IAO 2017, Теория, α+β — β-4. Eclipse in USA.

4.1.

At first, we have to calculate the speed of the shadow on the Earth's surface. The speed of the Moon in its synodic movement around the Earth is

The same speed would have the speed of the Moon’s shadow on perpendicular to the Sun direction unmoving surface. To take into account the inclination of the territory of Oregon during the eclipse to the perpendicular unmoving surface, we should divide $V_1$ by the sine of some angle $h_E$, which is defined as

where χ is the angle between the direction to the South and the azimuth of the Sun, in our case it is 59,5°.

And to take into account moving of the territory of Oregon due to sidereal rotation of the Earth we should substrate synodic (!) speed $V_\varphi$ of a point at latitude $\varphi=45^\circ$ from $V_2$

As we consider the pass of the eclipse was going along the geographical parallel, the other corrections should be not taken into account. So the speed of the Moon’s shadow on the territory of Oregon was about 985 m/s, and it is the answer to the first part of the problem.

Note. The real value of the speed of the shadow in that area was 1001 m/s.

4.2.

The size of the eclipsing shadow on the Earth is determined by the projection of the Moon’s shade diameter D of the on Earth's surface. This is usually an ellipse with axes equal to:

in the azimuth direction, where there are the Moon and the Sun: $A=D/\sin h$,

in the perpendicular to this azimuth direction: $B=D$.

It is more complicated to calculate the size of the shadow in other directions, but for ellipses with small eccentricities it is possible to use the approximation formula:

where α is the angle between the direction of measuring and the semi-major axis of the ellipse. The semi-major axis of the ellipse is directed to the Sun and the Moon. Therefore, it is also at an angle χ = 59.5° to the direction of the South in this configuration. One can write the approximate relations for the size of the shadow on the Earth's surface:

…

Substituting the numerical values of the angles, we get:

We need the width of track that is the size of the shadow in the direction perpendicular to the movement of the shadow that is approximately the direction North-South in our case, $L_\mathrm{NS}$. And the size of the shadow in the direction of its movement, $L_\mathrm{WE}$, approximately West-East, can be calculated as the speed of the shadow multiplied by the duration of the total phase of the Eclipse:

The duration of the totality can be obtained by analyzing pictures 2 and 4. Both of them demonstrate the so-called "diamond ring", a unique phenomenon, lasting a few seconds only. It is seen that images of approximately symmetrical in shape, and were shorted about 3-4 seconds before the second contact and after 3-4 seconds of the third one. In addition, from the fact that the points of the second and the third contacts are almost symmetric, it can be concluded that the Bulgarian-Russian group observed the central eclipse and, therefore, was located about in the middle of the totality path. Thus, it is possible to obtain the totality duration at the middle of the path $\tau=2^{\mathrm m}04^{\mathrm s}$ by calculating the time difference of the dating of these images.

Note. The above few seconds before the second contact and after the third one is a very delicate moment. It should not be considered an error if a participant will not take it into account and use the value $\tau=2^{\mathrm m}11^{\mathrm s}$.

Thus, the size of the shadow on the Earth's surface in the direction of its motion is equal to:

and the size in the perpendicular direction:

Thus, the width of the path of the totality was equal to approximately 105 km.

Note. The real value of the width of the path in that area was 102 km.

Basic criteria