Astronomy Answers: Time of Least Distance

# Astronomy AnswersTime of Least Distance

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This page explains how you can estimate the time and value of the least distance from a series of measurements or calculations of how the distance varies with time, of something that moves in a straight line.

## 1. The Method

Fig. 1: Elongation of Saturn (15 days)
Figure 1 shows the elongation (in degrees) of Saturn between November 17 and December 2 of 2018, with one data point per day. The x axis shows the number of days since January 1st, 2018. It is clear that the elongation has its lowest value for $$t$$ somewhere between 328 and 330, but where exactly? In this figure the distance is least for $$t = 329$$, but it would be an unusual coincidence if the elongation happened to reach its lowest value exactly at one of the times for which we have a measurement.

If you look at the position of Saturn in the sky relative to that of the Sun, then Saturn passes the Sun along a roughly straight line, and the elongation is the distance between the Sun and Saturn.

If you know the distance $$r$$ at three times $$t_0 − τ$$, $$t_0$$, and $$t_0 + τ$$, then you can estimate the smallest distance $$d$$ and the time $$t_\text{min}$$ at which it is attained, from

\begin{eqnarray} v & = & \frac{\sqrt{r(t_0 − τ)^2 − 2 r(t_0)^2 + r(t_0 + τ)^2}}{τ\sqrt{2}} \\ l_0 & = & \frac{r(t_0 + τ)^2 − r(t_0 − τ)^2}{4vτ} \\ d & = & \sqrt{r(t_0)^2 − l_0^2} \\ t_\text{min} & = & t_0 − \frac{l_0}{v} \end{eqnarray}

For example, the measurements from the figure for $$t$$ from 328 to 330 are

$${t}$$ $${r(t)}$$
328 0.8990879
329 0.6809476
330 1.1668921

With these values we find

\begin{eqnarray*} t_0 & = & 329 \\ v & = & \frac{\sqrt{0.8990879^2 − 2 × 0.6809476^2 + 1.1668921^2}}{1\sqrt{2}} = \frac{\sqrt{1.12426171}}{\sqrt{2}} = 0.7882313 \\ l_0 & = & \frac{1.1668921^2 − 0.8990879^2}{4 × 0.7882313 × 1} = \frac{0.5532783}{3.1529251} = 0.1754889 \\ d & = & \sqrt{0.6809476^2 − 0.1754889^2} = \sqrt{0.4328960} = 0.6579483 \\ t_\text{min} & = & 329 − \frac{0.1754889}{0.7882313} = 328.7773738 \end{eqnarray*}

so the estimate is that the smallest distance is 0.6579483° and is reached when $$t$$ is 328.7773738.

Fig. 2: Elongation of Saturn (2 days)
Figure 2 shows the elongation of Saturn for the two days around the time of the smallest distance. The crosses show the distances at an interval of a day ($$τ = 1$$), like in the previous figure. The solid line connects the distances with an interval of 0.1 day ($$τ = 0.1$$). The little square marks the time and value of the smallest distance as estimated from the three crosses. This fits very well with the lowest point of the solid curve.

The estimate of the time and value of the smallest distance is accurate if the motion can be described as having a constant speed along a straight line. If the speed varies slowly or if the trajectory is nearly but not exactly a straight line, then the estimate will be reasonable but may not be exactly right.

For comparison: The estimate of the smallest distance based on the three nearest data points with an interval of 0.1 days ($$τ = 0.1$$) instead of 1 day is that the smallest distance is 0.6579986° and is reached when $$t$$ is 328.7775848, so that differs only 0.00005° in the distance and only 0.0002 dagen or 18 seconds in time. This gives an impression of the accuracy of the estimate based on the interval of 1 day.

## 2. The Derivation

Suppose that a straight line runs past but not through point A. The point where the line comes closest to point A is called point B. The distance between A and B is $$d$$: That is the shortest distance that a point on the line can have to A. We turn the line into a number line, with the 0 at point B and with the same unit of measurement as for $$d$$. A point at number $$l$$ on that line then has a distance $$r$$ to point A that is equal to

$$r = \sqrt{d^2 + l^2}$$

An object moves along the line at fixed speed $$v$$. We have three successive measurements of the distance $$r = r(t)$$ for times $$t$$ equal to $$t_0 − τ$$, $$t_0$$ en $$t_0 + τ$$. The corresponding positions on the line are

$$l = l(t) = l_0 + v(t − t_0)$$

Then

$$\begin{split} r(t) & = & \sqrt{d^2 + l(t)^2} \\ & = & \sqrt{d^2 + (l_0 + vt)^2} \end{split}$$

so

\begin{eqnarray} r(t_0 − τ) & = & \sqrt{d^2 + (l_0 − vτ)^2} \\ r(t_0) & = & \sqrt{d^2 + l_0^2} \\ r(t_0 + τ) & = & \sqrt{d^2 + (l_0 + vτ)^2} \end{eqnarray}

From this we derive $$d$$, $$l_0$$, and $$v$$. We find

\begin{eqnarray} 2 v^2τ^2 & = & r(t_0 + τ)^2 − 2 r(t_0)^2 + r(t_0 − τ)^2 \\ 4 l_0vτ & = & r(t_0 + τ)^2 − r(t_0 − τ)^2 \\ d^2 & = & r(t_0)^2 − l_0^2 \end{eqnarray}

so

\begin{eqnarray} v & = & \frac{\sqrt{r(t_0 − τ)^2 − 2 r(t_0)^2 + r(t_0 + τ)^2}}{τ\sqrt{2}} \\ l_0 & = & \frac{r(t_0 + τ)^2 − r(t_0 − τ)^2}{4vτ} \\ d & = & \sqrt{r(t_0)^2 − l_0^2} \end{eqnarray}

The moment $$t_\text{min}$$ at which the smallest distance is attained has by definition $$l = 0$$, so

\begin{eqnarray} 0 & = & l_0 + vt_\text{min} \\ t_\text{min} & = & −\frac{l_0}{v} \end{eqnarray}

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Last updated: 2020-07-18