BAAO 2017, Теория — Задача 5. Imaging an Exoplanet

• a. 0.038 arcsec
• b. L = 1.95 × 10²² W; f_star = 9.69 × 10⁻¹³ W m⁻²; f_planet = 1.12 × 10⁻¹⁹ W m⁻²; CR = 8.64 × 10⁶; m_planet = 28.5
• d. $1.62 \times 10^{9}$ s
• e. 4.96 s

Q5 - Imaging an Exoplanet [Total = 20]

a. Calculate the maximum angular separation between the star and the planet, assuming a circular orbit. Give your answer in arcseconds (where 3600 arcseconds = 1°).

$a = \sqrt[3]{\frac{G\mathrm{M}_*T^2}{4\pi^2}} = \sqrt[3]{\frac{6.67\times10^{-11}\times(0.123\times1.99\times10^{30})\times(11.186\times24\times60\times60)^2}{4\pi^2}}$ [1]

$= 7.28\times10^{9}\ \mathrm{m} = (0.049\ \mathrm{AU})$ [1]

$\theta_{\max} = \frac{a}{d} = \frac{7.28\times10^{9}}{1.295\times3.09\times10^{16}} = 1.82\times10^{-7}\ \mathrm{radians}$ [0.5]

$= 0.038\ \mathrm{arcseconds}$ [0.5] [3]

b. Determine the luminosity of the star and hence calculate the flux received on the Earth (in W m⁻²) from both the star and the planet. Use them to work out the contrast ratio and thus the apparent magnitude of the planet. Assume the planet reflects half of the incident light and that $\mathcal{M}_{\odot}$ = 4.83.

$\mathcal{M} = m - 5\log\left(\frac{d}{10}\right) = 11.13 - 5\log\left(\frac{1.295}{10}\right) = 15.57$ [1]

$L = 10^{\left(\frac{\mathcal{M}-\mathcal{M}_{\odot}}{-2.5}\right)}\mathrm{L}_{\odot} = 10^{\left(\frac{15.57-4.83}{-2.5}\right)}\times3.85\times10^{26} = 1.95\times10^{22}\ \mathrm{W}$ [1] [2]

$(= 5.07\times10^{-5}\ \mathrm{L}_{\odot})$

$f_{\mathrm{star}} = \frac{L}{4\pi d^2} = \frac{1.95\times10^{22}}{4\pi\times(1.295\times3.09\times10^{16})^2} = 9.69\times10^{-13}\ \mathrm{W\,m^{-2}}$ [1] [1]

$f_{\mathrm{planet}} = \frac{f_{\mathrm{incident}}\times\text{fraction reflected}\times\text{cross-sectional area of planet}}{4\pi d^2}$

$= \frac{\frac{L}{4\pi a^2}\times0.5\times\pi r_p^2}{4\pi d^2} = \frac{\frac{1.95\times10^{22}}{4\pi\times\left(7.28\times10^{9}\right)^2}\times0.5\times\pi\times\left(1.1\times6.37\times10^{6}\right)^2}{4\pi\times(1.295\times3.09\times10^{16})^2}$ [1]

$= 1.12\times10^{-19}\ \mathrm{W\,m^{-2}}$ [1] [2]

$CR = \frac{f_{\mathrm{star}}}{f_{\mathrm{planet}}} = \frac{9.69\times10^{-13}}{1.12\times10^{-19}} = 8.64\times10^{6}$ [1] [1]

$\Delta m = 2.5\log(CR) = 2.5\log(8.64\times10^{6}) = 17.3$ [1]

$\therefore m_{\mathrm{planet}} = m_{\mathrm{star}} + \Delta m = 11.13 + 17.3 = 28.5$ [1] [2]

[Since we only know the minimum radius of the exoplanet it could be larger and hence brighter, however it may also reflect less than half the incident light from the star and so be fainter – in practice the numbers used here are an optimistic estimate and it is more likely to be fainter.]

(Accepted alternative methods: if they use Stephan-Boltzmann's Law then L = 5.88 × 10²³ W, f$_{\mathrm{star}}$ = 2.92 × 10⁻¹¹ W m⁻², and f$_{\mathrm{planet}}$ = 3.39 × 10⁻¹⁸ W m⁻², though the contrast ratio and magnitude should be the same. Also accept if they assume only the day side is able to reflect and hence the apparent magnitude brightens to 27.9)

c. Verify that the HST (which is diffraction limited since it's in space) would be sensitive enough to image the planet in the visible, but is unable to resolve it from its host star (take λ = 550 nm).

$\theta_{\mathrm{HST}} = \frac{1.22\lambda}{D} = \frac{1.22\times550\times10^{-9}}{2.4} = 2.80\times10^{-7}\ \mathrm{radians}\ (= 0.058\ \mathrm{arcseconds})$ [0.5]

Since $\theta_{\mathrm{HST}} > \theta_{\max}$ then the HST can't resolve it [0.5] [1]

Apparent magnitude of planet (28.5) is brighter (greater) than limiting magnitude (31) [1] [1]

d. Calculate the exposure time needed for a Keck II image of the exoplanet to have an SNR of 3. Assume that the telescope has perfect AO, is observed at the longest wavelength for which the planet can still be resolved from the star, all the received flux from the planet consists of photons of that longest wavelength , ε = 0.1 and b = $10^9$ photons s⁻¹ (so $b \gg f$). Comment on your answer.

$\lambda_{\mathrm{Keck}} = \frac{\theta_{\max}D}{1.22} = \frac{1.82\times10^{-7}\times10}{1.22} = 1.49\times10^{-6}\ \mathrm{m}$ [1]

Photon flux:

$f = \frac{f_{\mathrm{planet}}}{E_{\mathrm{photon}}} = \frac{f_{\mathrm{planet}}}{\frac{hc}{\lambda_{\mathrm{Keck}}}} = \frac{1.12\times10^{-19}}{\frac{6.63\times10^{-34}\times3.00\times10^{8}}{1.49\times10^{-6}}} = 0.84\ \mathrm{photons\ m^{-2}\ s^{-1}}$ [1]

(SB-Law f$_{\mathrm{planet}}$ gives f = 25.4 photons m⁻² s⁻¹, and day side reflection increases f by a factor of 2)

Since $b \gg f$ we can simplify the denominator of the SNR formula by ignoring the first term, so

$t = \frac{SNR^2\times b}{f^2A^2\epsilon} = \frac{3^2\times10^9}{0.84^2\times(\pi\times5^2)^2\times0.1} = 1.62\times10^{9}\ \mathrm{s}$ [1] [3]

This is really long ($>$ 50 years!) so it is unlikely Keck will ever be able to directly image it [1] [1]

(alternative f gives 22.7 ks, and hence the conclusion is that it is feasible)

e. How long an exposure would JWST need in order to get the same SNR as Keck II, again if observed at the longest wavelength for which the planet can still be resolved from the star by the telescope? (Make similar assumptions about the received flux and use the same value of ε).

(using similar reasoning to the previous part of the question)

$\lambda_{\mathrm{JWST}} = 9.70\times10^{-7}\ \mathrm{m}$ [1]

$f = 0.55\ \mathrm{photons\ m^{-2}\ s^{-1}}$ [1]

(alternative method gives f = 16.5 photons m⁻² s⁻¹)

Since $b \ll f$ we can simplify the denominator of the SNR formula by ignoring the second term, so

$t = \frac{SNR^2}{fA\epsilon} = \frac{3^2}{0.55\times(\pi\times3.25^2)\times0.1} = 4.96\ \mathrm{s}$ [1] [3]

(alternative method gives t = 0.16 s, and day side reflection decreases t by a factor of 2)

[This is much more reasonable – the optimistic assumptions we have made throughout this question mean this is just a lower limit and so the actual exposure time will be longer, perhaps several minutes to get a much higher SNR and to compensate for the fact for that the sensitivity is lower at 970 nm than we have quoted since it varies with wavelength. The main thing that would prevent direct observation of the exoplanet by JWST would be if the contrast ratio was much higher than calculated here as that would mean the star's light would need to be blocked out by the Near Infrared Camera's coronagraph – unfortunately this can only be used at wavelengths too long to resolve the system.]

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