Съдържание

Условие Part I: Geographic Coordinates (25 т.)Part II: Duration of the Totality (50 т.)Решение

IOAA 2024, Теория — Задача 10. T10. Greatest Eclipse

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IOAA 2024

IOAA 2024 — Theory Examination, English (Official) · 75 т.

Внимание: Бележка към темата

The printed instructions refer to blank Detailed Worksheets and a summary Answer Sheet. Those forms are absent from the available 2024 archive packet. The complete question booklet, solution booklet and separate marking scheme are included. Printed source inconsistencies are identified in the relevant tasks and retained without silently changing the source.

T2: The solution first gives $R=2.3143\times10^{23}\ \mathrm m$ and later substitutes $3.857\times10^{23}\ \mathrm m$ , while retaining its boxed $3.9\times10^{44}\ \mathrm{kg}\simeq2\times10^{14}M_\odot$ . The main double-sum potential expression also omits the factor $1/2$ used for unique pairs in the appendix. These printed expressions are retained; the marking scheme allows valid alternative estimates.

T6: The question names Palomar 4; the solution's repeated question names Palomar 6. The solution gives $V_{sky}=12.9$ mag but substitutes $12.8$ in the following surface-brightness line. The native question and printed calculation are retained.

T7: The printed expansion on solution page 13 has $-\sin(a)\cos(\phi)$ where the preceding cosine-law equation would give $-\sin(a)\sin(\phi)$ . It is retained as printed; the subsequent sunrise calculation sets $a=0$ .

T10: Solution page 25 prints the unit of $|\vec M|^2-R_\oplus^2=1.288\times10^{17}$ as m rather than $\mathrm m^2$ . In (j), the final vector prints 77.59 m/s after previously using 77.76 m/s; the magnitude uses 77.76 m/s. These printed values are retained.

T11: The question gives Maceió's longitude as $35.7^\circ W$ , while solution (e) uses $36.4^\circ W$ . The solution's printed margin marks total 11 points in (g), versus 12 points in the question and separate marking scheme; its (j) heading and marks total 7 points, versus 6 in the question and separate marking scheme. All three sources are retained with their original point values.

Common solar-system table: The native solar-system table labels the semi-major-axis column AU, while explicitly giving the Moon's entry in km; its printed unit is retained.

Указания

The theoretical competition will be 5 hours in duration and is marked out of a total of 300 points.
There are Detailed Worksheets for carrying out detailed work / rough work. On each of the Detailed Worksheets, please fill in Student Code, Question No., Page no. and total number of pages.
Start each problem on a new page of the Detailed Worksheets. Please write only on the printed side of the sheet. Do not use the reverse side. If you have written something on any sheet which you do not want to be marked, cross it out.
There is a summary Answer Sheet with your student ID code for your final answers.
Please remember that the graders may not understand your language. As far as possible, write your solutions only using mathematical expressions and numbers. If it is necessary to explain something in words, please use short phrases (if possible in English).
You are not allowed to leave your exam desk without permission. If you need any assistance (malfunctioning calculator, need to visit a restroom, need more Detailed Worksheets, etc.), please put up your hand to signal the invigilator.
The beginning and end of the competition will be indicated by a long sound signal. Additionally, there will be a short sound signal fifteen minutes before the end of the competition (before the final long sound signal).
Wait at your table until your envelope is collected. Once all envelopes are collected, your student guide will escort you out of the competition room.
A list of constants for this competition is given on the next page.

Fundamental Constants

Constant	Value
Speed of light in a vacuum	$c=2.998\times10^8\ \mathrm{m\,s^{-1}}$
Planck constant	$h=6.626\times10^{-34}\ \mathrm{J\,s}$
Boltzmann constant	$k_B=1.381\times10^{-23}\ \mathrm{J\,K^{-1}}$
Stefan-Boltzmann constant	$\sigma=5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$
Elementary charge	$e=1.602\times10^{-19}\ \mathrm C$
Universal Gravitational constant	$G=6.674\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$
Universal gas constant	$R=8.315\ \mathrm{J\,mol^{-1}\,K^{-1}}$
Avogadro constant	$N_A=6.022\times10^{23}\ \mathrm{mol^{-1}}$
Wien's displacement law	$\lambda_mT=2.898\times10^{-3}\ \mathrm{m\,K}$
Mass of the electron	$m_e=9.11\times10^{-31}\ \mathrm{kg}$
Mass of the proton	$m_p=1.67\times10^{-27}\ \mathrm{kg}$
Mass of the neutron	$m_n=1.67\times10^{-27}\ \mathrm{kg}$

Astronomical Data and Solar Magnitudes

Astronomical Data	Value
1 parsec	$1\ \mathrm{pc}=3.086\times10^{16}\ \mathrm m=206\,265\ \mathrm{AU}=3.262\ \mathrm{ly}$
1 Astronomical Unit (AU)	$1\ \mathrm{AU}=1.496\times10^{11}\ \mathrm m$
1 Jansky	$1\ \mathrm{Jy}=10^{-26}\ \mathrm{W\,m^{-2}\,Hz^{-1}}$
Hubble constant	$H_0=70\ \mathrm{km\,s^{-1}\,Mpc^{-1}}$
Solar luminosity	$L_\odot=3.826\times10^{26}\ \mathrm W$
Angular diameter of the Sun (from Earth)	$32'$
Effective temperature of the Sun	$T_{\mathrm{eff}}=5778\ \mathrm K$
Obliquity of the ecliptic	$23.5^\circ$
Inclination of the Moon's orbit (to ecliptic)	$5^\circ08'43''$
Magnitude of full Moon	$-12.74$
Coordinates of North Ecliptic Pole	$(18^{\mathrm h}00^{\mathrm m}00^{\mathrm s},+66^\circ33'39'')$
Coordinates of North Galactic Pole	$(12^{\mathrm h}51^{\mathrm m}26^{\mathrm s},+27^\circ07'42'')$
Sidereal day	$23^{\mathrm h}56^{\mathrm m}04^{\mathrm s}$
Tropical year	$365.2422$ solar days
Sidereal year	$365.2564$ solar days

Solar Magnitudes

Apparent Visual	Absolute Visual	Apparent Bolometric	Absolute Bolometric
$-26.75$	$+4.82$	$-26.83$	$+4.74$

Solar System, Spherical Trigonometry and Calculus

Solar System Data

Object	Radius (km)	Mass (kg)	Semi-major axis (AU)	Orbital eccentricity	Albedo
Sun	695500	$1.988\times10^{30}$	n/a	n/a	n/a
Mercury	2440	$3.301\times10^{23}$	0.387	0.206	0.088
Venus	6052	$4.867\times10^{24}$	0.723	0.007	0.76
Earth	6378	$5.972\times10^{24}$	1.000	0.016710	0.31
Moon	1737	$7.346\times10^{22}$	$3.844\times10^5\ \mathrm{km}$	0.054900	0.11
Mars	3390	$6.417\times10^{23}$	1.524	0.093	0.25
Jupiter	69911	$1.898\times10^{27}$	5.203	0.048	0.51
Saturn	58232	$5.683\times10^{26}$	9.537	0.054	0.34
Uranus	25362	$8.681\times10^{25}$	19.189	0.047	0.30
Neptune	24622	$1.024\times10^{26}$	30.070	0.009	0.29

Spherical Trigonometry

Cosine formula:

\cos a=\cos b\cos c+\sin b\sin c\cos A

Sine formula:

\frac{\sin a}{\sin A}=\frac{\sin b}{\sin B}=\frac{\sin c}{\sin C}

Four parts formula:

\cot b\sin a=\cos a\cos C+\sin C\cot B

Native common spherical-trigonometry reference sphere and triangle A, B, C with sides a, b, c.

Calculus

Chain Rule:

\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}

Product Rule:

\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}

\int x^r\,dx=\frac{x^{r+1}}{r+1}+C\qquad(r\ne-1)

\frac{d}{dx}\ln(x)=\frac1x,\qquad\frac{d}{dx}\tan(x)=\sec^2(x)

Условие

The greatest eclipse is defined as the instant when the axis of the Moon's shadow cone gets closest to the centre of the Earth in a solar eclipse. This problem explores the geometry of this phenomenon, using the solar eclipse of 29th May 1919 as an example, as it has great historical significance for being the first time astronomers were able to observationally verify general relativity. One of the scientific expeditions to observe this eclipse took place in the Brazilian city of Sobral.

The two following tables show the Cartesian and spherical coordinates of the Sun and the Moon at the time of the greatest eclipse. The system used for these coordinates is right-handed and has the origin at the centre of the Earth, the positive $x$ -axis pointing towards the Greenwich meridian, and the positive $z$ -axis pointing towards the North Pole. For the rest of this problem, this will be referred to as System I.

Spherical Coordinates:

	Centre of the Sun	Centre of the Moon
Radial Distance ( $r$ )	$1.516 \times 10^{11}\ \mathrm{m}$	$3.589 \times 10^{8}\ \mathrm{m}$
Polar Angle ( $\theta$ )	$68^\circ 29' 44.1''$	$68^\circ 47' 41.6''$
Azimuthal Angle ( $\varphi$ )	$-1^h 11^m 28.2^s$	$-1^h 11^m 22.9^s$

Cartesian Coordinates:

	Centre of the Sun	Centre of the Moon
$x$	$1.342 \times 10^{11}\ \mathrm{m}$	$3.185 \times 10^{8}\ \mathrm{m}$
$y$	$-4.327 \times 10^{10}\ \mathrm{m}$	$-1.025 \times 10^{8}\ \mathrm{m}$
$z$	$5.557 \times 10^{10}\ \mathrm{m}$	$1.298 \times 10^{8}\ \mathrm{m}$

For this problem, assume that the Earth is a perfect sphere.

Note: The spherical coordinates of a point $P$ are defined as follows:

Radial distance ( $r$ ): distance between the origin ( $O$ ) and $P$ (range: $r \geq 0$ ).
Polar angle ( $\theta$ ): angle between the positive $z$ -axis and the line segment $OP$ (range: $0^\circ \leq \theta \leq 180^\circ$ )
Azimuthal angle ( $\varphi$ ): angle between the positive $x$ -axis and the projection of the line segment $OP$ onto the $xy$ -plane (range: $-12^h \leq \varphi < 12^h$ )

Cartesian and spherical coordinate axes x, y, z and point P with radial distance r, polar angle theta and azimuthal angle phi.

Part I: Geographic Coordinates (25 т.)

(a) Determine the declination of the Sun and the Moon during the greatest eclipse for a geocentric observer. [3 т.]

(b) Determine the right ascension of the Sun and the Moon at the time of the greatest eclipse for a geocentric observer. The local sidereal time at Greenwich at that same moment was $5^h 32^m 35.5^s$ . [3 т.]

(c) Find a unit vector that indicates the direction of the axis of the Moon's shadow cone. This vector should point from the Moon to the vicinity of the centre of the Earth. [4 т.]

(d) Determine the latitude and the longitude of the point where the axis of the Moon's shadow cone crosses the surface of the Earth during the greatest eclipse. [15 т.]

Part II: Duration of the Totality (50 т.)

Precisely determining the duration of totality of a solar eclipse involves complex calculations that would be beyond the scope of this problem. However, it is possible to obtain a reasonable approximation for this value using the two following assumptions:

The size of the umbra on the surface of the Earth remains roughly constant throughout totality at a given location.
The velocity of the umbra on the surface of the Earth remains roughly constant throughout totality at a given location.

(e) Estimate the radius of the umbra during the greatest eclipse. In order to simplify the calculations, assume that the umbra is small enough that it can be considered approximately flat and that the axis of the Moon's shadow cone is extremely close to the centre of the Earth during the greatest eclipse. [10 т.]

(f) Calculate the velocity of the Earth's rotation at the latitude of the centre of the umbra. [3 т.]

(g) Determine the orbital velocity of the Moon at the instant of the greatest eclipse. Neglect the changes in the semi-major axis of the Moon's orbit. [4 т.]

For the remaining items of this problem, assume that the tangential velocity of the Moon is roughly the same as the orbital velocity and neglect its radial component.

In order to calculate the velocity of the umbra, it is convenient to define two new additional right-handed coordinate systems. System II is defined as follows:

Origin ( $O_{\mathrm{II}}$ ): position of the Moon at the instant of the greatest eclipse.
Positive $x$ -axis: Tangent to the declination circle. Points eastwards.
Positive $y$ -axis: Tangent to the meridian of right ascension. Points northwards.

System III is defined as follows:

Origin ( $O_{\mathrm{III}}$ ): centre of the umbra at the instant of the greatest eclipse.
Positive $x$ -axis: Tangent to the latitude circle. Points eastwards.
Positive $y$ -axis: Tangent to the meridian of longitude. Points northwards.

Note that in both systems, the $xy$ -plane is tangent to the celestial sphere at the position of the origin.

System III is similar to System II, with the only difference being that the origin ( $O_{\mathrm{III}}$ ) is at the centre of the umbra at the moment of the greatest eclipse.

(h) Using System II, determine the velocity vector of the Moon during the greatest eclipse. Note that the intersection between the Celestial Equator and the lunar orbit that is closer to the position of the eclipse has a right ascension of $23^h 07^m 59.2^s$ . [14 т.]

(i) Write the velocity vector of the Moon in System III. Note that in System I, the azimuthal angle difference between the positions of the origins $O_{\mathrm{II}}$ and $O_{\mathrm{III}}$ is negligible, so you should only take into account the difference in the polar angles. [10 т.]

(j) Calculate the speed of the centre of the umbra along the surface of the Earth at the instant of the greatest eclipse. [6 т.]

(k) Estimate the duration of the totality of the eclipse at the location with the coordinates found in part (d). [3 т.]

Решение

Part I: Geographic Coordinates (25 points)

(a) Solution: The declination is simply the complement of the polar angle:

\delta=90^\circ-\theta

[1.0]

\therefore\delta_\odot=21^\circ30'15.9''

[1.0]

\delta_{Moon}=21^\circ12'18.4''

[1.0]

(b) Solution: The local sidereal time at Greenwich corresponds by definition to the meridian of right ascension that is right above Greenwich. Since Greenwich corresponds to an azimuthal angle of $0^{\mathrm h}$ , the right ascension at any given point corresponds simply to the sum of the local sidereal time at Greenwich and the azimuthal angle:

\alpha=\varphi+LST_{Greenwich}

[1.0]

\therefore\alpha_\odot=4^{\mathrm h}21^{\mathrm m}7.3^{\mathrm s}

[1.0]

\alpha_{Moon}=4^{\mathrm h}21^{\mathrm m}12.6^{\mathrm s}

[1.0]

(c)

Solution: By subtracting the Cartesian coordinates of the Sun from the Cartesian coordinates of the Moon, it is possible to obtain a vector for the direction of the axis of the Moon's shadow cone: [1.0]

\vec{u} = \langle -1.339 \times 10^{11},\ 4.317 \times 10^{10},\ -5.544 \times 10^{10} \rangle\ \mathrm{m}

[1.0]

|\vec{u}| = 1.512 \times 10^{11}\ \mathrm{m}

[1.0]

\therefore \hat{u} = \langle -0.8855, 0.2855, -0.3667 \rangle

[1.0]

(d) Solution: If one draws a vector in the same direction as $\hat u$ up to the earth's surface, then its magnitude can be found by the formula, where $\vec M$ is the position vector of the Moon, $k$ is a constant, and $R_\oplus$ is the radius of the Earth:

|\vec M+k\hat u|^2=R_\oplus^2

[5.0]

Solving for $k$ :

0=k^2(\hat u\cdot\hat u)+2k(\hat u\cdot\vec M)+\vec M\cdot\vec M-R_\oplus^2

=k^2+2k(\hat u\cdot\vec M)+|\vec M|^2-R_\oplus^2

Now, $|\vec M|^2-R_\oplus^2=1.288\times10^{17}\ \mathrm m$

2(\hat u\cdot\vec M)=-7.178\times10^8\ \mathrm m

\therefore0=k^2-k(7.178\times10^8)+(1.288\times10^{17})

k=\frac{7.178\pm\sqrt{(-7.178)^2-4\times12.88}}{2}\times10^8

The vector will intersect the earth's surface at two points. Hence two solutions. However, only the first intersection with the sphere is relevant in this case, so only the smallest solution to the equation should be considered:

k=\frac{7.178-\sqrt{(-7.178)^2-4\times12.88}}{2}\times10^8=3.528\times10^8

[4.0]

The vector of location of the greatest eclipse corresponds to the position vector of the Moon plus $k$ times the direction vector $\hat u$ :

\vec p=\vec M+k\hat u=\langle6.091\times10^6,-1.828\times10^6,4.854\times10^5\rangle

[2.0]

It is possible to use the following expressions to convert this vector to the spherical system:

r = \sqrt{x^2+y^2+z^2} = 6.378\times 10^6 m

\theta = \arctan\left(\frac{\sqrt{x^2+y^2}}{z}\right) (valid for z>0) = 1.495 rad = 85^\circ 38'

\varphi = \arctan\left(\frac{y}{x}\right) (valid for x>0) = -0.2915 rad = -16^\circ 42'

[2.0]

The latitude corresponds to the complement of the angle $\theta$ , and the longitude corresponds to the angle $\varphi$ :

\boxed{4^\circ 22'\mathrm{N},\ 16^\circ 42'\mathrm{W}}

[2.0]

Part II: Duration of the Totality (50 points)

Precisely determining the duration of the totality of a solar eclipse involves complex calculations that would be beyond the scope of this problem. However, it is possible to obtain a reasonable approximation for this value using the two following assumptions:

The size of the umbra on the surface of the Earth remains roughly constant throughout the totality for a given location.
The velocity of the umbra on the surface of the Earth remains roughly constant throughout the totality for a given location.

(e) (10 points) Estimate the radius of the umbra during the greatest eclipse. In order to simplify the calculations, assume that the umbra is small enough that it can be considered approximately flat and that the axis of the Moon's shadow cone is extremely close to the center of the Earth during the greatest eclipse.

Solution: The approximate geometry described in the problem statement is shown below: Complete solar and lunar shadow geometry, both beta angles and radii, umbra diameter and both distance brackets. [3.0]

The angle $\beta$ can be calculated as follows:

\beta = \arccos\left(\frac{R_\odot - R_{Moon}}{d_\odot - d_{Moon}}\right) = \arccos\left(\frac{6.955\times 10^8 - 1.737\times 10^6}{1.516\times 10^{11} - 3.589\times 10^8}\right) = 89.74^\circ

[2.0] [1.0]

Using this angle, it is possible to determine the radius of the the umbra:

r_{umbra} = \frac{R_{Moon} - (d_{Moon} - R_\oplus)\cdot\cos(\beta)}{\sin(\beta)}

[3.0]

\boxed{r_{umbra} = 1.196\times 10^5\ \mathrm{m}}

[1.0]

(f) (3 points) Calculate the velocity of the Earth's rotation at the latitude of the center of the umbra.

Solution: The Rotational velocity of the Earth at the latitude of the center of the umbra is the following:

v_{rot} = \frac{2\pi R_\oplus}{23^h 56^m 04^s}\cos(\phi_{umbra}) = \frac{2\pi\times 6.378\times 10^6}{23^h 56^m 04^s}\cos(4^\circ 22')

[2.0]

\boxed{v_{rot} = 463.7\ \mathrm{m/s}}

[1.0]

(g) (4 points) Determine the orbital velocity of the Moon at the instant of the greatest eclipse. Neglect the changes in the semi-major axis of the Moon's orbit.

Solution: Using the vis-viva equation:

v_{Moon} = \sqrt{GM_\oplus\left(\frac{2}{d_{Moon}} - \frac{1}{a_{Moon}}\right)} = \sqrt{6.674\times 10^{-11}\times 5.972\times 10^{24}\left(\frac{2}{3.589\times 10^8} - \frac{1}{3.844\times 10^8}\right)}

[2.0]

\boxed{v_{Moon} = 1088\ \mathrm{m/s}}

[2.0]

For the remaining items of this problem, assume that the tangential velocity of the Moon is roughly the same as the orbital velocity and neglect its radial component.

In order to calculate the velocity of the umbra, it is convenient to define two new additional right-handed coordinate systems. System II is defined as follows:

Origin ( $O_{II}$ ): position of the Moon at the instant of the greatest eclipse.
Positive $x$ -axis: Tangent to the declination circle. Points eastwards.
Positive $y$ -axis: Tangent to the meridian of right ascension. Points northwards.

System III is defined as follows:

Origin ( $O_{III}$ ): center of the umbra at the instant of the greatest eclipse.
Positive $x$ -axis: Tangent to the latitude circle. Points eastwards.
Positive $y$ -axis: Tangent to the meridian of longitude. Points northwards.

Note that in both systems, the $xy$ -plane is tangent to the celestial sphere at the position of the origin.

System III is similar to system II, with the only difference being that the origin ( $O_{III}$ ) is at the center of the umbra at the moment of the greatest eclipse.

(h) (14 points) Using system II, determine the velocity vector of the Moon during the greatest eclipse. Note that the intersection between the Celestial Equator and the lunar orbit that is closer to the position of the eclipse has a right ascension of $23^h 07^m 59.2^s$ .

Solution: It is possible to use the following diagram to determine the value of the angle $\kappa$ between the $x$ -axis and the velocity vector of the Moon: Moon velocity spherical triangle with declination, right ascensions, gamma, lunar orbit and kappa. [6.0]

Note that the great circle that intersects the Moon's meridian of right ascension at a right angle has a tangent line that coincides with the $x$ -axis, so the angle $\kappa$ on the figure is also the angle between the velocity vector of the Moon and the $x$ -axis. Also note that the Moon is moving from west to east in the celestial sphere (to the left in the figure), so its velocity vector is on the first quadrant of the $xy$ -plane on this system.

Using the four parts formula:

\cos(\delta_{Moon})\cos(90^\circ) = \sin(\delta_{Moon})\cot(\alpha_{Moon} + 24^h - \alpha_{intersection}) - \sin(90^\circ)\cot(90^\circ - \kappa)

[3.0]

\therefore 0 = \sin(\delta_{Moon})\cot(\alpha_{Moon} - \alpha_{intersection}) - \tan(\kappa)

\tan(\kappa) = \frac{\sin(\delta_{Moon})}{\tan(\alpha_{Moon} - \alpha_{intersection})}

\kappa = \arctan\left[\frac{\sin(\delta_{Moon})}{\tan(\alpha_{Moon} - \alpha_{intersection})}\right] = \arctan\left[\frac{\sin(21^\circ 12' 18.4'')}{\tan(4^h 21^m 12.6^s - 23^h 07^m 59.2^s)}\right]

\kappa = 4^\circ 17'

[2.0]

Now it is possible to break down the velocity of the Moon into the $x$ and $y$ components:

\mathbf{v_{II}} = \begin{bmatrix} v_{Moon}\cos(\kappa) \\ v_{Moon}\sin(\kappa) \\ 0 \end{bmatrix} = \begin{bmatrix} 1085\ \mathrm{m/s} \\ 81.25\ \mathrm{m/s} \\ 0 \end{bmatrix}

[3.0]

The radial component of the velocity was neglected, so the velocity on the $z$ -axis is equal to zero.

(i) (10 points) Write the velocity vector of the Moon in system III. Note that in system I, the azimuthal angle difference between the positions of the origins $O_{II}$ and $O_{III}$ is negligible, so you should only take into account the difference in the polar angles.

Solution: It is possible to obtain system III by rotating system II from north to south by an angle of $\theta_{umbra} - \theta_{Moon}$ around the $x$ -axis. The angles $\theta_{umbra}$ and $\theta_{Moon}$ correspond to the polar angles of the umbra and the Moon on system I.

The following two figures illustrate this rotation. Both original panels illustrating rotation of System II into System III, including every coordinate axis and rotation angle.

Since the rotation is around the $x$ -axis, the $x$ component remains unchanged. The component that was originally in the $y$ -axis is broken down into two components on the $y$ -axis and the $z$ -axis of the new system, which results in the following vector: [3.0]

\mathbf{v_{III}} = \begin{bmatrix} v_{Moon}\cdot\cos(\kappa) \\ v_{Moon}\cdot\sin(\kappa)\cdot\cos(\theta_{umbra} - \theta_{Moon}) \\ -v_{Moon}\cdot\sin(\kappa)\cdot\sin(\theta_{umbra} - \theta_{Moon}) \end{bmatrix}

[6.0]

\mathbf{v_{III}} = \begin{bmatrix} 1085\ \mathrm{m/s} \\ 77.76\ \mathrm{m/s} \\ -23.54\ \mathrm{m/s} \end{bmatrix}

[1.0]

(j) (6 points) Calculate the speed of the center of the umbra along the surface of the Earth at the instant of the greatest eclipse.

Solution: The velocity of the umbra corresponds to the vector sum of the component caused by the rotation of the Earth and the component caused by the velocity of the Moon.

Note that all points on the axis of the Moon's shadow cone have the same velocity, $\mathbf{v_{III}}$ is also the velocity of the axis of the Moon's shadow cone at the center of the umbra. Since the umbra moves along the surface of the Earth, so the $z_{III}$ -component should be set to zero.

\mathbf{v_{umbra}} = \begin{bmatrix} v_{III,x} - v_{rot} \\ v_{III,y} \\ 0 \end{bmatrix} = \begin{bmatrix} 1085 - 463.7 \\ 77.76 \\ 0 \end{bmatrix} = \begin{bmatrix} 621.4\ \mathrm{m/s} \\ 77.59\ \mathrm{m/s} \\ 0 \end{bmatrix}

[4.0]

The modulus of this vector is the following:

v_{umbra} = |\mathbf{v_{umbra}}| = \sqrt{621.4^2 + 77.76^2} = 626.3\ \mathrm{m/s}

[2.0]

(k) (3 points) Estimate the duration of the totality of the eclipse at the location with the coordinates found on item (d).

Solution: Considering the assumptions listed at the beginning of part II, it is possible to estimate the length of the totality by dividing the diameter of the umbra by its speed:

\Delta t_{totality} = \frac{2\cdot r_{umbra}}{v_{umbra}} = \frac{2\times 1.196\times 10^5}{626.3}

[2.0]

\boxed{\Delta t_{totality} = 6\ \mathrm{min}\ 21.4\ \mathrm{s}}

[1.0]

Separate Marking Scheme — T10

(a)

1 point: understanding that the declination is the complement of the polar angle.
1 point: calculating the declination of the Moon.
1 point: calculating the declination of the Sun.

(b)

1 point: understanding that the right ascension is the sum of the azimuthal angle and the local sidereal time at Greenwich.
1 point: calculating the right ascension of the Moon.
1 point: calculating the right ascension of the Sun.

(c)

1 point: understanding that the vector corresponds to the subtraction of the Cartesian coordinates of the Sun from the Cartesian coordinates of the Moon.
1 point: correct vector.
1 point: correct modulus.
1 point: correct unit vector.

(d)

5 points: understanding that adding the vector position of the Moon with a product of the unit vector by a constant will result in the position vector of the intersection between the axis of the Moon’s shadow cone and the surface of the Earth, whose module equals the center of the Earth.
3 points: finding an expression for the constant factor k.
1 point: correctly determining the numerical value of k.
2 points: writing down the position vector of the intersection between the axis of the Moon’s shadow cone and the surface of the Earth.
2 points: finding the spherical coordinates of the intersection between the axis of the Moon’s shadow cone and the surface of the Earth.
2 points: finding the geographic coordinates of the intersection between the axis of the Moon’s shadow cone and the surface of the Earth.

(e)

3 points: demonstrating a correct understanding of the geometry of the situation. Note that it is also acceptable to consider that the angle $\beta$ is roughly equal to $90^\circ$ . Using this alternative approximate geometry, the angles marked as right angles in the official solution would actually be acute angles and their numerical values would need to be calculated.
2 points: setting up a correct equation to calculate the angle $\beta$ .
1 point: correctly calculating the value of the angle $\beta$ .
3 points: setting up a correct equation to calculate the radius of the umbra.
1 point: correctly calculating the value of the radius of the umbra.

(f)

2 points: setting up a correct equation to calculate the velocity of the Earth’s rotation at the latitude of the center of the umbra.
1 point: correctly calculating velocity of the Earth’s rotation at the latitude of the center of the umbra.

(g)

2 points: setting up a correct equation to calculate the velocity of the Moon. 1 point should be deducted if the student assumes a circular orbit.
2 points: correctly calculating velocity of the Moon.

(h)

6 points: correctly drawing or describing the geometry of the situation.
3 points: setting up a correct equation to calculate $\kappa$ .
2 points: determining the value of $\kappa$ .
3 points: correct velocity vector. 1 point per component.

(i)

3 points: understanding that it is possible to obtain system III by rotating system II by the difference in polar angles
6 points: correct algebraic expressions for the components of the vector in system III. 2 points per component.
1 point: correct numerical values of the components of the vector. 1/3 of a point per component.

(j)

2 points: correctly determining the x component of the vector.
1 point: correctly determining the y component of the vector.
1 point: correctly determining the z component of the vector. It is also acceptable to say that the z-component of the vector is equal to the z component of the velocity of the Moon on system III, as long as the student understands that only the x and y components of the vector are relevant to calculate the modulus of the velocity of the umbra on the surface of the Earth.
1 point: correctly setting up an equation to calculate the modulus of the velocity of the umbra.
1 point: calculating the modulus of the velocity of the umbra.

(k)

2 points: understanding that the duration of the totality can be approximated as the ratio of the diameter of the umbra and the velocity of the umbra.
1 point: correctly estimating the duration of the totality.

General Rules for Marking

These are general guidelines for the grading of all problems. All of the rules listed below refer to deductions on specific items in the marking scheme.

Case	Marking
Using absurd physical or mathematical concepts (despite correct answers)	No points given
Giving correct answer but the method is unclear	Deduct 50% of the marks for final answer
Minor mistakes in the calculations, e.g., wrong signs, numerical constants	Deduct 20% of the marks for that part, unless there is a physical incoherence
Too many or too few significant figures ( $n\pm1$ where $n$ is the expected number of significant figures)	Deduct 50% of the marks for final answer
Correct answer but missing or wrong units	No points given for final answer
For numerical values, the answer is not within a 5% precision range	No points given for final answer, unless differently stated in the problem or marking scheme
Error resulting from an earlier part error, for which the student already lost marks, if the answer is physically reasonable	No double penalization, full points given
Error resulting from an earlier part error, if the answer is physically unreasonable	No points given

First example: If due to an error in an earlier item, the student calculates the speed of a satellite as $4.0\times10^4\ \mathrm{m/s}$ instead of $4.5\times10^4\ \mathrm{m/s}$ , they will only lose marks for the earlier item. However, if, for the same reason, a student calculates the velocity of the satellite as $4.5\times10^{12}\ \mathrm{m/s}$ , they should realize this is wrong (higher than the speed of light) and thus should lose some marks for this item as well.

Second example: Consider a question that requires a student to determine the temperature of a planet that orbits a star with luminosity $L$ at a distance $r$ . To answer correctly, the student must calculate the star's radiant flux, the power incident on the planet, the power radiated by the planet, and then, using the condition of thermal equilibrium, the planet's temperature. If the student only calculates the radiant flux and arrives at the correct final answer without clearly explaining the other steps, they will receive marks only for the flux calculation, but not for the other steps, neither for the final answer.

Оригинал в Архива: theory_examination_304b6eb66a8940d79fccc404cf1ba9ab_preview.pdf · официални решения: theory solutions.pdf · допълнителни данни: theory marking scheme.pdf