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OPhO 2020, Теория — Задача 2. Solar Sails

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Всички материали от OPhO 2020

2020 Online Physics Olympiad: Invitational Contest v1.4 · 26 т.

Внимание: Бележка към темата

The v1.4 question paper, native Partial Solutions and separately attributed organizer clarifications are retained. The key is incomplete: Problem 1 is only marked TBD, Problem 8 has no key for parts f–h or an explicit critical power in e, and Problem 10 has no key for Part B f–h. Problems 2 and 9 also have source gaps noted below. Printed mathematical defects and differences between repeated questions in the key and the question original are not silently repaired. Problem2(e) gives a time-dependent spin result but no explicit answer in orbital angle; key2(c) refers to an absent diagram. Problem9(a) gives its ODE and initial conditions without an explicit speed, and 9(b) does not state whether the pressure increases or decreases. Native source typography is retained, including the Planck-constant value, missing electron-charge unit, missing kappa in key3(a), mismatched labels in key4, inconsistent theta indices and ring factors in key6, the negative probability/energy signs and r_B/K_B/k variants in key7, radiation expressions and fusion-rate/proton-rate difference in key8, and image/flux signs and field factors in key10. All these are printed-source issues, not repaired answers. Question2 omits the word define while the repeated key question includes it; the repeated key question7 adds For simplicity, ignore spin. The 4-page organizer clarification contradicts question1(d)’s frictionless table and also contains both only one final submission and multiple submissions instructions. The native original and the separately attributed clarifications remain visible.

Original credits

2020 Online Physics Olympiad (OPhO): Invitational Contest v1.4

OPhO Lead Organizers and Writers: Adithya Balachandran, Eddie Chen, Ashmit Dutta, Ethan Hu, Viraj Jayam, Rishab Parthasarathy, Kushal Thaman, Max Wang, Qilin Xue

OPhO Website Developer: Ben Oommen

Advisors: Dr. Branislav Kisačanin, Kye Shi, Kevin Zhou

July 3 - 6, 2020

Указания

If you wish to request a clarification, please use this form. To see all clarifications, see this document.

This test contains 10 long answer questions.
The total base score for the exam is 300 points; a factor incorporating the number of teams who solved the question will be added in the marking scheme. The final scores of all the teams will be available a few days after the contest ends.
The team leader should submit their final solution document in this here. We don’t anticipate a tie, but in the rare circumstance that there is one, the time you submit will be used to break it.
You can write on paper, type it up online, or a mix of both. If you wish to use a pre-made LaTeX template, we made one which you can choose to use here.
Since this is a long answer response, you will be judged on the quality of your work. To receive full points, you need to show your work, including deriving equations. As a general rule of thumb, any common equations (such as the ones in this formula sheet) can be cited without proof. Remember to state any approximations made, which system of equations were solved after every step. Explicitly showing every step of algebra is not necessary.
You may leave all final answers in symbolic form (in terms of variables) unless otherwise specified.2
Be sure to state all assumptions.

Sponsors

WOLFRAM · AwesomeMath (making x,y,z as easy as a,b,c) · Physics Olympiad · AoPS.com Art of Problem Solving

List of Constants

Proton mass, $m_p=1.67\cdot10^{-27}\ \mathrm{kg}$
Neutron mass, $m_n=1.67\cdot10^{-27}\ \mathrm{kg}$
Electron mass, $m_e=9.11\cdot10^{-31}\ \mathrm{kg}$
Avogadro’s constant, $N_0=6.02\cdot10^{23}\ \mathrm{mol^{-1}}$
Universal gas constant, $R=8.31\ \mathrm{J/(mol\cdot K)}$
Boltzmann’s constant, $k_B=1.38\cdot10^{-23}\ \mathrm{J/K}$
Electron charge magnitude, $e=1.60\cdot10^{-19}$
1 electron volt, $1\ \mathrm{eV}=1.60\cdot10^{-19}\ \mathrm J$
Speed of light, $c=3.00\cdot10^8\ \mathrm{m/s}$
Universal Gravitational constant, $G=6.67\cdot10^{-11}\ \mathrm{(N\cdot m^2)/kg^2}$
Acceleration due to gravity, $g=9.81\ \mathrm{m/s^2}$
1 unified atomic mass unit, $1\ \mathrm u=1.66\cdot10^{-27}\ \mathrm{kg}=931\ \mathrm{MeV}/c^2$
Planck’s constant, $h=6.63\cdot10^{-34}\ \mathrm{J\cdot s}=4.41\cdot10^{-15}\ \mathrm{eV\cdot s}$
Permittivity of free space, $\epsilon_0=8.85\cdot10^{-12}\ \mathrm{C^2/(N\cdot m^2)}$
Coulomb’s law constant, $k=\frac1{4\pi\epsilon_0}=8.99\cdot10^9\ \mathrm{(N\cdot m^2)/C^2}$
Permeability of free space, $\mu_0=4\pi\cdot10^{-7}\ \mathrm{T\cdot m/A}$
Magnetic constant, $\frac{\mu_0}{4\pi}=1\cdot10^{-7}\ \mathrm{(T\cdot m)/A}$
1 atmospheric pressure, $1\ \mathrm{atm}=1.01\cdot10^5\ \mathrm{N/m^2}=1.01\cdot10^5\ \mathrm{Pa}$
Wien’s displacement constant, $b=2.9\cdot10^{-3}\ \mathrm{m\cdot K}$
Stefan-Boltzmann constant, $\sigma=5.67\cdot10^{-8}\ \mathrm{W/m^2/K^4}$

Native key title

2020 Online Physics Olympiad (OPhO): Invitational Contest Partial Solutions

The same organizer, writer, developer and advisor credits appear on this key cover.

Organizer clarification document — major notes

Separately quoted from the organizer clarification document, titled 2020 OPhO Invitational Clarifications:

MAJOR NOTES

Hi everyone! All team members are welcome to fill out this (optional) feedback form. Everyone who submits (and consents) will be entered in a draw to receive a $25 Amazon gift card.
Use this link to submit clarification requests. The problem writers may or may not deem it necessary to answer all clarification requests.
Link to the overleaf template (the link has been fixed) for answer submission.
Reminder: There is only one final submission at the end of the competition via the submission form only. Do NOT email us a copy of your submission. In case of internet connectivity issues, any team member may email us the file.
Make sure to read over the instructions. You are expected to give symbolic answers unless explicitly otherwise stated. If the numerical values for a symbol are stated, you do not need to solve the problem for a general case.
Uploading a .jpeg on the overleaf for a graph is allowed.
Please check the newest version v1.4 for revisions.
Please provide clear reasoning or derivations of any obscure results. Directly citing them would not necessarily fetch full points.
We will be awarding a special prize for whichever team produces the most creative and/or elegant solution to certain problems. We will not be disclosing which problems until after the competition has concluded.

Note on submission of solutions:

It is highly recommended that the solutions to the invitational round be submitted in LaTeXed form. We recommend using Overleaf. It is preferred that the solutions are quite detailed; this means that while you do not need to explicitly write all your algebra, all the relevant equations must be clearly written. A brief wording on how you solve a given system of equations is also optimal. Clearly state any approximations you make. Solutions typed in .doc/.docx, and ones neatly written by hand will also be accepted. Some students have a handwriting that’s hard to read; if that’s you, please read this. You can submit solutions multiple times. Only the latest submission will count.

Условие

Much research is being done on the possibility of using solar sails to reach far away reaches in out galaxy. This is a method of propulsion that use light from the sun to exert a pressure on what is usually a large mirror. The sails themselves are often made of a thin reflective film. Compared to traditional spacecraft, while these sails have very limited payloads, they offer long operational lifetimes and are relatively low cost. The most significant advantage is speed: since solar sails do not depend on onboard propellant, they can travel much faster than a standard rocket, with the possibility of reaching a significant percentage of the speed of light. For all parts, assume the solar sails are only under the influence of gravity from the sun only. In addition, you may necessary variables such as the mass and power output of the sun.

a) (2 pt) Assume the solar sails are not revolving around the sun. Assume they are very reflective thin discs and that all of the mass is in the sail. What is the maximum area density so that the sail does not fall towards the sun? Does the distance matter (find the general density to distance equation if it does matter)?

b) (3 pts) Assume the solar sails are a thin spherical shell and made of a perfectly absorbent material instead. The area density is exactly half the maximum area density so that it does not fall towards the sun. What is its final speed if it starts at a stationary position 1 Au from the sun?

c) (5 pts) Assume the solar sails are a thin spherical shell and made of material having reflectance $r$ . With the same area density as in part (b), what is the final speed (it can reach this speed either when crashing into the sun, escaping the solar system, or remain in orbit)? This time, assume it starts in circular orbit 1 Au away from the sun as the starting condition. What is the orbital shape?

d) (6 pts) Simply not falling into the sun is insufficient for solar sails. Most are planned to reach relativistic speeds. A way to achieve this is to fire large Earthbound lasers at the sail. How powerful must the lasers be to accelerate the sail up to $v=0.2c$ in 50 days? Since this far exceeds the value obtained in part 3, neglect the effects of the sun.

e) (10 pts) Suppose a perfectly reflecting solar sail in the shape of a thin disk (with mass $m$ and radius $r$ ) orbiting around the star in a circular orbit of radius $d\gg r$ . It is also spinning around itself, such that the spin angular velocity is in the same direction as the orbital angular velocity, and its axis of symmetry always remains parallel to the plane of the orbit. If its initial spin velocity is $\omega_0$ find its spin velocity after it revolves an angle $\theta$ around the star.

Published organizer clarifications (separate source)

Separately quoted from the organizer clarification document:

Problem 2: Feel free to define any necessary variables such as power output of the sun, mass of the sun, etc.

Problem 2: b) Spherical shell means spherical full shell (hollow sphere), half the density refers to that as calculated from b
Problem 2: c) The initial conditions are picked such that if there were no radiation effects, it would be in a perfect circular orbit. The more accurately you can describe the orbit, the better. (mathematical representation would be better than qualitative)
Problem 2: d) use the sail from part c).
Problem 2: e) assume solar sail maintains circular orbit around the sun; its spin angular speed is much greater than the orbital angular speed; express your answer in terms of sail radius r, sail mass m, orbital radius d, and fundamental constants. ( $d\gg r$ )

Решение

Внимание: Непълно решение

The complete available printed key is retained. Part e gives a time-dependent spin result but does not explicitly express the requested spin in orbital angle theta; part c refers to a following diagram absent from the original key.

Solution 2:

(a) Let the sail have an area of $A$ and area density $\sigma$ and the sun have mass $M$ . The gravitational force acting on this sail is:

F_G=\frac{GM(A\sigma)}{R^2}

The photons emitted from the sun have a momentum of $\frac Ec$ . On the reflective sail, the change in momentum is:

\frac{2E}{c}

The force imparted from this change is:

F=\frac{dp}{dt}=\frac2c\frac{dE}{dt}

Now taking the projection area of the disk which is $A$ in this case, we can find the fraction of the sun’s power (denoted by $P$ ) reaching the sail:

\frac{PA}{4\pi R^2}

So the force from the photons emitted by the sun is

F_P=\frac{PA}{2\pi cR^2}

Now force balancing and solving algebraically we get,

\sigma=\frac P{2\pi GMc}

(b) The change in momentum for the photons is reduced by a factor of two because of the new perfectly absorbing sails:

F=\frac1c\frac{dE}{dt}

If we let the spherical shell have a surface area of $A$ , the projection area of the spherical shell is $\frac14A$ . Using the same analysis in part a we find that

\sigma_{\max}=\frac{PA}{16\pi cR^2}

and

F_P=\frac{PA}{16\pi cR^2}

Now using $\frac12\sigma_{\max}$ for our force balance equation we get,

\begin{aligned}(A\sigma)\ddot r&=F_P-F_G\\(A\sigma)\ddot r&=-\frac{PA}{32\pi cR^2}\end{aligned}

We find that the acceleration is $\frac{GM}{R^2}$ which means the magnitude of the gravitational potential field is $\frac{GM}{R}$ . Now using conservation of energy (the initial kinetic energy and the final potential energy are zero):

\frac12mv_f^2=\frac{GMm}{R}

v_f=\sqrt{\frac{2GM}{R}}

(c) The absorbance is $1-r$ , so the force contribution from absorbed photons is:

F_{P1}=(1-r)\frac{PA}{4\pi cR^2}

where $A$ is the area of projection of the sphere. For reflected photons, we only consider the change of momentum in the same direction of the original travelling direction since any impulse in the perpendicular directions cancel due to spherical symmetry. Incoming photons that are reflected at an incident angle $\theta$ have a change in momentum of

\frac Ec(1+\cos(2\theta))

in the direction of travel. This is better illustrated in the following diagram:

It is important to note that at different angles there are varying frequencies of photons reflecting off at that angle due to a varying area of projection. It is easy to show that the area of projection at an angle $\theta$ is

2\pi r^2\sin(\theta)\cos(\theta)d\theta=A\sin(2\theta)d\theta.

Taking the sum over all contributions from rings at all angles, we get the expression

F_{P2}=r\frac{PA}{4\pi cR^2}\int_0^{\pi/2}(1+2\cos(2\theta))(\sin(2\theta))d\theta

After evaluating, we get that

F_{P2}=r\frac{PA}{4\pi cR^2}

We can conclude that the force is independent of the reflectance value $r$ because the force from absorbance and reflectance contributions sum to $\frac{PA}{4\pi cR^2}$ .

Now to solve for $v$ , we can use conservation of energy (similar to part b, except this time we have an initial kinetic energy as well) to get:

\begin{aligned}\frac12mv_f^2&=\frac{GMm}{R}+\frac12mv_o^2\\v_o^2&=\frac{GM}{R}\\v_f&=\sqrt{\frac{3GM}{R}}\end{aligned}

It is well known that the eccentricity of an orbit in an inverse square relationship with two masses is:

e=\sqrt{1+\frac{2L^2E}{m_rk^2}}

With $E$ being the total energy, $L$ the angular momentum, $m_r$ the reduced mass, and $k$ a constant. Because the energy is positive, the eccentricity is greater than one. Therefore it is a hyperbolic orbit. (Note: points for describing the orbit were given to teams that went through the complete derivation of the eccentricity or at least provided some sort of mathematical analysis)

(d) We need to take into account both relativistic effects as well as the doppler effect. The force of radiation is given by

F_{\mathrm{rad}}=\frac{2P}{c}\sqrt{\frac{1-\beta}{1+\beta}}

which gives rise to the net force

\gamma^3mc\frac{d\beta}{dt}=\frac{2P}{c}\sqrt{\frac{1-\beta}{1+\beta}}.

Solving this differential equation gives us

P=\frac{0.113mc^2}{t}.

(e) Assume that the intensity of the solar radiation at the solar sail is $I$ . Let the solar sail have mass $m$ uniformly distributed across a disc of radius $r$ , and be perfectly reflecting. Suppose it is rotating with angular speed $\omega$ , about an axis on its plane (so by perpendicular axis theorem, it has moment of inertia $\frac14mr^2$ about that axis). We break the solar sail into many tiny pieces of area $dA$ and analyze the momentum transferred to each piece by the solar radiation.

Suppose a piece of area $dA$ has its normal oriented an angle $\theta$ from the oncoming radiation, and it is moving a speed $v$ directly along its normal. Consider what happens after a small time $dt$ . Assume that the total momentum of the photons it intercepts is $dp$ . The total momentum imparted to the area by those photons can be calculated by doppler shift, resulting in $2(\cos\theta+\frac vc)dp$ . Next, drawing a picture will show that the total momentum of the photons it intercepts is $(1+\frac v{c\cos\theta})\frac Ic\cos\theta dAdt$ .

Thus, we know $2(\cos\theta+\frac vc)(1+\frac v{c\cos\theta})\frac Ic\cos\theta dAdt$ gives the total momentum imparted to the piece of area by the radiation.

Now, we integrate all those small pieces of area of find the total angular impulse in time $dt$ . Let $\alpha$ be the angular position of the small piece of area with respect to the axis of rotation of the solar sail. Then we get

\begin{aligned}\tau={}&\int_0^\pi2\left(\cos\theta+\frac{\omega r\sin\alpha}{c}\right)\left(1+\frac{\omega r\sin\alpha}{c\cos\theta}\right)\frac Ic\cos\theta*2(r\cos\alpha)^2r\sin\alpha d\alpha-\\&\int_0^\pi2\left(\cos\theta-\frac{\omega r\sin\alpha}{c}\right)\left(1-\frac{\omega r\sin\alpha}{c\cos\theta}\right)\frac Ic\cos\theta*2(r\cos\alpha)^2r\sin\alpha d\alpha\end{aligned}

where $\tau$ is the angular impulse delivered per unit time. Note that the first integral represents the side of the disc that moves toward the sun, while the second integral represents the side of the disc that moves away from the sun. Also, notice we replaced $v$ with $\omega r\sin\alpha$ in the first integral and $v$ with $-\omega r\sin\alpha$ in the second integral. We also included the total area of the small pieces that were a distance $r\sin\alpha$ from the axis of rotation, which is $2(r\cos\alpha)^2d\alpha$ , and the lever arm is $r\sin\alpha$ . After simplification, we get

\tau=16\frac I{c^2}\omega r^4\cos\theta\int_0^\pi\cos^2\alpha\sin^2\alpha d\alpha

Finally, we get the average value of $|\tau|$ over $\theta$ is $\overline\tau=\frac4{c^2}I\omega r^4=-\frac14mr^2\dot\omega$ , since the torque is against the direction of rotation. Solving the differential equation gives $\omega(t)=\omega_0e^{-\frac{16Ir^2}{mc^2}t}$ .

Organizer clarification document — other notes

Separately quoted from the organizer clarification document:

Other Notes

Uploading images: View this link to find how to upload images on overleaf.
You are allowed to mix LaTeX and handwritten PDFs for your answer sheet. Note that you need to write your solution to the problem alongside your final answers.
You can use basic formulae related to topics on the IPhO syllabus without derivation but all other formulae which go beyond that must be derived.
If you find that space is running out on your LaTeX file, try to use $\verb|\newpage|$ and see if that fixes it.

Оригинал в Архива: invitational.pdf · официални решения: invitational_sol.pdf · допълнителни данни: Clarifications.pdf