BAAO 2019, Теория — Задача 1. Parker Solar Probe

• a. (i) -33.44 mag
• a. (ii) 1550 K
• b. 191 km/s
• c. $2.35 \times 10^{11}$ J
• d. $r = a(1 - e\cos E)$
• e. 9.84 days

Q1 – Parker Solar Probe [25 marks]

a. (i)

ratio of brightnesses, $\frac{b_1}{b_0} = \frac{(1\ \mathrm{au})^2}{(9.86\ R_{\odot})^2} = \frac{(1.50\times10^{11})^2}{(9.86\times6.96\times10^{8})^2} = 478$ [1]

$\therefore m_{\mathrm{new}} = m_{\odot} - 2.5\log\left(\frac{b_1}{b_0}\right) = -26.74 - 6.698$

$= -33.44$ [1] [2]

[Allow calculating the solar absolute magnitude, $\mathcal{M}_{\odot} = 4.829$, as an alternative first mark]

a. (ii)

For thermal balance, power absorbed must equal power emitted, so

$\frac{L_{\odot}}{4\pi r^2} \times A_{abs} = \sigma A_{emit} T^4$

But since $A_{\mathrm{emit}} = 2 \times A_{\mathrm{abs}}$ then

$\frac{L_{\odot}}{4\pi r^2} = 2\sigma T^4 \quad \therefore \quad T = \sqrt[4]{\frac{L_{\odot}}{8\pi\sigma r^2}}$ [1]

$= \sqrt[4]{\frac{3.85\times10^{26}}{8\pi\times5.67\times10^{-8}\times(9.86\times6.96\times10^{8})^2}}$ [1]

$= 1550\ \mathrm{K}$ [1] [3]

b.

Using Kepler’s third law to find the semi-major axis of the final orbit,

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

$= \sqrt[3]{\frac{6.67\times10^{-11}\times1.99\times10^{30}}{4\pi^2}\times(88\times24\times3600)^2}$ [1]

$= 5.79\times10^{10}\ \mathrm{m}$ [1]

$v = \sqrt{GM\left(\frac{2}{r} - \frac{1}{a}\right)} = \sqrt{6.67\times10^{-11}\times1.99\times10^{30}\left(\frac{2}{9.86\times6.96\times10^{8}} - \frac{1}{5.79\times10^{10}}\right)}$ [1]

$(= 190766\ \mathrm{m\,s^{-1}}) = 191\ \mathrm{km\,s^{-1}}$ [Must be in km s$^{-1}$] [1] [4]

[This will mean that the Parker Solar Probe will become the fastest spacecraft (relative to the Sun) ever flown – at this speed you could travel from New York to Tokyo in less than a minute!]

c.

Using a similar method to the previous question to work out the semi-major axis and perihelion speed for the first orbit,

$a_{initial} = 8.27\times10^{10}\ \mathrm{m}$ [1]

$v_{initial} = 95.3\ \mathrm{km\,s^{-1}}$ [1]

The total energy is the sum of the potential and kinetic energies,

$E_{tot} = E_P + E_K$ and $E_P = -\frac{GMm}{r}$

At the first perihelion,

$E_{tot} = -\frac{6.67\times10^{-11}\times1.99\times10^{30}\times685}{35.7\times6.96\times10^{8}} + \frac{1}{2}\times685\times(95279)^2$

$= -3.66\times10^{12} + 3.11\times10^{12}$

$= -5.50\times10^{11}\ \mathrm{J}$ [Allow 1 mark if only have a correct $E_{\mathrm{P}}$ or $E_{\mathrm{K}}$ term] [2]

At the final perihelion,

$E_{tot} = -\frac{6.67\times10^{-11}\times1.99\times10^{30}\times685}{9.86\times6.96\times10^{8}} + \frac{1}{2}\times685\times(190766)^2$

$= -1.32\times10^{13} + 1.25\times10^{13}$

$= -7.85\times10^{11}\ \mathrm{J}$ [Allow 1 mark if only have a correct $E_{\mathrm{P}}$ or $E_{\mathrm{K}}$ term] [2]

$\therefore \Delta E_{tot} = 2.35\times10^{11}\ \mathrm{J}$ [Ignore a minus sign] [1] [7]

d.

In this diagram:

cx = cz = a

SP = r

cS = ae

cd = U

Pd = V

therefore Sd = cd – cs = U – ae

For an ellipse, the general equation is $\frac{U^2}{a^2} + \frac{V^2}{b^2} = 1$, so the points (U,V) on the ellipse are

$\cos E = \frac{U}{a}$ and $\sin E = \frac{V}{b}$ [1]

We are also given that $e = \sqrt{1 - \frac{b^2}{a^2}} \quad \therefore \quad b^2 = a^2(1 - e^2)$

Using Pythagoras’ theorem,

$(\boldsymbol{SP})^2 = (\boldsymbol{Pd})^2 + (\boldsymbol{Sd})^2$

$\therefore r^2 = (b\sin E)^2 + (a\cos E - ae)^2$ [1]

$= a^2(1 - e^2)\times(1 - \cos^2 E) + a^2(\cos^2 E - 2e\cos E + e^2)$ [1]

$= a^2 - 2a^2e\cos E + a^2e^2\cos^2 E$

$= a^2(1 - e\cos E)^2$

$\therefore r = a(1 - e\cos E)$ [1] [4]

[Third mark is for eliminating b and writing all expressions in terms of only one trigonometric function. A reasonable attempt at a derivation (allowing alternative methods) must be present to get the marks for this question (simply writing the answer only scores 1 mark)]

e.

With the perihelion distance we can find the eccentricity,

$r_{peri} = a(1 - e) \quad \therefore \quad e = 1 - \frac{r_{peri}}{a} = 1 - \frac{9.86\times6.96\times10^{8}}{5.79\times10^{10}} = 0.882$ [1]

Using the formula just derived we can find E for r = 0.25 au,

$E = \cos^{-1}\left(\frac{1}{e}\left(1 - \frac{r}{a}\right)\right) = \cos^{-1}\left(\frac{1}{0.882}\left(1 - \frac{0.25\times1.50\times10^{11}}{5.79\times10^{10}}\right)\right) = 1.16\ \mathrm{rad}\ (= 66.4^{\circ})$ [1]

Using the formula given we can find M,

$M = E - e\sin E = 1.16 - 0.882\times\sin 1.16 = 0.351\ \mathrm{rad}\ (= 20.1^{\circ})$ [1]

Allowing for both sides of the ellipse, $\Delta M = 0.703\ \mathrm{rad}\ (= 40.3^{\circ})$ [1]

Movement through the circular orbit has constant angular velocity, meaning

$\frac{\Delta M}{\Delta t} = \frac{2\pi}{T} \quad \therefore \Delta t = \frac{T\Delta M}{2\pi} = \frac{88\times0.703}{2\pi} = 9.84\ \mathrm{days}\ (= 9\ \mathrm{days}\ 20\ \mathrm{hours}\ 12\ \mathrm{mins})$ [1] [5]

[If they forget the factor of two (so $\Delta t$ = 4.92 days) allow 4 marks. Allow ±1 hour on the final answer to account for intermediate rounding errors. Allow full ecf for using their value of the semi-major axis from part b. Must be given in days for the final mark]