USAAAO 2023, I кръг — Задача 1

• 1. B
• 2. B
• 3. E
• 4. A
• 5. D
• 6. B
• 7. D
• 8. B
• 9. D
• 10. C
• 11. B
• 12. D
• 13. A
• 14. D
• 15. C
• 16. E
• 17. E
• 18. A
• 19. D
• 20. C
• 21. C
• 22. B
• 23. B
• 24. A
• 25. C
• 26. D
• 27. A
• 28. B
• 29. E
• 30. A

1. Solution: The difference is exactly 54 days, so JD 2460000.0 corresponds to 24 February at 12:00 UT. Poland follows Central European Time, so we must add 1 hour to the Universal Time.

Answer: B

2. Solution: The Sun-like star has absolute magnitude 4.83 (from the table of constants). Then, using

we get $D = 108$ parsecs.

Then we notice this means the true FOV of the cluster is $\frac{2*R}{D} = 2.1$ degrees.

So $\frac{FOV_{Telescope}}{FOV_{Actual}} = magnification = \frac{f_{obj}}{f_{eye}}$

This gives magnification=14.2 so $f_{obj}$= 283 mm

Answer: B

3. Solution: The brightness of each object is proportional to its surface area and its albedo. The required ratio is therefore $4^2 \times 3 = 48$.

Answer: E

4. Solution: Wien's law $\lambda_{max}T = 2.9 \times 10^{-3}\,\mathrm{m\,K}$

Answer: A

5. Solution: The parsec is defined as $d(pc) = \frac{1}{\alpha}$ where $\alpha$ is the stellar parallax in arc-seconds.

Answer: D

6. Solution: The black body luminosity is $4\pi\sigma R^2T^4$. After unit conversion, $7 \times 10^{32}\ \frac{erg}{s} \sim 0.2\ \mathrm{L}_{\odot}$.

Answer: B

7. Solution: At $n = 2$ state of hydrogen, the excitation energy of approximately 10 eV, which rarely occurs in low-temperature gas.

Answer: D

8. Solution: Since the comet is in a parabolic orbit, its total orbital energy is 0 and it's exactly at the threshold to be gravitationally bound to the sun. This means that any fragments moving faster than the original comet will escape, and any fragments moving slower than the original comet will be gravitationally bound to the Sun. If a fragment is launched at an angle $\theta$ relative to the comet's direction of motion around the Sun, then its final velocity is $|v_f| = \sqrt{v^2 + V^2 + 2vV\cos\theta}$.

Assuming $v \ll V$, we can expand this in to first order in $v/V$, giving $|v_f| = V + v\cos\theta$, which is larger than $V$ only for $\theta < \frac{\pi}{2}$. This means that the fragments will escape for $\theta < \frac{\pi}{2}$ but not for $\theta > \frac{\pi}{2}$, so 50% of the fragments escape.

Answer: B

9. Solution: Note that spiral galaxies are uniformly rotating, so we can simply write

However, the velocity distribution in an elliptical galaxy is uniform and random. Then, we can use the Virial theorem ($T = -U/2$) to calculate the velocity at the edge of the galaxy.

Where $v_{tot}$ is the total magnitude of the velocity (in tangential and nontangential directions). However, we also note that $v_{\mathrm{tot}}^2 = v_x^2 + v_y^2 + v_z^2 = 3v_r^2$ (from the assumption that velocities are randomly distributed). Then, $M = 5\frac{v^2R}{G}$.

Answer: D

10. Solution: It is important to note that the orbital radius of the satellite is much larger than the radius of a rocky planet, so it is possible to consider that the time to reach the surface of Venus is approximately the same as the time to reach the center of Venus.

An easy method to calculate the time that the satellite takes to reach Venus is to assume that as it falls to the ground, it is in an elliptical orbit with a semi-minor axis tending to zero and a semi-major axis equal to half of the orbital radius of the satellite. Venus is one of the foci of the ellipse. In that case, the satellite would take half of the orbital period to reach Venus. Therefore, it is possible to use Kepler's Third Law to determine the time that the satellite would take to reach Venus:

Answer: C

11. Solution: There is a total hydrogen mass of $f_{\mathrm{H}}M$ for fusion (proton-proton chain reaction uses hydrogen, not helium). Then, the total energy released over the lifetime of the star is $\epsilon f_{\mathrm{H}}Mc^2$. The lifetime of the star is then $\epsilon f_{\mathrm{H}}Mc^2/L_{\odot} \approx 1.1 \times 10^{18}$ s.

Answer: B

12. Solution: We recall that if $e$ is the eccentricity and $a$ is the semi-major axis for the orbit, then $r_{\min} = a(1 - e)$ and $r_{\max} = a(1 + e)$. Thus,

Now, we have $r_{\min} = \frac{2}{5}a$, so $a = \frac{5}{2}r_{\min}$. The points where the semiminor axis intersects the ellipse are both a distance $a$ from the foci. Thus, the semi-minor axis precisely cuts off the portion of the elliptical orbit where the distance is more than $\frac{5}{2}r_{\min}$.

To calculate the percentage of the orbit spent on this part of the orbit, we use Kepler's second law. The area swept during this time is half of the ellipse plus a triangle with base $2b$ and height $c = \frac{3}{5}a$. Thus, the proportion of area swept is

Answer: D

13. Solution: The change in flux occurs when the exoplanet is transiting. The flux during transit is given by

where $F$ is the maximum flux of the star. Thus, the change in apparent magnitude is

Solving for $r_{\mathrm{ex}}$ yields $0.075R_{\odot}$.

Answer: A

14. Solution: 400 million years after the Big Bang was roughly 13.3 billion years ago. If we set $T$ equal to $1.33 \cdot 10^{10}$ years and solve for $M$, we get

Answer: D

Since the lifetime of the Sun is 10 billion years which is somewhat less than the age of the first stars, it makes sense that the answer is a bit less than the mass of the Sun.

15. Solution: The first step is to calculate the velocity of the light sail:

The Lorentz factor associated with this velocity is the following:

Considering that the time for the Earth reference frame is dilated with respect to the time for the light sail reference frame:

Therefore, this trip would take 19.5 years in the reference frame of the light sail.

Answer: C

16. Solution:

Corona Borealis is the only constellation from the given options that is not on the image.

Answer: E

17. Solution: Let $N$ denote the north pole, $B$ denote Baia, and $C$ denote Caia. It is well known that the area of $\triangle NBC$ is equal to its spherical excess multiplied by the squared radius of earth. If $f = [\triangle NBC]/(4\pi R_{\oplus}^2)$, then this implies that

Substituting $\angle N = 67.55^{\circ} - 18.95^{\circ} = 48.6^{\circ} = 0.848\ rad$ and $f = 0.0675$ gives $\angle B + \angle C = \pi$, so $\sin\angle B = \sin\angle C$. Applying the spherical law of sines in $\triangle NBC$, this gives us $\sin(90^{\circ} - 66.56^{\circ}) = \sin NC$. Since $\angle B$ and $\angle C$ are supplementary, we thus have $NC = 180^{\circ} - 23.44^{\circ} = 156.56^{\circ}$, so Caia is in the southern hemisphere and $\delta = 66.56^{\circ}$ S.

Alternatively, you can also have used the answer choices to help deduce the answer. Given $\delta = 66.56^{\circ}$ S, the area of our desired triangle is exactly half of the area of the wedge of earth from $\lambda = 18.95^{\circ}$ to $\lambda = 67.55^{\circ}$. Since this wedge covers $48.6^{\circ}/360^{\circ} = 13.5\%$ of the Earth's surface, $\triangle NBC$ covers 6.75%, so $\delta = 66.56^{\circ}$ is our answer.

Answer: E

18. Solution: To determine the shape of each path, consider the cone whose base is the declination circle of the Sun and whose vertex is located at the head of either Lucas or Justin. Each path is then equal to the intersection of the great circle determined by the ground and the cone.

First we consider the summer solstice from each point. From the previous problem, the latitude of Baia is $90^{\circ} - \varepsilon$, so the declination circle of the sun touches the ground at one point. This implies that ground is parallel to a generator of our cone, so the path of Lucas's shadow at this time is a parabola. At the equator, the declination circle intersects the ground in two points, so the path of Justin's shadow is a non-parallel cross section of our cone, i.e., a hyperbola.

Now, consider what happens at the vernal equinox. In both locations, the head of Lucas and Justin coincides with the declination circle, so the “cone” has height zero. But then the intersection of the ground with our flat cone is just a line, so both shadows are lines at the vernal equinox.

Answer: A

19. Solution: If an object is viewed with its orbit edge-on, its radial velocity semi-amplitude $v$ is just its speed along its circular orbit. This problem can be solved by combining Kepler's Third with the knowledge that $m_sr_s = m_pr_p$ for two orbiting bodies, but we'll use conservation of momentum instead. Because the two orbiting bodies will always have velocity vectors pointing in opposite directions, in the center of mass frame, $m_sv_s - m_pv_p = 0$. In addition, because the each body orbits in a circle, we know that $v = 2\pi R/T$. Using these, we can rewrite Kepler's Third:

Now, applying conservation of momentum, $m_p = m_s \cdot v_s/v_p = 1.1M_{\odot} \cdot \frac{56\ \mathrm{m/s}}{1.357 \cdot 10^5\ \mathrm{m/s}} = 0.47M_{\text{♃}}$.

Answer: D

20. Solution: Since the intensity of an electromagnetic wave just depends on its amplitude, both waves have the same intensity. This result can be seen by the Poynting Vector, which represents the energy flux of the wave:

Answer: C

21. Solution: From the definition of parallax, the distance of the system from us, in parsecs, is given by:

Hence, the distance between the stars projected on the sky in the moment of maximum separation is

Since they are on an edge-on orbit, the projected distance is equal to the physical distance of the stars. Therefore, via Kepler's Third Law, we have that

Comparing to the Earth-Sun system,

Answer: C

22. Solution: Since we are dealing with an ideal black body, its albedo is equal to zero and its emissivity is 1. So, we can calculate the planet's equilibrium temperature setting the absorbed power to be equal to the emitted power.

Here, $Pot$ refers to Power, $F$ to Flux, and $A$ to area. Using Stefan-Boltzmann's Law, the power emitted by a perfect black body is given by:

Besides that, we have that the effective area is equal to $\pi R^2$. So,

Therefore, since the temperature is inversely proportional to the distance squared, we have the lowest possible temperature on the perihelion and the highest possible on the aphelion. For the planet to be on the habitable zone, $T_{\min} \geqslant T_1 = 273$ K and $T_{\max} \leqslant T_2 = 373$ K So, we can write that:

In the extreme condition, we have the maximum possible aphelion and the minimum possible perihelion. So, since $d_{\max} = a(1 + e)$ and $d_{\min} = a(1 - e)$,

Hence,

Answer: B

23. Solution: The Sun's declination varies over the course of the year due to the Earth's axial tilt. At latitude $\phi$, the Sun will be observed to never set if it has a declination $\delta$ greater than $90^{\circ} - \phi$. Thus, we can calculate the beginning and end of the polar day by calculating when the Sun reaches this declination.

We draw a spherical triangle with the vernal equinox forming one corner, two sides formed by the celestial equator and the ecliptic, and a third side perpendicular to the celestial equator which represents the declination of the Sun. The lengths of these sides are $\lambda$ (the ecliptic longitude of the sun), $\alpha$ and $\delta$ (the right ascension and declination of the sun) respectively.

Using the spherical law of sines on this triangle,

Thus,

For an observer at $40^{\circ}$ North, the Sun never sets if $\delta > 50^{\circ}$, so this period begins and ends when $\delta = 50^{\circ}$ exactly. Solving for $\lambda$, we get $\lambda = 62.2^{\circ}$ or $117.8^{\circ}$. The Sun never sets when it's between these two points on the ecliptic, which are $55.6^{\circ}$ apart. Since the Sun traverses a full 360 degrees around the ecliptic in 365 days, this means that the polar day is 56 days long.

Answer: B

24. Solution: First, we solve $0 = x^2/36 - 1$ to find that the mirror ends at $x = \pm 6\mathrm{m}$, thus the diameter of the telescope is 12m. The equation for an axial parabola in terms of its focal length is $y = x^2/4f + c$ where $f$ is the focal length, so the focal length is $f = 9\mathrm{m}$; this focal length can also be derived using the geometric definition of the parabola. The focal ratio (f-number) is defined as $f/d$, and thus the ratio is $9/12 = 0.75$. We write the f-number as f/0.75.

Answer: A

25. Solution: The Sun travels up and down (its declination changes) by $i$ degrees from the equator over the course of a year. Since the zenith and horizon (the point with the highest altitude on a path that just doesn't rise about the horizon is on the horizon) are 90 degrees off, $i$ needs to be 45 degrees at least for the sun to be able to reach both.

Answer: C

26. Solution: One particularly short solution uses the relation $p = l^2/\mu$ where $l$ is the specific angular momentum, $\mu = MG$ is the standard gravitational parameter, and $p = a(1 - e^2)$ is the semi-latus rectum. With $l = vr_{\perp} = R_{\odot} \cdot 10000$ m/s, combining $a(1 - e^2) = l^2/\mu$ with $v^2/2 = -\mu/2a$ from vis-viva ($r = \infty$ so $-\mu/r = 0$) yields:

A more elementary solution to this question uses only relations about conics and the vis-viva equation: it can be shown by congruent triangles that the perpendicular distance from an asymptote to a focus is equal to the semi-minor axis $b = a\sqrt{1 - e^2}$ of the hyperbola, so we find that $R_{\odot} = a\sqrt{1 - e^2}$. This can be combined with $v^2/2 = -\mu/2a$ from vis-viva.

Answer: D

27. Solution: Since this is a low redshift, it is possible to use the following expression to estimate the velocity of the galaxy:

Using Hubble's Law, it is possible to calculate the distance:

Answer: A

28. Solution: It is important to notice that this ratio will be minimal when the orbits are at limit of touching each other. If the orbits weren't about to touch each other, it would be possible to shrink the larger orbit to obtain a smaller ratio between the semi-major axes.

Considering that the major axes of the orbits are aligned, there are two possible scenarios in which the orbits are the limit of touching each other. The first one is when the apoapsis of the smaller orbit is about to touch the periapsis of the larger one. In this scenario, the ratio between the major axes would be the following:

The second scenario is when the apoapsis of the smaller orbit is at the limit of touching the apoapsis of the larger one. In that case, the ratio would be the following:

The smallest of the two ratios is $\frac{7}{6}$, which is therefore the minimal possible ratio between the semi-major axes of the orbits.

Answer:B

29. Solution: There are two pieces of information that matter here:

1. MIT is in the Northern Hemisphere, and

2. The Sun is above the horizon north of due East.

These two pieces of information imply the declination of the Sun must be greater than 0. Therefore, we must be after the spring equinox and before the fall equinox, so the only possible answer is late April-early May.

Answer: E

30. Solution:

As the Sun was north of East in the morning, it is now south of West. By symmetry across the meridian, the altitude of the Sun now is the same as the altitude of the Sun when it was south of East by the same angle. But as the Sun rises in altitude in the morning before it passes the meridian, it therefore must've been higher than it was in the morning.

Answer: A