is then $$q =
prod_{i=2}^n q_i = left( frac{δ}{360°} right)^{n-1}$$
To translate that probability to an average period, we need to
estimate for how much time all planets are aligned (to within $δ$)
each time they are all aligned.
The first two planets to lose their mutual alignment are the fastest
and slowest of the planets. If their synodic period is $P_*$, then
they'll be in alignment for an interval $$A = P_* frac{δ}{360°}$$ and
then out of alignment for some time before coming into alignment
again. So, each alignment of all planets lasts about an interval $A$,
and all of those alignments together cover a fraction $q$ of all time.
If the average period after which another alignment of all planets
occurs is $P$, then we must have $qP = A$, so $$P = frac{A}{q} = P_*
left( frac{360°}{δ} right)^{n-2}$$
If there are only two planets, then $P = P_*$ regardless of $δ$, which is as expected.
If there are many planets, then the fastest planet is a lot faster
than the slowest one, so then $P_*$ is very nearly equal to the
orbital period of the fastest planet.
Here, too, the estimate for the average time between successive
alignments is very sensitive to the chosen deviation limit (if there
are more than two planets involved), so it is meaningless to quote
such a combined period if you don't also mention what deviation was
allowed.
It is also important to remember that (if there are more than two
planets) these (near-)alignments of all of them do not occur at
regular intervals.
Now let's plug in some numbers. If you want all 8 planets to be
aligned to within 1 degree of longitude, then the average time between
two such alignments is roughly equal to $P = 360^6 = 2.2×10^{15}$
orbits of the fastest planet. For the Solar System, Mercury is the
fastest planet, with a period of about 0.241 years, so then the
average time between two alignments of all 8 planets to within 1
degree of longitude is about $5×10^{14}$ years.
If you are satisfied already with an alignment to within 10 degrees
of longitude, then the average period between two such alignments is
roughly equal to $P = 36^6 = 2.2×10^9$ orbits of Mercury, which is
about 500 million years.
What is the best alignment that we can expect during the coming 1000
years? 1000 years are about 4150 orbits of Mercury, so $(360°/δ)^6
approx 4150$, so $δ approx 90°$. In an interval of 1000 years
chosen at random, there is on average one alignment of all 8 planets
to within a segment of 90°.
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