Съдържание

OPhO 2020, Открит кръг — Задача 55. When Rocket Scientists Play Catch

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от OPhO 2020

2020 Online Physics Olympiad (OPhO): Open Contest v1.5 · 25 май 2020 г.

2020 Online Physics Olympiad (OPhO): Open Contest v1.5

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

Указания

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

Use $g = 9.81\ \mathrm{m/s^2}$ in this contest. See the constants sheet on the following page for other constants.
This test contains 55 short answer questions. Each problem will have three possible attempts.
The weight of each question depends on our scoring system found here. Put simply, the later questions are worth more, and the overall amount of points from a certain question decreases with the number of attempts and days that you take to solve a problem as well as the number of teams who solve it. This means that your score decreases with the number of tries and days you take to solve a given problem.
Any team member is able to submit an attempt. Choosing to split up the work or doing each problem together is up to you. Note that after you have submitted an attempt, your teammates must refresh their page before they are able to see it.
Answers should contain at least three significant figures, unless otherwise specified. All answers within the 1% range will be accepted.
When submitting a response using scientific notation, please use exponential form. In other words, if your answer to a problem is $A \times 10^B$ , please type $AeB$ into the submission portal.
A standard scientific or graphing handheld calculator may be used. Technology and computer algebra systems like Wolfram Alpha or the one in the TI nSpire will not be needed. Attempts to use these tools will be classified as cheating.
You are allowed to use Wikipedia or books in this exam. Asking for help on online forums or your teachers will be considered cheating and may result in a possible ban from future competitions.
Top scorers from this contest will qualify to compete in the Online Physics Olympiad Invitational Contest, which is an olympiad-style exam. More information will be provided to invitational qualifiers after the end of the Open Contest.
In general, answer in SI units (meter, second, kilogram, watt) unless otherwise specified. Please answer all questions in degrees unless otherwise specified.
If the question asks to give your answer as a percent and your answer comes out to be “ $x\%$ ”, please input the value $x$ into the submission form.
Do not put letters in your answer on the submission portal! If your answer is “ $x$ meters” , input only the value $x$ into the submission portal.
Do not communicate information to anyone else apart from your team-members before the exam ends on May 29, 2020 at 11:59 PM UTC.

List of Constants

Proton mass, $m_p = 1.67 \cdot 10^{-27}\ \mathrm{kg}$
Neutron mass, $m_n = 1.67 \cdot 10^{-27}\ \mathrm{kg}$
Electron mass, $m_e = 9.11 \cdot 10^{-31}\ \mathrm{kg}$
Avogadro’s constant, $N_0 = 6.02 \cdot 10^{23}\ \mathrm{mol^{-1}}$
Universal gas constant, $R = 8.31\ \mathrm{J/(mol \cdot K)}$
Boltzmann’s constant, $k_B = 1.38 \cdot 10^{-23}\ \mathrm{J/K}$
Electron charge magnitude, $e = 1.60 \cdot 10^{-19}$
1 electron volt, $1\ \mathrm{eV} = 1.60 \cdot 10^{-19}\ \mathrm{J}$
Speed of light, $c = 3.00 \cdot 10^{8}\ \mathrm{m/s}$
Universal Gravitational constant,
$G = 6.67 \cdot 10^{-11}\ \mathrm{(N \cdot m^2)/kg^2}$
Acceleration due to gravity, $g = 9.81\ \mathrm{m/s^2}$
1 unified atomic mass unit,
$1\ \mathrm{u} = 1.66 \cdot 10^{-27}\ \mathrm{kg} = 931\ \mathrm{MeV/c^2}$
Planck’s constant,
$h = 6.63 \cdot 10^{-34}\ \mathrm{J \cdot s} = 4.41 \cdot 10^{-15}\ \mathrm{eV \cdot s}$
Permittivity of free space,
$\epsilon_0 = 8.85 \cdot 10^{-12}\ \mathrm{C^2/(N \cdot m^2)}$
Coulomb’s law constant,
$k = \frac{1}{4\pi\epsilon_0} = 8.99 \cdot 10^{9}\ \mathrm{(N \cdot m^2)/C^2}$
Permeability of free space, $\mu_0 = 4\pi \cdot 10^{-7}\ \mathrm{T \cdot m/A}$
Magnetic constant,
$\frac{\mu_0}{4\pi} = 1 \cdot 10^{-7}\ \mathrm{(T \cdot m)/A}$
1 atmospheric pressure,
$1\ \mathrm{atm} = 1.01 \cdot 10^{5}\ \mathrm{N/m^2} = 1.01 \cdot 10^{5}\ \mathrm{Pa}$
Wien’s displacement constant, $b = 2.9 \cdot 10^{-3}\ \mathrm{m \cdot K}$
Stefan-Boltzmann constant,
$\sigma = 5.67 \cdot 10^{-8}\ \mathrm{W/m^2/K^4}$

Условие

During the cold war, there was tension between the USSR and the U.S. But now, contrary to popular belief, American and Russian astronauts pass time by hanging out, enjoying the view from the moon, and even playing catch by launching projectiles at each other:

A projectile is launched with a speed $v_0 = 2200\ \mathrm{m/s}$ from the North Pole to the South Pole of a moon with radius $r_0 = 1.7 \times 10^{6}\ \mathrm{m}$ and $M = 7.4 \times 10^{22}\ \mathrm{kg}$ .

How long does the flight take? Answer in seconds.

A dark grey moon with a projectile launched from its top point along an arrow up and to the left; a dotted elliptical trajectory loops around to the left and returns to the bottom of the moon.

Отговори

Решение

Solution: The projectile will follow an elliptical path. It is easiest to represent this path in terms of an ellipse:

r = \frac{a(1 - e^2)}{1 - e\cos\theta}

At $\theta = 90^{\circ}$ , the numerator becomes $a(1 - e^2) = r_0$ . We will now show that the angular momentum is given by:

L = m\sqrt{GMr_0}.

If the apoapsis is $r_a$ and the speed of the projectile at apoapsis is $v_a$ , then conservation of angular momentum and energy gives:

L = mv_f r_f

\frac{1}{2}mv_0^2 - \frac{GMm}{r_0} = \frac{1}{2}mv_f^2 - \frac{GMm}{r_f}

Solving these equations, we get:

r_a = GM\cdot\frac{-1 + \sqrt{\frac{v^2 r}{GM} - 1}}{2E}

where $E$ is the energy per unit mass of the projectile. By setting $\theta = 180^{\circ}$ , we can write the location of the apoapsis to be

r_a = a(1 + e) = 9274582m

Therefore, to determine the apoapsis $a$ , we just need to determine $e$ . The eccentricity is given by:

e = \sqrt{1 + \frac{2Er_0}{GM}} = 0.8167.

Plugging in numbers, we find the semi-major axis to be $a = 461{,}670\ \mathrm{m}$ and the orbital period to be

T = 2\pi\sqrt{\frac{a^3}{GM}} = 32622\ \mathrm{s}

However, we are only interested in the section of the orbit that occurs above the surface. If we are able to determine the area the center of the moon subtends with the curve inside the moon, then we can apply Kepler’s second law to determine the orbital period.

This area can be determined via polar integration to be:

\frac{1}{2}\int_{\frac{5\pi}{2}}^{\frac{7\pi}{2}}\left(\frac{a\left(1 - e_c^2\right)}{1 - e_c\cos\theta}\right)^2 d\theta = 2.1633\times 10^{12}\ \mathrm{m^2}

The fraction of area covered by the projectile while outside the moon will be:

f = \frac{\pi ab - 2.1633\times 10^{12}}{\pi ab} = \frac{\pi a^2\sqrt{1 - e^2} - 2.1633\times 10^{12}}{\pi a^2\sqrt{1 - e^2}} = 0.9542

where $b = c^2 - a^2$ and $c = e/a$ were used to simplify the expression. Applying Kepler’s second law, the time of flight is then:

t = fT = \boxed{31128.8\ \mathrm{s}}

Оригинал в Архива: open.pdf · официални решения: open_sol.pdf