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OWAO 2023, Теория — Задача 3. Apparently Invisible

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ΩWAO 2023 ◇ Theoretical Round ◇ Problems

Условие

A planet orbits a main sequence star with apparent bolometric magnitude $m = +7^{\mathrm{m}}$ in a circular orbit with period $T = 570$ years. The star is 385 light years from the Sun.

a) Estimate the maximum angular distance between the planet and the star for an observer on the Earth.

b) Compare the obtained value with the angular resolution of the James Webb Space Telescope (JWST) at wavelength $\lambda = 3.8$ µm.

The effective diameter of JWST is $D = 6.5$ m. Neglect the interstellar extinction.

Отговори

Решение

a) In order to determine the maximum angular distance between a star and a planet, we need to know the radius of the planet’s orbit, and for this we need to estimate the mass of the star. Recalling $^{3}$ that 1 pc = 3.26 light years, we estimate the luminosity based on the apparent magnitude and distance, namely

M = m + 5 - 5\lg r = 7 + 5 - 5\lg\frac{385}{3.26} = 1.6^{\mathrm{m}},

M - M_{\odot} = -2.5\lg\frac{L}{L_{\odot}} \quad\Rightarrow\quad L = 10^{0.4(M_{\odot} - M)}L_{\odot} = 10^{0.4(4.7 - 1.6)}L_{\odot} = 17L_{\odot}.

According to the mass–luminosity relation for main sequence stars, for such luminosities there is a power law relationship between mass $\mathfrak{M}$ and luminosity $L$ :

L \propto \mathfrak{M}^4.

Therefore, the mass of the star is

\mathfrak{M} = \left(\frac{L}{L_{\odot}}\right)^{1/4}\mathfrak{M}_{\odot} \approx 2\mathfrak{M}_{\odot}.

Next, we estimate the radius of the orbit:

\frac{T^2}{a^3} = \frac{4\pi^2}{G\mathfrak{M}} \quad\Rightarrow\quad a = \sqrt[3]{\frac{G\mathfrak{M}T^2}{4\pi^2}};

a = \sqrt[3]{\frac{6.67 \cdot 10^{-11} \times (2 \times 2.0 \cdot 10^{30}) \times (570 \times 365.25 \cdot 24 \cdot 60 \cdot 60)^2}{4\pi^2}} = 1.3 \cdot 10^{13}\ \mathrm{m} = 87\ \mathrm{au}.

The maximum angular distance is

\alpha = \frac{a}{r} = \frac{87\ \mathrm{au}}{\frac{385}{3.26} \cdot 206265\ \mathrm{au}} \approx 4 \cdot 10^{-6}\ \mathrm{rad} \approx 0.7''.

b) Next, we estimate the angular resolution of JWST:

\beta \sim \frac{\lambda}{D} = \frac{3.8 \cdot 10^{-6}}{6.5} \approx 6 \cdot 10^{-7}\ \mathrm{rad} \approx 0.1''.

It can be noticed that the maximum angular distance between the planet and the star exceeds the angular resolution of the telescope.

The problem discusses the planetary system HIP 65426, the planet was observed directly (Carter et al., 2023, arXiv:2208.14990).

Marking Scheme:

Absolute magnitude of the star $M$ — 1.5 pt.
Luminosity of the star $L$ — 1.5 pt.
Mass estimation: $L \propto \mathfrak{M}^{\alpha}$
- $\alpha \in [3.5;\ 4.0]$ — 1.5 pt.
- $\alpha \in [3.0;\ 3.5)$ — 1 pt.
Radius of the orbit $a$ — 1.5 pt.
Maximum angular distance $\alpha$ — 1 pt.
Angular resolution estimation $\beta$ — 2 pt.
Conclusion about the resolution — 1 pt.

$^{3}\ 1\ \mathrm{pc} = 206\,265\ \mathrm{au} = 3.086 \cdot 10^{13}\ \mathrm{km} = 3.086 \cdot 10^{16}\ \mathrm{m} = 1.03 \cdot 10^{8}\ \mathrm{s} \times c = 3.26 \times c \times 365.25\ \mathrm{d} := 3.26\ \mathrm{ly}.$

Константи

Universal
Speed of light	$c = 3.00 \cdot 10^8\ \mathrm{m/s}$
Planck constant	$h = 6.63 \cdot 10^{-34}\ \mathrm{J \cdot s}$
Hubble constant	$H_0 = 70\ (\mathrm{km/s})/\mathrm{Mpc}$
Astronomical unit	$1\ \mathrm{au} = 149.6 \cdot 10^6\ \mathrm{km}$
Parsec	$1\ \mathrm{pc} = 206\,265\ \mathrm{au}$

Earth
Radius	$R_{\oplus} = 6371\ \mathrm{km}$
Obliquity	$\varepsilon = 23.4^{\circ}$
Surface gravity	$g = 9.81\ \mathrm{m/s^2}$
Orbital period	$T_{\oplus} = 365.26^{\mathrm{d}}$
Orbital eccentricity	$e_{\oplus} = 0.0167$

Moon
Radius	$R_{\text{☾}} = 1737\ \mathrm{km}$
Orbital period	$T_{\text{☾}} = 27.32^{\mathrm{d}}$
Orbital inclination	$i_{\text{☾}} = 5.1^{\circ}$

Sun
Radius	$R_{\odot} = 6.96 \cdot 10^5\ \mathrm{km}$
Absolute magnitude	$M_{\odot} = 4.74^{\mathrm{m}}$ (bol.)
Effective temperature	$T_{\odot} = 5.8 \cdot 10^3\ \mathrm{K}$
Luminosity	$L_{\odot} = 3.828 \cdot 10^{26}\ \mathrm{W}$

Emission constants
Stefan–Boltzmann	$\sigma = 5.67 \cdot 10^{-8}\ (\mathrm{W/m^2})/\mathrm{K^4}$
Wien’s displacement	$b = 2898\ \mu\mathrm{m} \cdot \mathrm{K}$

UBV system	Mean wavelengths
U band	$\lambda_U = 364\ \mathrm{nm}$
B band	$\lambda_B = 442\ \mathrm{nm}$
V band	$\lambda_V = 540\ \mathrm{nm}$

Hydrogen spectrum
Lyman L $\alpha$	$\lambda_{L\alpha} = 1215.7$ Å
Balmer H $\alpha$	$\lambda_{H\alpha} = 6562.8$ Å

Оригинал в Архива: OWAO_2023_Theoretical_Round_Problems.pdf · официални решения: OWAO_2023_Theoretical_Round_Solutions.pdf