IOAA 2023, Теория — Задача 10. Aldebaran

(a) Aldebaran is far enough away that the light rays from it can be considered to be parallel, thus the problem can be modelled by the following diagram:

where the light ray through $C$ passes through the centre of the Earth and thus $\delta_{\mathrm{A}}$ is the declination of Aldebaran; the ray through $D$ and $B$ (Bologna) represents the situation observed by Copernicus, with Aldebaran behind the Moon and $0.32'$ above the south edge; and the ray through $M$ and $P$ represents the situation where Aldebaran is exactly behind the centre of the Moon. $\varphi_{\mathrm{B}}$ is the geocentric latitude of Bologna, $\varphi_1$ is the latitude of the place $P$ and $r$ is the radius of the Earth (assume the Earth is spherical).

Calculating the latitude $\varphi_1$ is thus a matter of simple geometry, where:

and the distance $|DM|$ is the fraction of the Moon’s radius given by:

Therefore, since

(6 points)

(b) Given the assumptions, the duration of the occultation will just depend on the difference between the tangential speeds of the observer, $v_{\mathrm{obs}}$, and Moon, $v_{\mathrm{Moon}}$, and on the diameter of the Moon.

For the observer, the tangential speed is given by the circumference of their path at latitude $\varphi_1$ divided by the sidereal day:

Similarly for the Moon, using the semi-major axis of the Moon’s orbit and the length of the sidereal month (from the table of constants):

The difference of speeds is then:

and the time taken to move across the lunar diameter ($2\times1737\,\mathrm{km}$) at that speed is:

(5 points)

(c) To determine the topocentric angular velocity, we need to find the distance $d$ between the observer (at place $P$ at latitude $\varphi_1$) and the Moon:

where $a$ is the semi-major axis of the Moon’s orbit, $\delta$ is the geocentric declination of the Moon, and $\delta'$ is the topocentric declination of the Moon as seen by the observer. We can rewrite $\varrho=\delta-\delta'$ and $\alpha=180^\circ-(\varphi_1-\delta)-\varrho=180^\circ-(\varphi_1-\delta)-(\delta-\delta')=180^\circ-(\varphi_1-\delta')$, then using the sine theorem:

Simplifying and substituting $\delta'=\delta_{\mathrm{A}}$:

We can now calculate $\delta$:

and $d$:

Thus the angular velocity is:

(7 points)

(d) The result in part (c), $27\,\mathrm{arcmin/h}$, was calculated for the situation when the Moon and observer are moving in the same direction, reducing their relative velocity, and thus represents the lower limit of the range of topocentric angular velocities. The upper limit is given by the situation when the Moon and observer are moving in the opposite direction, i.e. when the Moon is due north.

Using the calculation from part (b), we find that the relative linear speed is:

Neglecting the effect of the slightly greater distance between the observer and the Moon when the Moon is on the northern side of the sky, the angular speed will be greater by the ratio of the linear speeds,

thus the range is approximately $27$–$40\,\mathrm{arcmin/h}$. (3 points)

If the additional distance is taken into account, the upper limit will be around 1% less.

Expressing the velocities as vectors,

the magnitude of the relative velocity, $v_{\mathrm{rel}}=|\vec{v}_{\mathrm{rel}}|$ is given by the scalar projection of $\vec{v}_{\mathrm{obs}}$ on $\vec{v}_{\mathrm{Moon}}$:

where $\alpha$ is the angle between the vectors. From this it can be seen that the relative velocity must be minimum for $\alpha=0^\circ$ and maximum for $\alpha=180^\circ$. These conditions are met on the local meridian, due south or due north, respectively. (4 points)