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The sidereal period of Mercury's revolution is 88 days and the synodic period — 116d.

my solution, but in the question featured "the greatest rapprochement." And this is no longer so easy. Because the orbit of Mercury is noticeably elongated and ... in turn, it experiences rotation around the Sun. The effect is called "precession of the perihelion of the orbit of Mercury". We must take it into account. The closest approach of Mercury to Earth will be when it is at this point, and it, in turn, is on the Sun-Earth line.

The time of Mercury's revolution around the Sun is 88 days,

Earth around the Sun -365 days,

Let's say we are on Mercury and today there is an Earth opposition, the next one will be when Mercury returns to its original place (having made a circle in its orbit) it is 88 days, but these 88 days the Earth does not stand still, it has shifted in its orbit by 88/365 by about a quarter => Mercury will have to "catch up" with the Earth for 22 days, but during these 22 days the Earth will still shift in orbit by 22/365 approximately one sixteenth => another +5 days. Total 88 days + 22 days + 5 days = 115 days

Answer: oppositions of the Earth for an observer with Mercury are repeated with a frequency of 115 days

I found the answer 7.26 years but how do get it?

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  • $\begingroup$ @fasterthanlight added more details $\endgroup$ – марат Nov 29 '20 at 17:31
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    $\begingroup$ I wouldn't expect Mercury's perihelion precession to have that large of an effect, since the total precession is only 574.10 ± 0.65 arc-seconds per century. But its (orbit) eccentricity is certainly important for this question. You can probably ignore Earth's eccentricity, since it's only around 0.0167. $\endgroup$ – PM 2Ring Nov 29 '20 at 21:31
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    $\begingroup$ This is a great question! :-) $\endgroup$ – uhoh Nov 30 '20 at 4:13
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The text you quoted (is that from your textbook?) uses a simple approach; it counts every closest approach of Mercury to Earth, not just the ones which occur when Mercury is relatively far from the Sun (and hence closer to Earth). That means you don't need to account for

  • the orbit of Mercury is noticeably elongated
  • precession of the perihelion of the orbit of Mercury

Especially the latter effect is so small that you don't need to account for it; only very careful measurements at the end of the 19th century revealed there was something wrong with Newtonian mechanics.

The calculation itself is quite simple: Mercury has an angular velocity of $\frac{1}{87.97} \text{day}^{-1}$ and Earth $\frac{1}{365.24} \text{day}^{-1}$. If their closest approach is at $t = 0\,\text{day}$, the next one occurs when Mercury did exactly one revolution more than Earth, so

$$\frac{t}{87.97} = 1 + \frac{t}{365.24}$$

$$t = 87.97 + 0.24 t$$

$$0.76t = 87.97$$

$$t = 115.75$$

so almost 116 days. I don't really like the quoted 'Achilles and the tortoise paradox'-style approach.

If you do want to take Mercury's rather elliptical orbit into account, you can note that after 3 times 115.75 = 347.25 days, the Earth (and hence Mercury) are at almost the same places in their orbit, so a 'relatively close' closest approach is followed by another one 347 days later.

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@Glorfindel's explanation is very clear. I wondered about how big of an effect "everything else" would have, but I couldn't figure out how to add an image to a comment. Here is a graph from 2018 to 2023 of the distance between Mercury and Earth. You can see every 3rd minimum is a bit deeper than the others just like Glorfindel said. The details of each local minimum change in a somewhat irregular way (about 10% of the closest distance variation), but the broad pattern fits the simple explanation well.

A sinusoidal graph with minimums every 88 days, but with deeper minimums every 3rd time.

Here is python code to generate such a graph yourself for a different time period. It uses skyfield to access JPL's data on the planetary positions and matplotlib to graph it.

from skyfield.api import load
import numpy as np
import matplotlib.pyplot as plt

planets = load("de421.bsp")
ts = load.timescale()

times = ts.utc(2018,1,np.linspace(0,365*5,10000))

distances = (planets["Mercury"] - planets["Earth"]).at(times).distance()

plt.plot(times.utc_datetime(), distances.km);
plt.title("Distance from Mercury to Earth in km versus time");

plt.savefig("Mercury-Earth-2018-2023.png");
plt.close()
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