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USAAAO 2022, II кръг — Задача 8. The Curious Orbit of James Webb

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Всички материали от USAAAO 2022

2022 National Astronomy Competition · 45 т.

Instructions (Please Read Carefully)

The top 5 eligible scorers on the NAC will be invited to represent USA at the next IOAA. In order to qualify for the national team, you must be a high school student with US citizenship or permanent residency.

This exam consists of 3 parts: Short Questions, Medium Questions and Long Questions.

The maximum number of points is 200.

The test must be completed within 2.5 hours (150 minutes).

Please solve each problem on a blank piece of paper and mark the number of the problem at the top of the page. The contestant’s full name in capital letters should appear at the top of each solution page. If the contestant uses scratch papers, those should be labeled with the contestant’s name as well and marked as “scratch paper” at the top of the page. Scratch paper will not be graded. Partial credit will be available given that correct and legible work was displayed in the solution.

This is a written exam. Contestants can only use a scientific or graphing calculator for this exam. A table of physical constants will be provided. Discussing the problems with other people is strictly prohibited in any way until the end of the examination period on March 26th. Receiving any external help during the exam is strictly prohibited. This means that the only allowed items are: a calculator, the provided table of constants, a pencil (or pen), an eraser, blank sheets of papers, and the exam. No books or notes are allowed during the exam. Exam is proctored and recorded. You are expected to have your video on at all times.

Условие

For his upcoming Astrophysics Club presentation, Will researches the recently launched James Webb Space Telescope (JWST), the next-generation telescope designed as the successor of the Hubble Space Telescope. The largest space telescope ever built, the JWST uses its large collecting area to observe in the infrared spectrum. It orbits around the $L_2$ Lagrange point of the Earth-Sun system. Lagrange points are equilibrium points for a small body in the Earth-Sun system; $L_2$ is the point on the Earth-Sun line located beyond Earth’s orbit.

In the problem, let $M$ and $m$ be the mass of the Sun and Earth, respectively, with $M \gg m$ . Additionally, consider the Sun’s and Earth’s radius to be $R_{\odot}$ and $R_{\oplus}$ respectively, and the Earth to orbit the Sun in a perfectly circular orbit of radius $R$ .

(a) The orbit of JWST was designed to circle around $L_2$ in a big enough orbit to avoid Earth’s shadow. What is the benefit of i) being at a Lagrangian point and ii) avoiding Earth’s shadow? [2 т.]

(b) Taking first order approximations, about how far is $L_2$ from Earth? Express your answer both in terms of the variables defined and numerically, in km. [5 т.]

(c) In the rotating reference frame in which the Earth and the Sun are stationary, JWST orbits $L_2$ in the plane perpendicular to the Earth-Sun line that passes through $L_2$ . If JWST orbits in a circle of radius $r$ around $L_2$ in this frame, what is the minimum $r$ that avoids the Earth’s shadow at all times? Express your answer both in terms of the variables defined and numerically, in km. [5 т.]

(d) Consider a scenario where JWST is stationary in the aforementioned rotating reference frame and has a small displacement $\boldsymbol{\delta r} = \delta x \hat{\boldsymbol{\imath}} + \delta y \hat{\boldsymbol{\jmath}}$ relative to $L_2$ , where $\hat{\boldsymbol{\imath}}$ is the unit vector along the Earth-Sun line away from the Sun and $\hat{\boldsymbol{\jmath}}$ is a unit vector perpendicular to $\hat{\boldsymbol{\imath}}$ . Both $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ are stationary in the rotating frame. To first order (i.e. assuming $|\boldsymbol{\delta r}| \ll x$ ), what is the acceleration of JWST in the rotating frame? [20 т.]

(e) The presence of the Coriolis force in the rotating reference frame destabilizes orbits around $L_1$ , $L_2$ , and $L_3$ while stabilizing orbits around $L_4$ and $L_5$ . Disregarding the Coriolis force for this part only, are orbits stable around $L_2$ when there is no Coriolis force? Is this result generalizable? In other words, what can be said about the stability of orbits around an arbitrary, stationary point where there are no masses within the orbit and no fictitious forces involved? [5 т.]

(f) Suppose JWST orbits in the circle described in part (c) with a constant speed and an orbital radius of 500,000 km. Suppose further that the jet propulsion of the JWST is programmed to counteract only the Coriolis force; the rest of JWST’s motion is due to the natural gravitational dynamics at $L_2$ . Using the assumption that the first order expression derived in (d) still applies, if JWST has a mass of 6500 kg, what is the average magnitude of the force over a long period of time? The following averages (calculated from 0 to $2\pi$ ) might be helpful:

\overline{|\sin x|} = \frac{2}{\pi} \qquad \overline{\sin^2 x} = \frac{1}{2} \qquad \overline{|\sin^3 x|} = \frac{4}{3\pi}

For reference, the magnitude of the Coriolis force is given as,

|\boldsymbol{F}| = 2m|\boldsymbol{\omega} \times \boldsymbol{v}|

[8 т.]

Отговори

Решение

(a) i) A Lagrangian point is stationary with respect to the Earth and the Sun, which is beneficial for a space telescope because it requires little fuel to maintain its position.

ii) Avoiding Earth’s shadow is beneficial for JWST because it always has access to the Sun’s energy, never to be interrupted by eclipses.

(b) This is a classic problem. $L_2$ is the Lagrangian point just beyond Earth’s orbit. Let us define the $L_2$ -Earth distance to be $x$ , where $x \ll R$ . Taking into account the forces of the Earth and the Sun yields:

\frac{GM}{(R + x)^2} + \frac{Gm}{x^2} = \omega^2 (R + x)

where $\omega = \sqrt{GM/R^3}$ is the rotational velocity of Earth’s orbit. Taking advantage of the binomial approximation $(1 + y)^{\alpha} \approx 1 + \alpha y$ when $y \ll 1$ , we can simplify our expression to the following:

\frac{GM}{R^2}\left(1 - \frac{2x}{R}\right) + \frac{Gm}{x^2} \approx \frac{GM}{R^2}\left(1 + \frac{x}{R}\right)

Solving the equation yields

\boxed{x = R\sqrt[3]{\frac{m}{3M}} = 1.50 \times 10^6\ \mathrm{km}}

(c) From geometry, JWST must lie outside of the cone that is defined by the common internal tangents of the Earth and the Sun. The tip of the cone has an angle

2\theta = 2\sin^{-1}\left(\frac{R_{\odot} + R_{\oplus}}{R}\right)

If you extend one of those internal tangents as follows (insert diagram), one can see that $r_{\mathrm{min}}$ satisfies the following equation:

\tan\theta = \frac{R_{\odot}/\cos\theta + r}{R + x}

Therefore:

\boxed{r_{\mathrm{min}} = \frac{(R + x)R_{\oplus} + xR_{\odot}}{\sqrt{R^2 - (R_{\odot} + R_{\oplus})^2}} = 1.34 \times 10^4\ \mathrm{km}}

Note that realistically, $R_{\oplus}, R_{\odot} \ll R$ and $x \ll R$ , so $r_{\mathrm{min}} \approx R_{\oplus} + xR_{\odot}/R$ , which is off by less than 1%.

(d) Because $L_2$ is a Lagrangian point, we know that in the rotating frame, the forces at $L_2$ balance. Therefore, we only need to find first order changes in force relative to $L_2$ . Suppose a mass $M$ is located at a distance $R$ to the left of $L_2$ . Then the change in gravitational field from $L_2$ to $L_2 + \boldsymbol{\delta r}$ is

\boldsymbol{\delta g} = -\frac{GM}{((R + \delta x)^2 + \delta y^2)^{3/2}}((R + x)\hat{\boldsymbol{\imath}} + \delta y \hat{\boldsymbol{\jmath}}) + \frac{GM}{R^2}\hat{\boldsymbol{\imath}}

The $\delta y^2$ in the denominator of the first term is of second order and can be discarded:

\boldsymbol{\delta g} \approx -\left(\frac{GM}{(R + \delta x)^3}(R + x) - \frac{GM}{R^2}\right)\hat{\boldsymbol{\imath}} - \frac{GM}{R^3}\delta y \hat{\boldsymbol{\jmath}}

The binomial approximation can be used once again:

\boldsymbol{\delta g} \approx \frac{GM}{R^2}\left(1 - \frac{2x}{R} - 1\right)\hat{\boldsymbol{\imath}} - \frac{GM}{R^3}\delta y \hat{\boldsymbol{\jmath}} = \frac{GM}{R^3}(2\delta x \hat{\boldsymbol{\imath}} - \delta y \hat{\boldsymbol{\jmath}})

You might notice that this expression looks similar to that of tidal forces, because what we’ve done here is exactly the same! Now, to find the acceleration of JWST, we simply need to add the $\boldsymbol{\delta g}$ of the Earth and the Sun, then add $\boldsymbol{\delta a}$ of the centrifugal force on top (the coriolis force is 0 since JWST is stationary). $\boldsymbol{\delta a}$ due to the centrifugal force is simply:

\boldsymbol{\delta a} = \delta(\omega^2 \boldsymbol{r}) = \omega^2 \boldsymbol{\delta r} = \frac{GM}{R^3}(\delta x \hat{\boldsymbol{\imath}} + \delta y \hat{\boldsymbol{\jmath}})

Therefore, the acceleration of JWST is

\boldsymbol{a} = \left(\frac{GM}{(R + x)^3} + \frac{Gm}{x^3}\right)(2\delta x \hat{\boldsymbol{\imath}} - \delta y \hat{\boldsymbol{\jmath}}) + \frac{GM}{R^3}(\delta x \hat{\boldsymbol{\imath}} + \delta y \hat{\boldsymbol{\jmath}})

Plugging in $x = R\sqrt[3]{m/3M}$ and sticking to first order in $\boldsymbol{\delta r}$ yields

\begin{aligned} \boldsymbol{a} &\approx \left(2\left(\frac{GM}{R^3} + \frac{3GM}{R^3}\right) + \frac{GM}{R^3}\right)\delta x \hat{\boldsymbol{\imath}} + \left(-\left(\frac{GM}{R^3} + \frac{3GM}{R^3}\right) + \frac{GM}{R^3}\right)\delta y \hat{\boldsymbol{\jmath}} \\ &= \boxed{\frac{GM}{R^3}(9\delta x \hat{\boldsymbol{\imath}} - 3\delta y \hat{\boldsymbol{\jmath}})} \end{aligned}

(e) From our expression in part (d), we see that a small displacement in the $\hat{\boldsymbol{\imath}}$ direction results in a force that points in the same direction, meaning any orbits are $\boxed{\text{unstable}}$ in that direction.

In fact, this result is general for all orbits around arbitrary points with no mass and no fictitious forces. Using Gauss’s law for gravitation, the flux through a surface surrounding such a point must be 0, indicating it is impossible for the gravitational force to be pointing inward for displacements in every possible direction.

(f) The acceleration in the direction perpendicular to $\hat{\boldsymbol{\imath}}$ was found in part (d) to be $-3GM\delta y/R^3$ . Setting this equal to the centripetal acceleration, we have

\frac{3GMr}{R^3} = \frac{v^2}{r}

v = r\sqrt{\frac{3GM}{R^3}}

The magnitude of the Coriolis force is

|\boldsymbol{F}| = 2m|\boldsymbol{\omega} \times \boldsymbol{v}| = 2m\sqrt{\frac{GM}{R^3}} r \sqrt{\frac{3GM}{R^3}} |\sin\theta|

where $\theta$ is the angle between $\boldsymbol{v}$ and $\boldsymbol{\omega}$ . Since it varies from 0 to $2\pi$ , we use the given value of $\overline{|\sin\theta|}$ to find the average magnitude to be

\boxed{|\boldsymbol{F}| = \frac{4\sqrt{3}GMmr}{\pi R^3} = 0.28\ \mathrm{N}}

Note just how small the force is!

Оригинал в Архива: usaaao_second_exam_2022_problems.pdf · официални решения: usaaao_second_exam_2022_solutions.pdf