USAAAO 2020, II кръг — Задача 12

• (a) III

• (b) 26° to 29° (full credit); 25° or 30° half credit

• (d) A = 61.11699°, B = 89.67051°, C = 29.21265°

• (e) $5.16 \times 10^{-7}$

  The official solution gives the sunspot-to-disk ratio r = 5.16 × 10⁻⁷ (5.13 × 10⁻⁷ with the flat-triangle approximation), i.e. the inverse of the ratio worded in the question.

• (f) Yes: α = 4.89 × 10⁻⁶ rad > θ = 2.58 × 10⁻⁷ rad

• (g) Δt = 3.213 × 10¹⁷ s ≈ 10 billion years

Solution:

(a) The correct answer is III.

(b) The acceptable range will be $26^{o}$ to $29^{o}$ for full credit, and $25^{o}$ or $30^{o}$ to half of the punctuation.

It is useful to know that sunspots always come in pairs and have the same absolute value of latitude (one is positive and the other negative).

(c) Every 11 years (the period of the Schwabe Cycle), the Solar magnetosphere flips its orientation. Therefore, the orientation of the sunspots on each hemisphere is reversed. The pictures on item A were likely taken while the Sun was going through this transition, which explains their anomaly.

(d) A, B, and C are the internal angles of the spherical triangle.

Using the Law of the Cosines for spherical triangles: $\cos(0.176^{o}) = \cos(0.0981^{o}) \times \cos(0.201^{o}) + \sin(0.0981^{o}) \times \sin(0.201^{o}) \times \cos(A)$

$A = 61.11699^{o}$

Using the Law of the Sines for spherical triangles:

$\sin(61.11699^{o})/\sin(0.176^{o}) = \sin(B)/\sin(0.201^{o})$

$B = 89.67051^{o}$

Using the Law of the Sines for spherical triangles to calculate the last internal angle:

$\sin(61.11699^{o})/\sin(0.176^{o}) = \sin(C)/\sin(0.0981^{o})$

$C = 29.21265^{o}$

(e) First, it is necessary to calculate the spherical excess:

$E = 61.1^{o} + 89.7^{o} + 29.2^{o} - 180^{o}$

$E = (1.51 \times 10^{-4})^{o}$

Then, it is possible to calculate the area of the spherical triangle:

$A = \pi \times R_{Sun}^2 \times 1.51 \times 10^{-4}/180$

$A = 8.41 \times 10^{-7} \times \pi \times R_{Sun}^2$

Since the question asks about the observed area it is necessary to correct it with the inclination of the tangential plane.

Inclination of the tangential plane (Law of the Cosines with a right spherical triangle containing both coordinates):

$\cos(i) = \cos(7.89^{o}) \times \cos(51.74^{o})$

$i = \arccos(0.613)$

$i = 52.2^{o}$

Now, it is possible to correct the observed area with the inclination:

$A_c = 8.41 \times 10^{-7} \times \pi \times R_{Sun}^2 \times \cos(52.2^{o})$

$A_c = 5.16 \times 10^{-7} \times \pi \times R_{Sun}^2$

Considering that the area of the Solar disk is $\pi \times R_{Sun}^2$, it is possible to calculate the ratio between the areas observed:

$r = 5.16 \times 10^{-7} \times \pi \times R_{Sun}^2/(\pi \times R_{Sun}^2)$

$r = 5.16 \times 10^{-7}$

It is important to consider that small rounding errors will lead to huge differences in the spherical excess, so students won’t be penalized if their answer is different than expected due to rounding errors. In fact, if students notice that the spherical excess is so small that the triangle is basically flat, they will also get full credit for this item. In this case, the calculations for the value of r would be the following (it is possible to use any pair of sides in the triangle and the angle in between them):

$A = 1/2 \times (2 \times \pi \times R_{Sun} \times 0.0981^{o}/360^{o}) \times (2 \times \pi \times R_{Sun} \times 0.201^{o}/360^{o}) \times \sin(61.1^{o})$

$A = 8.37 \times 10^{-7} \times \pi \times R_{Sun}^2$

$A_c = 8.37 \times 10^{-7} \times \pi \times R_{Sun}^2 \times \cos(52.2^{o})$

$A_c = 5.13 \times 10^{-7} \times \pi \times R_{Sun}^2$

$r = 5.13 \times 10^{-7} \times \pi \times R_{Sun}^2/(\pi \times R_{Sun}^2)$

$r = 5.13 \times 10^{-7}$

(f) First, it is necessary to calculate the diameter of the telescope:

$f/D = 5$

$13/D = 5$

$D = 2.6m$

Then, it is possible to calculate the angular resolution of the telescope:

$\theta = 1.22\lambda/D$

$\theta = 1.22 \times 5.5 \times 10^{-7}/2.6$

$\theta = 2.58 \times 10^{-7}rad$

It is possible to use the smallest side of the triangle to see if it will be resolved by the telescope. Length of the smallest size of the triangle:

$l = 2 \times \pi \times R_{Sun} \times 0.0981^{o}/360^{o}$

$l = 1.19 \times 10^{6}m$

Correcting it for the inclination:

$l_c = 1.19 \times 10^{6}m \times cos(52.2^{o})$

$l_c = 7.31 \times 10^{5}m$

Angle that corresponds to this length:

$\alpha = 7.31 \times 10^{5}/(1.496 \times 10^{11})$

$\alpha = 4.89 \times 10^{-6}\mathrm{rad}$

Since $\alpha > \theta$, it is possible to resolve the sunspot.

(g) Variation in mass per reaction:

Fraction of mass converted:

Mass converted into energy while the Sun is in the Main Sequence:

Total energy converted:

Time in the main sequence: