IAO 2016, Практически тур, Група α — Задача 6. Наблюдатель кометы

6.1

We start off by drawing the Earth’s orbit. Using the tabular data we find that the suitable value for the radius is 2 cm. Because there are 3 observations per year, as noted in the text, there are just 3 positions of the Earth, where the observations take place: one for 30 September, one for 29-30 January and one for 31 May-1 June. We mark these positions on the graph paper. Because 30 Sep 2004 is when Delta takes the largest tabular value (6.485 AU), it is convenient to place 30 Sep on the x-axis. The other two observational points are separated by an angle of $120^{\circ}$ and are easily marked using simple geometry.

Next we draw all 20 points from the table on the graphing paper, where each point represents the comet’s position for a particular observation with given $N$. To this end, we use only columns Date(UT), Delta and S-O-T and disregard the rest. Note that Delta is the distance between the comet and the Earth. Therefore some care must be exercised to measure Delta from the correct position of the Earth defined by Date(UT).

Next we draw a smooth closed curve around the points, which resembles an ellipse.

Source note: the solution names the column S-O-T, while the supplied table labels it S-O-C.

6.2

First by observing the curve we identify the direction of the semi-major axis. Then we draw a line along this direction, which approximately halves the ellipse. Next we mark the so-obtained perihelion and aphelion and measure the distance between them, which yields $2a$. We also measure the perihelion distance, which yields $a(1-e)$. Bearing in mind that 2 cm = 1 AU, we find that

In fact, the real values are $a \approx 3.44$ AU and $e \approx 0.57$.

6.3

The first point ($N=1$) and the last point ($N=20$) almost lie on the x-axis meaning that for the time interval separating these observations, the comet has done a full cycle. Therefore we obtain

while the real value is $T \approx 6.38$ years.

6.4

From Kepler’s second law we know that a line segment joining an orbiting body and the Sun sweeps out equal areas $\Delta S$ during equal intervals of time $\Delta t$. This means that $\Delta S / \Delta t = S / T$, where $S$ is the area inside the ellipse, and $T$ is the comet’s orbital period. Note that at the perihelion and at the aphelion the following identities hold for very small $\Delta t$:

For the derivation consider the area of an isosceles triangle of height $a(1-e)$ and base $v_p \Delta t$.

Thus we find

We can express the speeds directly by $a$ and $T$ if we use the formula $S = \pi ab = \pi a^2 \sqrt{1-e^2}$:

Source discrepancy: the first numerical line swaps the perihelion/aphelion labels; the next line changes the square-root factors but retains the same numerical values. These expressions are reproduced as printed. The following 6.6 derivation uses the perihelion expression with $(1+e)/(1-e)$.

6.5

The solar mass $M$ can be obtained directly from the identity (neglecting the comet’s mass): $T^2 = \dfrac{4\pi^2 a^3}{\gamma M}$. Thus we have

6.6

From conservation of energy we have $\dfrac{mv^2}{2} - \dfrac{\gamma M m}{r} = \dfrac{m v_p^2}{2} - \dfrac{\gamma M m}{a(1-e)}$, which gives

In solution 6.4. we found $v_p = \dfrac{2\pi a}{T}\sqrt{\dfrac{1+e}{1-e}}$. From $T^2 = \dfrac{4\pi^2 a^3}{\gamma M}$ we obtain $v_p^2 = \dfrac{\gamma M}{a}\dfrac{1+e}{1-e}$.

Thus we arrive at

For the observation with $N = 7$ we measure that $r = 3.25$ AU, so we get $v_7 \approx 17.2$ km/s.

The escape velocity at this point is obtained by taking the limit as $a$ approaches $\infty$: