Calculating A Retrograde Of Mercury - Total Beginner Question

Question

I am a total beginner to doing this kind of math, so there is a mean learning curve.

There is need to calculate the number of seconds from when mercury is direct to the next direct phase. My thoughts about doing this:

speed of mercury = 170496, speed of earth = 107226

days for mercury to complete 87.9691, days for earth to complete 365.25

distance mercury travels = speed * days, distance earth travels = speed * days

Since mercury completes in 87 days and earth completes in 365 days, need to calculate the difference of earths travel.

offset distance = earth speed * days for mercury.

difference + offset distance

since the two circles (earths orbit & mercury orbit) are different sizes, need to calculate the difference between the two circles.

Would I want to multiply the distance mercury travels with the difference of the two orbits?

and here is where I get lost. My formula comes out with a wrong offset of six days.

Is there a better way to calculate what I want?

If you could be clear about how to solve the problem, that would be great.

Answer 1

One can compute this time referring only to orbital periods and not to distances, at least to a first-order approximation that assumes that both orbits are circular and coplanar.*

Let $\omega_M$ and $\omega_E$ denote the angular frequencies of Mercury and Earth, respectively. Then, the (constant) rate of change of the Mercury-Sun-Earth angle $\theta$ is $\omega_M - \omega_E$, and so at time $t$ after an opposition (where all three bodies are in a line, with the Sun in the middle) that angle is $$\theta = (\omega_M - \omega_E) t.$$ The next opposition after the one at time $t = 0$ occurs at the time $T$ when $\theta = 2\pi$, that is, at time $$T = \frac{2\pi}{\omega_M - \omega_E} .$$

More or less by definition, the angular velocity $\omega_A$ of the (circular) orbit of a body $A$ and its orbital period $T_A$ are related by $\omega_A = \frac{2\pi}{T_A}$. Substituting in the above formula, simplifying, and substituting the values $$T_M=87.97\,\textrm{d} \qquad \textrm{and} \qquad T_E=365.26\,\textrm{d}$$ gives that the time between successive oppositions is $$\boxed{T = \frac{T_M T_E}{T_E - T_M} \approx 115.88\,\textrm{d}} .$$

*The orbit of Mercury is noticeably noncircular: Its eccentricity is $e_M = 0.206$---by far the largest among the eight planets. This makes Mercury's speed at perihelion (its closest point to the sun) $$\frac{1 + e_M}{1 - e_M} \approx 1.53$$ times its speed at aphelion (the farthest point). So both Mercury's linear and angular velocity are far from constant, causing some variation in the duration of retrograde: It's shorter when Mercury is in opposition near perihelion and longer when it is in opposition near aphelion. For example, the three retrogrades of Mercury in 2021 have durations $21.38\,\textrm{d}$, $23.98\,\textrm{d}$, and $21.42\,\textrm{d}$. As a consequence, the time between the starts of Mercury's successive retrogrades varies some, too.