Astronomy Answers: Transits of the Sun

# Astronomy AnswersTransits of the Sun

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If a planet passes in front of the Sun, as seen from Earth, then that is called a transit of that planet (across the Sun), for example a transit of Venus. When do those happen?

The geocentric ecliptic latitude of the Sun is always equal to zero (if we use the equinox of the date), so for a transit the geocentric ecliptic latitude of the planet must be close to zero, too, so the planet must then be close to one of the two nodes of its orbit. In other words, then $$z_\text{planet}$$ from the Sky Position Page must be close to zero. The planet's coordinates are, according to that page (with an extra subscript "planet" thrown in here and there):

\begin{align} x_\text{planet} \| = r_\text{planet} (\cos Ω \cos(ω + ν_\text{planet}) - \sin Ω \cos i \sin(ω + ν_\text{planet})) \\ y_\text{planet} \| = r_\text{planet} (\sin Ω \cos(ω + ν_\text{planet}) + \cos Ω \cos i \sin(ω + ν_\text{planet})) \\ z_\text{planet} \| = r_\text{planet} \sin i \sin(ω + ν_\text{planet}) \end{align}

$$z_\text{planet}$$ is equal to zero if $$ν_\text{planet} + ω = 0 \bmod 180°$$. We call $$ν_\text{planet} + ω ≡ α_0 + α_\text{planet}$$ and approximate the above formulas (to first order in $$α_\text{planet}$$) for the case that $$α_0$$ is equal to 0° or to 180°. For convenience, we set $$Ω$$ equal to 90°, so the nodes have $$x_\text{planet} = 0$$. For these kinds of approximations, you must measure the angles in radians, so the formulas below have all angles measured in radians, unless it explicitly says otherwise.

\begin{align} x_\text{planet} \| ≈ α_\text{planet} r_\text{planet} \cos i \\ y_\text{planet} \| ≈ r_\text{planet} \\ z_\text{planet} \| ≈ α_\text{planet} r_\text{planet} \sin i \end{align}

For $$α_0 = 180°$$ you should multply the right-hand side of the above equations (and hence also for quite a few formulas below) by −1, but that does not change any of the conclusions, so we'll not indicate this all the time.

In the same fashion, we find for the Earth (which here has $$i = 0$$ by definition):

\begin{align} x_\text{Earth} \| ≈ α_\text{Earth} r_\text{Earth} \\ y_\text{Earth} \| ≈ r_\text{Earth} \\ z_\text{Earth} \| = 0 \end{align}

For the geocentric ecliptic coordinates of the planet we now find

\begin{align} β_\text{planet} \| ≈ \frac{z_\text{planet} - z_\text{Earth}}{r_\text{Earth} - r_\text{planet}} = \frac{z_\text{planet}}{r_\text{Earth} - r_\text{planet}} \\ λ_\text{planet} \| ≈ \frac{x_\text{planet} - x_\text{Earth}}{r_\text{Earth} - r_\text{planet}} \end{align}

and for the geocentric ecliptic coordinates of the Sun

\begin{align} β_\text{Sun} \| = 0 \\ λ_\text{Sun} \| ≈ \frac{x_\text{Sun} - x_\text{Earth}}{r_\text{Earth} - r_\text{Sun}} = -\frac{x_\text{Earth}}{r_\text{Earth}} \end{align}

Relative to the middle of the solar disk, the coordinates of the planet are

\begin{align} β \| = β_\text{planet} - β_\text{Sun} = \frac{z_\text{planet}}{r_\text{Earth} - r_\text{planet}} = f α_\text{planet} \sin i \\ λ \| = λ_\text{planet} - λ_\text{Sun} = \frac{x_\text{planet} - x_\text{Earth}}{r_\text{Earth} - r_\text{planet}} + \frac{x_\text{Earth}}{r_\text{Earth}} \notag \\ \| = \frac{x_\text{planet} - x_\text{Earth} \frac{r_\text{planet}}{r_\text{Earth}}}{r_\text{Earth} - r_\text{planet}} = f (α_\text{planet} \cos i - α_\text{Earth}) \end{align}

where

$$f = \frac{r_\text{planet}}{r_\text{Earth} - r_\text{planet}}$$

and the angular distance $$δ$$ from the center of the solar disk satisfies

$$δ^2 = λ^2 + β^2 = f^2 (α_\text{planet}^2 + α_\text{Earth}^2 - 2 α_\text{planet} α_\text{Earth} \cos i)$$

To have a transit, the angular distance $$δ$$ from the center of the solar disk must become less than the radius of the solar disk, which we call $$ρ_\text{Sun}$$. We define

\begin{align} α_\text{planet} \| = n_\text{planet} t \\ α_\text{Earth} \| = n_\text{Earth} (t + ∆t) \end{align}

where $$n_\text{planet}$$ is the daily angular motion of the planet in its orbit around the Sun (i.e., the time derivative of the true anomaly) near the node passage, $$n_\text{Earth}$$ that of the Earth, $$t$$ the time, in days, since the node passage of the planet, and $$∆t$$ how many days earlier the Earth reached the heliocentric ecliptic longitude of the node of the orbit of the planet than the planet itself did. With this, we find

\begin{align} \frac{δ^2}{f^2} \| = (n_\text{planet}^2 + n_\text{Earth}^2 - 2 n_\text{planet} n_\text{Earth} \cos i) t^2 \notag \\ \| + 2 n_\text{Earth} ∆t (n_\text{Earth} - n_\text{planet} \cos i) t + n_\text{Earth}^2 ∆t^2 \end{align}

A second-degree polynomial equation $$y = a t^2 + b t + c$$ has its greatest (if $$a \lt 0$$) or least (if $$a \gt 0$$) value of $$y$$ for $$t = -\frac{b}{2a}$$ and that greatest or least value is then equal to $$c - \frac{b^2}{4a}$$. In our case, the least value of $$δ$$ is attained for

$$t = t_\min = n_\text{Earth} ∆t \frac{n_\text{planet} \cos i - n_\text{Earth}}{n_\text{planet}^2 + n_\text{Earth}^2 - 2 n_\text{planet} n_\text{Earth} \cos i}$$

and that least value $$δ_\min$$ is then equal to

$$δ_\min = \frac{f n_\text{Earth} n_\text{planet} ∆t \sin i}{\sqrt{n_\text{Earth}^2 + n_\text{planet}^2 - 2 n_\text{Earth} n_\text{planet} \cos i}}$$

For a transit we must have $$δ_\min \lt ρ_\text{Sun}$$, so

$$|∆t| \lt ∆t_\max = \frac{ρ_\text{Sun} \sqrt{n_\text{Earth}^2 + n_\text{planet}^2 - 2 n_\text{Earth} n_\text{planet} \cos i}}{f n_\text{Earth} n_\text{planet} \sin i}$$

Now we should calculate what $$n_\text{Earth}$$, $$n_\text{planet}$$, $$r_\text{Earth}$$ and $$r_\text{planet}$$ are near both nodes of the orbit of the planet, but I don't feel like doing that at the moment. Those values are different not just for each planet and each node, but also slowly change with time, because the nodes slowly move along the orbit of the planet.

To still get a quick impression of the size of this boundary, we assume that all planets have circular orbits around the Sun, we measure distances in AU, and we assume that the inclination $$i$$ is small. Then we have

$$n_\text{planet} = n_\text{Earth} \left( \frac{r_\text{Earth}}{r_\text{planet}} \right)^{3/2} = n_\text{Earth} r_\text{planet}^{−3/2}$$

and then

$$∆t_\max ≈ \frac{(1 - r_\text{planet}^{3/2}) (1 - r_\text{planet})}{r_\text{planet} n_\text{Earth} \sin i}$$

For Venus with $$r_\text{planet}$$ equal to 0.7233 and $$i$$ equal to 3.4° (and with $$n_\text{Earth}$$ = 360 degrees per year) we find $$∆t_\max$$ ≈ 2.5 days. For Mercury with $$r_\text{planet}$$ equal to 0.3872 and $$i$$ equal to 7.0° we find $$∆t_\max$$ ≈ 10.0 days. If the Earth is at the same node of the orbit of the planet as the planet itself not more than that many days before or after the planet itself, then there is a transit of that planet across the Sun (according to our simplified calculations). The same boundary holds for the difference between the time of the inferior conjunction of the planet and the time of the passage of the planet through the node.

Because the boundary is wider for Mercury than for Venus, Mercury has about 10.0/2.5 = 4 times more transits than Venus.

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Last updated: 2017-12-28