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OWAO 2022, Теория — Задача 7. Polar Wi-Fi

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ΩWAO 2022 ◇ Theoretical Round ◇ Problems

Условие

A satellite is moving in a polar geocentric orbit with semi-major axis $a = 15\,400$ km, eccentricity $e = 0.55$ , and argument of pericenter $\omega = 270^{\circ}$ . Imagine that an observer is at the North Pole of the Earth. The satellite emits a signal with a frequency of 2.4 GHz. At some point in time, the satellite is observed on the horizon.

Estimate

a) the distance from the observer to the satellite,

b) the total velocity of the satellite at that moment,

c) the shift in the observed signal frequency.

Отговори

Решение

Elliptical orbit with apocentre α at the top and pericentre π at the bottom; the Earth (circle with centre C, radius R⊕) sits near the lower focus with the observer P at its top. The horizontal line through P meets the orbit at M and N. From M the radius r goes to C; angles τ and θ̃ are marked at M, the velocity vector V points down along the orbit; angles ν̃ and ν are marked at C.

a) The observer is at point $P$ . We should determine the geocentric distance $r \equiv MC$ .

\begin{cases} R_\oplus = r\cos\tilde{\nu} = -r\cos\nu, \\ r = \dfrac{a(1-e^2)}{1+e\cos\nu} \end{cases} \quad\Rightarrow\quad r = \frac{a(1-e^2)r}{r - eR_\oplus} \quad\Rightarrow\quad r = a(1-e^2) + eR_\oplus;

r = 15\,400\ \mathrm{km}\cdot(1 - 0.55^2) + 6\,371\ \mathrm{km}\cdot 0.55 = 14\,246\ \mathrm{km}.

Therefore, the distance from the observer to the satellite is

MP = \sqrt{r^2 - R_\oplus^2} = \sqrt{(14\,246\ \mathrm{km})^2 - (6\,371\ \mathrm{km})^2} = 12\,742\ \mathrm{km}.

b) Let us write the vis-viva equation. Here $V$ is the total velocity of the satellite:

V^2 = GM_\oplus\cdot\left(\frac{2}{r} - \frac{1}{a}\right) = gR_\oplus^2\cdot\left(\frac{2}{r} - \frac{1}{a}\right); \qquad (2)

V = \sqrt{9.81\ \mathrm{m/s}\cdot 6\,371\ \mathrm{km}\cdot\left(\frac{2}{1.4246\cdot 10^7\ \mathrm{m}} - \frac{1}{1.5400\cdot 10^7\ \mathrm{m}}\right)} = 5.48\ \mathrm{km/s}.

c) First, we should determine the line-of-sight velocity of the satellite. Let us find the angle between the direction of the velocity vector of the satellite and the observer’s line of sight.

Let $\theta$ be the angle between the velocity vector and the radius vector. In the figure, the symbol $\tilde{\theta} \equiv 180^{\circ} - \theta$ . We may write the conservation of angular momentum law for the current point and the pericenter:

V^2r^2\sin^2\theta = V_\pi^2r_\pi^2;

V^2r^2\sin^2\theta = \frac{GM_\oplus}{a}\cdot\frac{1+e}{1-e}\cdot a^2(1-e)^2 = GM_\oplus a(1-e^2). \qquad (3)

Next, we substitute the expression for the velocity (2) into the formula (3) above:

gR_\oplus^2\cdot\left(\frac{2}{r} - \frac{1}{a}\right)\cdot r^2\sin^2\theta = gR_\oplus^2\cdot a(1-e^2) \quad\Rightarrow\quad \sin^2\theta = \frac{a(1-e^2)}{r^2\left(\frac{2}{r} - \frac{1}{a}\right)}.

In our case

\sin^2\theta = \frac{1.5400\cdot 10^7\ \mathrm{m}\cdot(1 - 0.55^2)}{(1.4246\cdot 10^7\ \mathrm{m})^2\cdot\left(\frac{2}{1.4246\cdot 10^7\ \mathrm{m}} - \frac{1}{1.5400\cdot 10^7\ \mathrm{m}}\right)} = 0.701 \quad\Rightarrow\quad \theta = 123.1^{\circ}.

Next, we estimate the angle $\tau$ between the radius vector of the satellite and the line of sight.

\begin{cases} \tau = 90^{\circ} - \tilde{\nu}, \\ \tilde{\nu} = \arccos\dfrac{R_\oplus}{r} = \arccos\dfrac{6\,371\ \mathrm{km}}{14\,246\ \mathrm{km}} = 63.4^{\circ} \end{cases} \quad\Rightarrow\quad \tau = 90^{\circ} - 63.4^{\circ} = 26.6^{\circ}.

The angle between the direction of the velocity vector of the satellite and the observer’s line of sight $\psi$ is $\theta + \tau = (180^{\circ} - 123.1^{\circ}) + 26.6^{\circ} = 83.5^{\circ}$ , or $180^{\circ} - 83.5^{\circ} = 96.5^{\circ}$ in the symmetrical case. Finally, the radial velocity of the satellite

V_r = V\cos\psi = 5.48\ \mathrm{km/s}\cdot\pm\cos 83.5^{\circ} = \pm 0.62\ \mathrm{km/s}.

The shift in the observed frequency due to radial motion of the satellite is

\Delta f = \frac{V_r}{c}f = 2.4\cdot 10^9\ \mathrm{Hz}\cdot\frac{\pm 0.62\ \mathrm{km/s}}{3\cdot 10^5\ \mathrm{km/s}} \approx \pm 5\ \mathrm{kHz}.

Marking Scheme:

Correct orbital diagram is given (aphelion at zenith) — 2 pt.
Distance:
- True anomaly — 1 pt,
- Geocentric distance — 1 pt,
- Distance from the observer to the satellite — 1 pt.
Total velocity:
- Correct formulae (vis-viva equation or similar) — 1 pt,
- Correct answer — 1 pt.
Shift in observed frequency:
- Correct formula and value of radial velocity — 1 pt,
- Doppler shift — 1 pt,
- Two possible cases of observation (rise/set) — 1 pt.

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	$H_0 = 70\ \mathrm{(km/s)/Mpc}$
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	$M_\odot = 4.74^{\mathrm{m}}$
Effective temperature	$T_\odot = 5.8\cdot 10^3\ \mathrm{K}$

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

Оригинал в Архива: OWAO_2022_Theoretical_Problems.pdf · официални решения: OWAO_2022_Theoretical_Solutions.pdf