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BAAO 2020, Теория — Задача 3. Inspirals of Exoplanets Near Tidal Destruction

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British Astronomy and Astrophysics Olympiad 2019-2020 — Astronomy & Astrophysics Competition Paper — Monday 20th January 2020 · 20 януари 2020 г. · 30 т.

Указания

This question paper must not be photographed or taken out of the exam room

Time: 3 hours (approximately 60 minutes per question).

Questions: All three questions should be attempted. The last question is worth slightly fewer marks than the rest.

Solutions: Answers and calculations are to be written on loose paper. Students should ensure their name and school is clearly written on all answer sheets and pages are numbered. EACH QUESTION ANSWERED must be started on a new page. A standard formula booklet with standard physical constants should be supplied.

Instructions: To accommodate students sitting the paper at different times, please do not discuss any aspect of the paper on the internet until 8 am Saturday 25th January.

Clarity: Solutions must be written legibly, in black pen (the papers are photocopied), and working down the page. Scribble will not be marked and overall clarity is an important aspect of this exam.

Eligibility: The International Olympiad will be held during September 2020; all sixth form students are eligible to participate, even if they will be attending university from October.

Calculators: Any standard calculator may be used, but calculators cannot be programmable and must not have symbolic algebra capability.

Training Dates and the IOAA (Bogota, Columbia, 13th to 22nd September 2020)

The best students taking this paper that are eligible to represent the UK at the IOAA will be invited to attend the Training Camp to be held in the Physics Department at the University of Oxford, (Friday 27th March to Tuesday 31st March 2020). Astronomy material will be covered; problem solving skills and observational skills (telescope and naked eye observations) will be developed. At the Training Camp a data analysis exam and a short theory paper will be sat. A team of five students (plus one reserve) will be selected for further training. From April there will be mentoring by email to cover some topics and problems, followed by additional training camps in the summer.

Important Constants

Constant	Symbol	Value
Speed of light	$c$	$3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$
Earth’s rotation period	1 day	24 hours
Earth’s orbital period	1 year	365.25 days
parsec	pc	$3.09 \times 10^{16}\ \mathrm{m}$
Astronomical Unit	au	$1.50 \times 10^{11}\ \mathrm{m}$
Semi-major axis of the Earth’s orbit		1 au
Radius of the Sun	$R_{\odot}$	$6.96 \times 10^{8}\ \mathrm{m}$
Radius of the Earth	$R_{\oplus}$	$6.37 \times 10^{6}\ \mathrm{m}$
Mass of the Sun	$M_{\odot}$	$1.99 \times 10^{30}\ \mathrm{kg}$
Mass of the Earth	$M_{\oplus}$	$5.97 \times 10^{24}\ \mathrm{kg}$
Luminosity of the Sun	$L_{\odot}$	$3.85 \times 10^{26}\ \mathrm{W}$
Stephan-Boltzmann constant	$\sigma$	$5.67 \times 10^{-8}\ \mathrm{J\,m^{-2}\,K^{-4}}$
Gravitational constant	$G$	$6.67 \times 10^{-11}\ \mathrm{m^{3}\,kg^{-1}\,s^{-2}}$
Boltzmann constant	$k_{\mathrm{B}}$	$1.38 \times 10^{-23}\ \mathrm{J\,K^{-1}}$
Permittivity of free space	$\varepsilon_0$	$8.85 \times 10^{-12}\ \mathrm{F\,m^{-1}}$
Permeability of free space	$\mu_0$	$4\pi \times 10^{-7}\ \mathrm{H\,m^{-1}}$
Planck’s constant	$h$	$6.63 \times 10^{-34}\ \mathrm{J\,s}$
Elementary charge	$e$	$1.60 \times 10^{-19}\ \mathrm{C}$
Proton rest mass	$m_{\mathrm{p}}$	$1.67 \times 10^{-27}\ \mathrm{kg}$
Electron rest mass	$m_{\mathrm{e}}$	$9.11 \times 10^{-31}\ \mathrm{kg}$
Wien’s displacement law	$\lambda_{\max} T$	$2.90 \times 10^{-3}\ \mathrm{m\,K}$

Basic calculus formulae:

Chain rule	$\frac{\mathrm{d}}{\mathrm{d}x} f(g(x)) = f'(g(x))g'(x)$
Product rule	$\frac{\mathrm{d}}{\mathrm{d}x}(uv) = \frac{\mathrm{d}u}{\mathrm{d}x}v + u\frac{\mathrm{d}v}{\mathrm{d}x}$
Quotient rule	$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{u}{v}\right) = \frac{\frac{\mathrm{d}u}{\mathrm{d}x}v - u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2}$
Integration by parts	$\int u\frac{\mathrm{d}v}{\mathrm{d}x}\,\mathrm{d}x = uv - \int v\frac{\mathrm{d}u}{\mathrm{d}x}\,\mathrm{d}x$
Standard integral	$\int \frac{1}{x}\,\mathrm{d}x = \ln\lvert x\rvert + C$

Important Formulae

You might find the diagram of an elliptical orbit below useful in solving some of the questions:

Elements of an elliptic orbit:

$a = \mathbf{OA}\ (= \mathbf{OP})$	semi-major axis
$b = \mathbf{OB}\ (= \mathbf{OC})$	semi-minor axis
$e = \sqrt{1 - \frac{b^2}{a^2}}$	eccentricity
F	focus
P	periapsis (point nearest to F)
A	apoapsis (point furthest from F)

Kepler’s Third Law: For an elliptical orbit, the square of the period, $T$ , of an object about the focus is proportional to the cube of the semi-major axis, $a$ (as defined above), such that

T^2 = \frac{4\pi^2}{GM}a^3\,,

where $M$ is the total mass of the system (typically dominated by the central object) and $G$ is the universal gravitational constant.

Vis-Viva Equation: For an elliptical orbit, the speed $v$ of an object at a distance $r$ from the focus is related to the semi-major axis, $a$ , total mass of the system, $M$ , and universal gravitational constant, $G$ , (as defined above), such that

v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right)\,.

Magnitudes: The apparent magnitudes of two objects, $m_1$ and $m_0$ , are related to their apparent brightnesses, $b_1$ and $b_0$ , via the formula

\frac{b_1}{b_0} = 10^{-0.4(m_1 - m_0)}\,.

The absolute magnitude of an object, $\mathcal{M}$ , is the same as its apparent magnitude when viewed from 10 pc, hence the relationship between apparent and absolute magnitude and distance is

m - \mathcal{M} = 5\log\left(\frac{d}{10}\right),

where $d$ is measured in parsecs.

Note for markers

BAAO 2019/20 Solutions and Marking Guidelines

Note for markers:

Answers to two or three significant figures are generally acceptable. The solution may give more in order to make the calculation clear. Units should be present on final answers when appropriate.
There are multiple ways to solve some of the questions; please accept all good solutions that arrive at the correct answer. If a candidate gets the final (numerical) answer then allow them all the marks for that part of the question (as indicated in red), so long as there are no unphysical / nonsensical steps or assumptions made.

Условие

Some of the very first exoplanets to be discovered in large surveys were dubbed ‘hot Jupiters’ as they were similar in mass to Jupiter (i.e. a gas giant) but were much closer to their star than Mercury is to the Sun (and hence are in a very hot environment). Planetary formation models suggest that they were unlikely to have formed there, but instead formed much further out from the star and migrated inwards, due to gravitational interactions with other planets in the system. Studies of ‘hot Jupiters’ show that there is an overabundance of them with periods of $\sim 3 - 4$ days, and very few with periods shorter than that. Since large, close-in planets should be the easiest to detect in all of the main methods of finding exoplanets, this scarcity is likely to be a real effect and suggests that exoplanets which are that close to their star are in a relatively rapid (by astronomical standards) inspiral towards destruction by their star.

Left: scatter plot with error bars of a/a_Roche (1.0 to 4.5) against M_star/M⊙ (0.6 to 1.6), a dotted horizontal line at 2.0; labelled points below it include WASP-43b, WASP-52b, WASP-4b, CoRoT-1b, OGLE-TR-56b, WASP-10b, HAT-P-32b, WASP-19b, WASP-12b and WASP-18b, with WTS-2b marked by a red square near 1.4. Right: logarithmic plot of T_shift / s against M_star / M⊙ (0.4 to 1.6) with points labelled WASP-18b, WASP-103b, WASP-19b, CoRoT-14b, OGLE-TR-56b, WASP-43b, WASP-78b, HAT-P-7b and WTS-2b, and a dotted horizontal detection line near 5 s. — Figure 6: Left: The orbital radius of several ‘hot Jupiters’ scaled by the Roche radius of the system (where tidal forces would destroy the planet). There is an expected pile up close to radii double the Roche radius (dotted line), and very few with radii smaller than that - those that are will inevitably spiral into the star and be destroyed by the tidal forces when they get too close. Credit: Birkby et al. (2014). Right: As the planets inspiral we should see a shift in when their transits occur. This figure shows the predicted size of the shift after a period of 10 years if the tidal dissipation quality factor $Q'_{\star} = 10^{6}$ , as well as the current detection limit of 5 seconds (dotted line). Therefore measuring if there is any shift in the transit times over the course of a decade of observations can put stringent limits on the value of $Q'_{\star}$ . Credit: Birkby et al. (2014).

The Roche radius, where a planet will be torn apart due to the tidal forces acting on it, is defined as

a_{\mathrm{Roche}} \approx 2.16R_P\left(\frac{M_{\star}}{M_P}\right)^{1/3}\,,

where $R_P$ is the radius of the planet, $M_P$ is the mass of the planet and $M_{\star}$ is the mass of the star. If a gas giant is knocked into a highly elliptical orbit (i.e. $e \approx 1$ ) that has a periapsis $r_{\mathrm{peri}} < a_{\mathrm{Roche}}$ then it will not survive. However, if the periapsis just grazes the Roche radius ( $r_{\mathrm{peri}} \approx a_{\mathrm{Roche}}$ ) then the orbit will rapidly circularise. By conserving angular momentum, it can be shown that the circular orbit will have a radius $a = 2a_{\mathrm{Roche}}$ (see the left panel of Fig 6). Exoplanets observed to be in an orbit with a radius less than that will be unstable and angular momentum will be transferred from the planet to the star, causing the star to spin more rapidly and the planet’s orbital radius to decrease. Eventually this will result in the planet’s orbit crossing the Roche radius and being destroyed by the tidal forces.

The duration of this inspiral will be dependent on how well the star can dissipate the orbital energy through frictional processes within the star, and can be parameterised by the tidal dissipation quality factor, $Q'_{\star}$ . By looking for changes in the orbital period of the planet, detectable by shifts in the timing of transits by the planet in front of the star, we can determine an estimate of $Q'_{\star}$ , which hence tells us about the internal structure of stars. These ‘hot Jupiters’ are the best laboratory we have for this, as they are the most likely to produce a measurable shift (i.e. $\sim 5$ s) in transit times within only $\sim 10$ years (see the right panel of Fig 6). We will try and reproduce these results in this question.

The WTS-2 system consists of a star of mass $M_{\star} = 0.820\,M_{\odot}$ , peak in its black-body spectrum at $\lambda_{\max} = 580$ nm, and distance from us of 1.03 kpc, with an orbiting planet (called WTS-2b) with a period $P = 1.0187$ days, mass of 1.12 $M_J$ and radius 1.36 $R_J$ . The mass and radius of Jupiter are $M_J = 1.90 \times 10^{27}$ kg and $R_J = 7.15 \times 10^{7}$ m respectively.

a. Show that WTS-2b has an orbital radius of $\sim 1.4\ a_{\mathrm{Roche}}$ . [4 т.]

b. (i) Given the apparent magnitude of WTS-2 in the visible is $m = 16.14$ and the absolute magnitude of the Sun in the same part of the EM spectrum is $\mathcal{M}_{\odot} = 4.83$ :

(i) Calculate the luminosity of the star, $L_{\star}$ . Give your answer in units of $L_{\odot}$ . [3 т.]

b. (ii) Hence work out the radius of the star, $R_{\star}$ . Give your answer in units of $R_{\odot}$ . [3 т.]

b. (iii) Show that the incident flux (in $\mathrm{W\,m^{-2}}$ ) on the planet is $\sim 10^{3}$ times larger than what we receive on Earth from the Sun. [3 т.]

The change in the semi-major axis of the planet, $a$ , due to tidal forces is given by

\left|\frac{\dot{a}}{a}\right| = 6k_2\Delta t\frac{M_P}{M_{\star}}\left(\frac{R_{\star}}{a}\right)^5 n^2

where the dot notation is used to indicate the differential with respect to time (i.e. $\dot{a} \equiv \mathrm{d}a/\mathrm{d}t$ ), $k_2$ is a constant related to the density structure of the star, $\Delta t$ is the (assumed constant) time lag between where the planet is in its orbit and the location of the tidal bulge on the star, and $n = 2\pi/P$ . By separating variables and integrating this equation, an expression can be derived for the time it takes for $a$ to decrease to zero. This is known as the inspiral time, $\tau$ , and even though the planet will be destroyed when $a = a_{\mathrm{Roche}}$ the time to go from $a = a_{\mathrm{Roche}}$ to $a = 0$ is negligible in comparison to the time to get to $a = a_{\mathrm{Roche}}$ , so $\tau$ is a good estimate of the remaining lifetime of the planet.

c. Derive an equation for $\tau$ , showing $\tau \propto a^8$ . Hint: $n$ is a function of $a$ due to Kepler’s third law. [4 т.]

The tidal dissipation quality factor is defined as

Q'_{\star} = \frac{3}{4k_2\Delta t n}

and so absorbs some of our unknown parameters about the stellar structure. Based upon studies of binary stars it seems a value of $Q'_{\star} \sim 10^{6} - 10^{7}$ is reasonable, although it is poorly constrained. The size of the shift in transit timings, $T_{\mathrm{shift}}$ , over an observation period of $T$ seconds is

T_{\mathrm{shift}} = \frac{1}{2n}T^2\left(\frac{\mathrm{d}n}{\mathrm{d}t}\right)\,.

d. (i) For the planet WTS-2b, assuming $Q'_{\star} = 10^{6}$ :

(i) Show that $\tau \propto \frac{Q'_{\star}}{n}\frac{M_{\star}}{M_P}\left(\frac{a}{R_{\star}}\right)^5$ , finding the constant of proportionality (as a fraction), and hence the remaining lifetime of the planet. Give your answer in Myr. [1 Myr = $10^{6}$ years] [5 т.]

d. (ii) Show that $T_{\mathrm{shift}} \propto \frac{T^2}{\tau}$ , finding the constant of proportionality, and hence verify that $T_{\mathrm{shift}}$ is measurable (i.e. $> 5$ s) if $T = 10$ years. Hint: Using the chain rule, $\frac{\mathrm{d}n}{\mathrm{d}t} = \frac{\mathrm{d}n}{\mathrm{d}a}\dot{a}$ . [8 т.]

Отговори

Решение

Q3 – Inspirals of Exoplanets Near Tidal Destruction [30 marks]

Using the given formula for $a_{\mathrm{Roche}}$ ,

$a_{\mathrm{Roche}} = 2.16R_P\left(\frac{M_{\star}}{M_P}\right)^{1/3} = 2.16 \times 1.36 \times 7.15 \times 10^{7} \times \left(\frac{0.820\times1.99\times10^{30}}{1.12\times1.90\times10^{27}}\right)^{1/3}$

$= 1.92 \times 10^{9}\ \mathrm{m}$ [1]

Using Kepler’s third law,

$P^2 = \frac{4\pi^2}{GM}a^3 \quad \therefore a = \sqrt[3]{\frac{GM}{4\pi^2}P^2}$

$= \sqrt[3]{\frac{6.67\times10^{-11}\times0.820\times1.99\times10^{30}}{4\pi^2} \times (1.0187 \times 24 \times 3600)^2}$ [1]

$= 2.77 \times 10^{9}\ \mathrm{m}$ [1]

$\therefore a = 1.44\ a_{\mathrm{Roche}}$ [since ‘show that’ must be at least 3 s.f.] [1] [4]

b. (i)

Since the absolute magnitude is the apparent magnitude at a fixed distance of 10 pc,

$\frac{b_1}{b_0} = 10^{-0.4(m_1 - m_0)}$ and $b_{10\ \mathrm{pc}} = \frac{L}{4\pi(10\ \mathrm{pc})^2} \quad \Rightarrow \frac{L_1}{L_0} = 10^{-0.4(\mathcal{M}_1 - \mathcal{M}_0)}$ [1]

$\mathcal{M}_{\star} = m_{\star} - 5\log\left(\frac{d}{10}\right) = 16.14 - 5\log\left(\frac{1.03\times10^{3}}{10}\right) = 6.08$ [1]

$\frac{L_{\star}}{L_{\odot}} = 10^{-0.4(\mathcal{M}_{\star} - \mathcal{M}_{\odot})} = 10^{-0.4(6.08 - 4.83)} \qquad \therefore L_{\star} = 0.317\ L_{\odot}$ [1] [3]

[Final answer must be in $L_{\odot}$ for the final mark]

b. (ii)

Using Wien’s displacement law,

$T = \frac{B}{\lambda_{\max}} = \frac{2.90\times10^{-3}}{580\times10^{-9}} = 5000\ \mathrm{K}$ [1]

Using the Stephan-Boltzmann Law,

$L_{\star} = 4\pi R_{\star}^2\sigma T^4 \therefore R_{\star} = \sqrt{\frac{L_{\star}}{4\pi\sigma T^4}} = \sqrt{\frac{0.317\times3.85\times10^{26}}{4\pi\times5.67\times10^{-8}\times5000^4}}$ [1]

$= 5.24 \times 10^{8}\ \mathrm{m} \quad = 0.753\ R_{\odot}$ [1] [3]

[Final answer must be in $R_{\odot}$ for the final mark]

b. (iii)

$b_{\mathrm{Earth}} = \frac{L_{\odot}}{4\pi(1\ \mathrm{au})^2} = \frac{3.85\times10^{26}}{4\pi\times(1.50\times10^{11})^2} = 1.36 \times 10^{3}\ \mathrm{W\,m^{-2}}$ [1]

$b_P = \frac{L_{\star}}{4\pi a^2} = \frac{0.317\times3.85\times10^{26}}{4\pi\times(2.77\times10^{9})^2} = 1.26 \times 10^{6}\ \mathrm{W\,m^{-2}}$ [1]

$\therefore b_P = 930\ b_{\mathrm{Earth}}$ [since ‘show that’ must be at least 2 s.f.] [1] [3]

[The consequence of such a large incident flux is that the surface of this planet is incredibly hot – given it is also tidally locked (so the same hemisphere always faces the star) it is modelled to have a maximum day-side surface temperature as high as ~2000 K]

$n^2 = \left(\frac{2\pi}{P}\right)^2 = \frac{GM_{\star}}{a^3}$ [1]

$\therefore \frac{\dot{a}}{a} = 6k_2\Delta t\frac{M_P}{M_{\star}}\left(\frac{R_{\star}}{a}\right)^5 n^2 = 6k_2\Delta t\frac{M_P}{M_{\star}}\left(\frac{R_{\star}}{a}\right)^5\frac{GM_{\star}}{a^3}$

$\therefore \dot{a} = 6k_2\Delta t G M_P R_{\star}^5\frac{1}{a^7}$ [1]

Separating variables and integrating,

$\int_0^a a^7\,\mathrm{d}a = \int_0^{\tau} 6k_2\Delta t G M_P R_{\star}^5\,\mathrm{d}t$ [1]

$\frac{1}{8}a^8 = 6k_2\Delta t G M_P R_{\star}^5\tau \qquad \therefore \tau = \frac{a^8}{48k_2\Delta t G M_P R_{\star}^5}$ [1] [4]

[We know that a is actually decreasing as time increases, so removing the modulus sign from the original given equation should give a negative sign, however the limits of the integral have been swapped around so that the value for τ evaluates to a positive number – do not penalise any student getting a negative expression for τ]

[This incredibly steep dependence on a is why we can treat the time to get from the initial value of a to zero as essentially the same as the time taken to get to $a_{\mathrm{Roche}}$ (when the planet is destroyed)]

d. (i)

$Q'_{\star} = \frac{3}{4k_2\Delta t n} \therefore k_2\Delta t = \frac{3}{4Q'_{\star}n}$

$\tau = \frac{a^8}{48k_2\Delta t G M_P R_{\star}^5} = \frac{a^8}{48\left(\frac{3}{4Q'_{\star}n}\right)GM_P R_{\star}^5} = \frac{a^8 Q'_{\star}n}{36GM_P R_{\star}^5} = \frac{Q'_{\star}na^3}{36M_P G}\left(\frac{a}{R_{\star}}\right)^5$ [1]

$= \frac{Q'_{\star}n\left(\frac{GM_{\star}}{n^2}\right)}{36M_P G}\left(\frac{a}{R_{\star}}\right)^5$ [1]

$= \frac{1}{36}\frac{Q'_{\star}}{n}\frac{M_{\star}}{M_P}\left(\frac{a}{R_{\star}}\right)^5$ [so const = 1/36] [1] [3]

[First mark is for correctly substituting in the expression for Q’ $_{\star}$ , the second for use of Kepler’s third law to cancel G and introduce $M_{\star}$ , and the third for the value of the constant. Simply stating the constant without showing the rest of the algebraic expression cancelling into the required form loses at least the second mark.]

So we can now calculate the remaining lifetime,

$\tau = \frac{1}{36}\frac{Q'_{\star}}{n}\frac{M_{\star}}{M_P}\left(\frac{a}{R_{\star}}\right)^5 = \frac{1}{36}\frac{10^{6}}{(2\pi/(1.0187\times24\times3600))}\frac{0.820\times1.99\times10^{30}}{1.12\times1.90\times10^{27}}\left(\frac{2.77\times10^{9}}{5.24\times10^{8}}\right)^5$ [1]

$= 1.24 \times 10^{15}$ s = 39.4 Myr [must be in Myr] [1] [2]

[Allow full ecf for the student’s values of a and $R_{\star}$ and be generous with errors that may have arose from rounding]

[This is unsurprisingly rather short on astronomical timescales – this planet really is close to destruction!]

d. (ii)

$\frac{\dot{a}}{a} = 6k_2\Delta t\frac{M_P}{M_{\star}}\left(\frac{R_{\star}}{a}\right)^5 n^2 = 6\frac{3}{4Q'_{\star}n}\frac{M_P}{M_{\star}}\left(\frac{R_{\star}}{a}\right)^5 n^2$

$\therefore \dot{a} = \frac{9}{2}\frac{n}{Q'_{\star}}\frac{M_P}{M_{\star}}\frac{R_{\star}^5}{a^4}$ [1]

$n = \sqrt{\frac{GM_{\star}}{a^3}} = \sqrt{GM_{\star}}\,a^{-3/2}$ [1]

$\therefore \frac{\mathrm{d}n}{\mathrm{d}a} = -\frac{3}{2}\sqrt{GM_{\star}}\,a^{-5/2} = -\frac{3}{2}\frac{n}{a}$ [2]

$\frac{\mathrm{d}n}{\mathrm{d}t} = \left(\frac{\mathrm{d}n}{\mathrm{d}t}\right)\dot{a} = -\frac{3}{2}\frac{n}{a} \times \frac{9}{2}\frac{n}{Q'_{\star}}\frac{M_P}{M_{\star}}\frac{R_{\star}^5}{a^4} = \frac{27}{4}n^2\frac{M_P}{M_{\star}}\left(\frac{R_{\star}}{a}\right)^5\frac{1}{Q'_{\star}}$ [1]

Putting this into the given formula,

$T_{\mathrm{shift}} = \frac{1}{2n}T^2\left(\frac{dn}{dt}\right) = \frac{1}{2n}T^2 \times \frac{27}{4}n^2\frac{M_P}{M_{\star}}\left(\frac{R_{\star}}{a}\right)^5\frac{1}{Q'_{\star}} = \frac{27}{8}T^2\frac{n}{Q'_{\star}}\frac{M_P}{M_{\star}}\left(\frac{R_{\star}}{a}\right)^5$ [1]

But we found $\tau = \frac{1}{36}\frac{Q'_{\star}}{n}\frac{M_{\star}}{M_P}\left(\frac{a}{R_{\star}}\right)^5$ so,

$T_{\mathrm{shift}} = \frac{3}{32}\frac{T^2}{\tau}$ [so constant = 3/32] [1] [7]

Evaluating the expression for T = 10 years,

$T_{shift} = \frac{3}{32}\frac{(10\times365\times24\times3600)^2}{1.24\times10^{15}} = 7.50\ \mathrm{s}$ [so > 5 s therefore measurable] [1] [1]

[Allow full ecf from their value of τ. Allow use of 365.25 days in a year]

[WTS-2b is one of a number of exoplanets that could produce measurable changes in their period over the next decade (see the right panel in Fig 6 in the paper), and hence are an incredibly useful way of constraining the value of Q’ $_{\star}$ , which in itself will greatly inform stellar models]

Questions proposed by

END OF PAPER

Questions proposed by:

Dr Alex Calverley (Royal Grammar School, Guildford)

Alan Sun (University of Cambridge)

Navonil Neogi (University of Cambridge)

Оригинал в Архива: BAAO_2020.pdf · официални решения: BAAO_2020_solutions.pdf