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IOAA 2019, Теория — Задача 12. Distance to a Near-Earth Asteroid

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Theory problem sheet · 25 т.

Условие

Assume that a Near-Earth Asteroid is observed by two astronomers, one from Nagykanizsa, Hungary and one from Windhoek, Namibia. The longitudes of the two cities are exactly 17° east of Greenwich. They observe the asteroid when it crosses their respective meridians. The Nagykanizsa-observer sees the asteroid 25° south of his zenith at this instant, while the Windhoek-observer sees it 45° north of his zenith at the same instant. The latitudes of the two cities are 46°27’ N and 22°34’ S, respectively. The sites of both astronomers are at sea level.

a) Draw a diagram of the geometric configuration. [5 т.]

b) What is the distance of this asteroid from the centre of the Earth, expressed in units of Earth-radii and the average Earth-Moon distance? Provide a solution which makes use of all available information. Neglect the effect of the atmospheric refraction. [20 т.]

Решение

a)

The geometric situation is the following: (5 p)

Geometry diagram: circle (Earth) with centre O, equator line, points N (Nagykanizsa) and W (Windhoek) on the surface with radii R and angles φ₁, φ₂ to the equator; zenith directions at N and W with zenith angles z₁ and z₂; asteroid A at top right connected by lines x₁, x₂ to N and W and D to O, with small angles p₁, p₂ and angle δ at O.

Fig. 12.1: $N$ : Nagykanizsa, $W$ : Windhoek, $O$ : centre of Earth, $A$ : asteroid, $D$ : distance from the centre of Earth to the asteroid $(D = \overline{OA})$ , $R$ : radius of the Earth

b)

One can recognize that the triangle $NOW$ is an isosceles triangle, two sides $(ON$ and $OW)$ are equal to each other in length:

(12.1)\quad ON = OW = R

(1 p)

Therefore the angles $\alpha = ONW\sphericalangle = OWN\sphericalangle$ . For the $NOW$ triangle we can write, that:

(12.2)\quad 2\alpha + \varphi_1 + \varphi_2 = 180^{\circ} \rightarrow \alpha = 90^{\circ} - \dfrac{\varphi_1 + \varphi_2}{2}

(1 p)

Let us denote the direct -- i.e. measured through the body of the Earth, not on the surface -- distance between Nagykanizsa and Windhoek by $d$ : $d = \overline{NW}$ . This distance can be obtained straightforward by the cosine theorem (one has to take care that the absolute values of the latitudes should be added inside the cosine function):

(12.3)\quad d = \sqrt{R^2 + R^2 - 2R^2\cos(\varphi_1 + \varphi_2)} = R\sqrt{2(1 - \cos(\varphi_1 + \varphi_2))}

(2 p)

With numerical values:

$d = 1.133\,05\,R$ (1 p)

Let us use the following notations:

(12.4)\quad x_1 = \overline{NA}, x_2 = \overline{WA}, p = p_1 + p_2

For the $NOWA$ quadrilateral we can write that:

(12.5)\quad \varphi_1 + \varphi_2 + (180^{\circ} - z_1) + p + (180^{\circ} - z_2) = 360^{\circ}

(1 p)

and therefore:

(12.6)\quad p = z_1 + z_2 - (\varphi_1 + \varphi_2)

(1 p)

For the $NWA$ triangle we can write that:

(12.7)\quad \dfrac{x_1}{d} = \dfrac{\sin(180^{\circ} - z_2 - \alpha)}{\sin p} \quad\text{and}\quad \dfrac{x_2}{d} = \dfrac{\sin(180^{\circ} - z_1 - \alpha)}{\sin p}

(2 p)

Thus:

(12.8)\quad x_1 = d\,\dfrac{\sin(z_2 + \alpha)}{\sin(z_1 + z_2 - (\varphi_1 + \varphi_2))} \quad\text{and}\quad x_2 = d\,\dfrac{\sin(z_1 + \alpha)}{\sin(z_1 + z_2 - (\varphi_1 + \varphi_2))}

(2 p)

$D$ can be obtained either from the triangle $NOA$ or $WOA$ by the cosine theorem:

(12.9)\quad D^2 = R^2 + x_1^2 - 2Rx_1\cos(180^{\circ} - z_1)

(12.10)\quad D^2 = R^2 + x_2^2 - 2Rx_2\cos(180^{\circ} - z_2)

(2 p)

Let us take the sum of them:

(12.11)\quad 2D^2 = 2R^2 + (x_1^2 + x_2^2) + 2R(x_1\cos z_1 + x_2\cos z_2),

(1 p)

where we have used that $\cos(180^{\circ} - x) = -\cos x$ .

(If someone determines $D$ only from one of the two $(NOA, WOA)$ triangles, 1 point can be deducted, because that means, we ignore the error bars of one of the measurements in a significant way.)

After substitution and dividing by two one gets:

(12.12)\quad D^2 = R^2 + d^2\,\dfrac{\sin^2(z_1 + \alpha) + \sin^2(z_2 + \alpha)}{2\sin^2(z_1 + z_2 - (\varphi_1 + \varphi_2))} + Rd\,\dfrac{\sin(z_2 + \alpha)\cos z_1 + \sin(z_1 + \alpha)\cos z_2}{\sin(z_1 + z_2 - (\varphi_1 + \varphi_2))}

(3 p)

This contains only known quantities. After substitution we get (notice that for alpha and $D$ we need the absolute values of the geographical latitudes):

$\alpha = 55.5^{\circ}$ (1 p)

$D = \boxed{65.8\,R} = \boxed{1.1\ d_{\mathrm{Moon}}}$ (2 p)

So, the asteroid is at 65.8 Earth-radii from the centre of the Earth. This is slightly more than the distance to the Moon.

Оригинал в Архива: TH-problem-final.pdf · официални решения: TH-solution-final.pdf · допълнителни данни: TH-anssheet.pdf