BAAO 2017, Теория — Задача 3. Starkiller Base

• (i) $5.91 \times 10^{14}$ s

  = 18.7 Myr

• c. 582000 s

  = 6.7 days

• f. $1 \times 10^{13}$ years

Q3 - Starkiller Base [Total = 20]

a. Assume the Sun was initially made of pure hydrogen, carries out nuclear fusion at a constant rate and will continue to do so until the hydrogen in its core is used up. If the mass of the core is 10% of the star, and 0.7% of the mass in each fusion reaction is converted into energy, show that the Sun’s lifespan on the main sequence is approximately 10 billion years.

Time on main sequence = total nuclear energy available / luminosity

$t_{\mathrm{MS}} = \frac{E_{\mathrm{nuc}}}{\mathrm{L}_{\odot}} = \frac{\Delta mc^2}{\mathrm{L}_{\odot}} = \frac{f\epsilon\mathrm{M}_{\odot}c^2}{\mathrm{L}_{\odot}} = \frac{0.1\times0.007\times1.99\times10^{30}\times\left(3.00\times10^{8}\right)^2}{3.85\times10^{26}}$ [1]

$= 3.26\times10^{17}\ \mathrm{s} = 1.03\times10^{10}\ \mathrm{years}$ [1] [2]

$(\approx 10\ \text{billion years})$

b. The Starkiller Base is able to stop nuclear fusion in the Sun’s core

i. At its current luminosity, how long would it take the Sun to radiate away all of its gravitational binding energy? (This is an estimate of how long it would take to drain a whole star when radiatively charging the superweapon.)

Time radiating energy = total gravitational binding energy / luminosity

$t_{\mathrm{rad}} = \frac{U_{\odot}}{\mathrm{L}_{\odot}} = \frac{3G\mathrm{M}_{\odot}^2}{5\mathrm{R}_{\odot}\mathrm{L}_{\odot}} = \frac{3\times6.67\times10^{11}\times\left(1.99\times10^{30}\right)^2}{5\times6.96\times10^{8}\times3.85\times10^{26}}$ [1]

$= 5.91\times10^{14}\ \mathrm{s}\ (= 18.7\ \mathrm{Myr})$ [1] [2]

ii. How does your value compare to the main sequence lifetime of the Sun calculated in part a.?

This is much shorter than $t_{\mathrm{MS}}$ [1] [1]

iii. Comment on whether there were (or will be) any events in the life of the Sun with a timescale of this order of magnitude.

An event in the Sun's life that happened on a timescale of this order of magnitude is the gravitational collapse of the protostar before it joined the main sequence [1] [1]

[The Sun will also be on the asymptotic giant branch (AGB) for a similar order of magnitude of time - this is when the core is completely carbon / oxygen (but no longer undergoing fusion) and there is a spherical shell of helium burning happening just outside the core (with a shell outside that of hydrogen burning). Credit this answer too if a student mentions it.]

c. In practice, the gravitational binding energy of the Earth is much lower than that of the Sun, and so the First Order would not need to drain the whole star to get enough energy to destroy the Earth. Assuming the weapon is able to channel towards it all the energy being radiated from the Sun’s entire surface, how long would it take them to charge the superweapon sufficiently to do this?

Time charging the weapon = total energy needed / rate of energy transfer

$t_{\mathrm{charge}} = \frac{U_{\oplus}}{\mathrm{L}_{\odot}} = \frac{3G\mathrm{M}_{\oplus}^2}{5\mathrm{R}_{\oplus}\mathrm{L}_{\odot}} = \frac{3\times6.67\times10^{11}\times\left(5.97\times10^{24}\right)^2}{5\times6.37\times10^{6}\times3.85\times10^{26}}$ [1]

$= 5.82\times10^{5}\ \mathrm{s}\ (= 6.7\ \mathrm{days})$ [1] [2]

(So it would only take a week to absorb enough energy from the Sun to destroy the Earth!)

d. Taking the Starkiller Base’s ice planet to have a diameter of 660 km, show that the Sun can be safely contained, even if it was fully drained.

Need to work out the Schwarzschild radius for the Sun, and compare it to the size of the base

$R_{\mathrm{S}} = \frac{2G\mathrm{M}_{\odot}}{c^2} = \frac{2\times6.67\times10^{-11}\times1.99\times10^{30}}{(3.00\times10^{8})^2} = 2.95\times10^{3}\ \mathrm{m}\ (= 2.95\ \mathrm{km})$ [1] [1]

(About 100 times) Smaller than the radius of the base $\therefore$ the Sun can be safely contained [1] [1]

e. The Starkiller Base wants to destroy all the planets in a stellar system on the far side of the galaxy and so drains 0.10 $\mathrm{M}_{\odot}$ from the Sun to charge its weapon. Assuming that the U per unit volume of the Sun stays approximately constant during this process, calculate:

i. The new luminosity of the Sun.

Need to use the mass-luminosity relation for main sequence stars ($\mathrm{L} \propto \mathrm{M}^4$) to work out the luminosity of a $0.9\ \mathrm{M}_{\odot}$ star (since that is the new mass of the Sun)

$L \propto M^4 \quad \therefore \quad \frac{L_{\mathrm{new}}}{\mathrm{L}_{\odot}} = \left(\frac{M_{\mathrm{new}}}{\mathrm{M}_{\odot}}\right)^4$

$\therefore \quad L_{\mathrm{new}} = 0.9^4\ \mathrm{L}_{\odot} = 0.66\ \mathrm{L}_{\odot}\ (= 2.53\times10^{26}\ W)$ [1] [1]

ii. The new radius of the Sun.

Energy density = U / V = constant

$\frac{U}{V} = \frac{\frac{3GM^2}{5R}}{\frac{4}{3}\pi R^3} = \frac{9GM^2}{20\pi R^4} \propto \frac{M^2}{R^4} = \text{constant} \quad \therefore \quad \frac{\mathrm{M}_{\odot}^2}{\mathrm{R}_{\odot}^4} = \frac{M_{\mathrm{new}}^2}{R_{\mathrm{new}}^4} \quad \Rightarrow \quad \frac{\mathrm{M}_{\odot}}{\mathrm{R}_{\odot}^2} = \frac{M_{\mathrm{new}}}{R_{\mathrm{new}}^2}$ [1]

$R_{\mathrm{new}} = \sqrt{\frac{M_{\mathrm{new}}}{\mathrm{M}_{\odot}}}\ \mathrm{R}_{\odot} = \sqrt{0.9}\ \mathrm{R}_{\odot} = 0.95\ \mathrm{R}_{\odot}\ (= 6.60\times10^{8}\ \mathrm{m})$ [1] [2]

iii. The new temperature of the surface of the Sun (current $\mathrm{T}_{\odot}$ = 5780 K), and suggest (with a suitable calculation) what change will be seen in terms of its colour.

Need to use the Stephan-Boltzmann Law to get the new temperature and then Wien's Law to determine the effect on the peak wavelength (and hence the colour)

$L = 4\pi R^2\sigma T^4 \quad \therefore \quad \frac{L_{\mathrm{new}}}{\mathrm{L}_{\odot}} = \left(\frac{R_{\mathrm{new}}}{\mathrm{R}_{\odot}}\right)^2\left(\frac{T_{\mathrm{new}}}{\mathrm{T}_{\odot}}\right)^4$

$\therefore \quad T_{\mathrm{new}} = \sqrt[4]{\frac{\left(\frac{L_{\mathrm{new}}}{\mathrm{L}_{\odot}}\right)}{\left(\frac{R_{\mathrm{new}}}{\mathrm{R}_{\odot}}\right)^2}}\ \mathrm{T}_{\odot} = \sqrt[4]{\frac{0.9^4}{0.9}}\ \mathrm{T}_{\odot}$ [1]

$= 5340\ \mathrm{K}\ (= 0.92\ \mathrm{T}_{\odot})$ [1] [2]

$\lambda_{\mathrm{peak,new}} = \frac{2.90\times10^{-3}}{T_{\mathrm{new}}} = \frac{2.90\times10^{-3}}{5340} = 5.43\times10^{-7}\ \mathrm{m}\ (= 543\ \mathrm{nm})$ [1] [1]

This is a longer wavelength than the current peak (500 nm) so the Sun is redder [1] [1]

f. Assume that at the moment of destruction of the Starkiller Base the mass of the new star formed is equal to the mass drained from the Sun (0.10 $\mathrm{M}_{\odot}$). Derive an expression for the main sequence lifetime in terms of stellar mass, and hence calculate the main sequence lifetime of this new star.

We can combine the mass-luminosity relation with the expression we used in part a.

$t_{\mathrm{MS}} \propto \frac{M}{L} \quad but \quad L \propto M^4 \quad \therefore \quad t_{\mathrm{MS}} \propto \frac{M}{M^4} \propto M^{-3}$ [1] [1]

$\therefore t_{\mathrm{MS,new}} = \left(\frac{M_{\mathrm{new}}}{\mathrm{M}_{\odot}}\right)^{-3} t_{\mathrm{MS},\odot} = 0.1^{-3}\times10^{10}\ \mathrm{years}$ [1]

$= 10^{13}\ \mathrm{years}$ [1] [2]