Съдържание

IOAA 2015, Теория, кратки задачи, английски — Задача 2

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IOAA 2015

SHORT PROBLEMS

Условие

One satellite of a planet has an orbital period of 7 days, 3 hours, 43 minutes, and the semi major axis is 15.3 times the mean radius of the planet. The Moon has an orbital period of 27 days, 7 hours, 43 minutes and the semi major axis is 60.3 times the Earth's mean radius. Assume that the mass of the moon and the satellite is negligible compared to the mass of the planet. Calculate the ratio of the planet’s mean density to that of the Earth.

Решение

Answer and Marking Scheme:

First Version

1 Kepler third Law for satellite and Moon

\frac{P_S^2}{a_S^3}=\frac{P_S^2}{(15.3R_P)^3}=\frac{4\pi^2}{GM_P};

\frac{P_M^2}{a_M^3}=\frac{P_M^2}{(60.3R_E)^3}=\frac{4\pi^2}{GM_E};

[30]

2 Express mass in term of mass density:

\frac{P_S^2}{(15.3R_P)^3}=\frac{4\pi^2}{GM_P}=\frac{4\pi^2}{G\rho_P\frac43\pi R_P^3}

\frac{P_M^2}{(60.3R_E)^3}=\frac{4\pi^2}{GM_E}=\frac{4\pi^2}{G\rho_E\frac43\pi R_E^3}

[30]

3 Evaluate ratio of mass-density:

\frac{\rho_E}{\rho_P}=\frac{60.3^3\times P_S^2}{15.3^3\times P_M^2}

[30]

\frac{\rho_E}{\rho_P}=0.238\approx0.24

[10]

Second Alternate Version

The period of the Satellite: $P_S=7\ days,3\ hours,43\ minutes=0.0196\ years$ ,
Mean orbital radius of the Satellite: $a_S=15.3R_P$ ; $R_P$ : radius of the Planet.
The period of the Moon: $P_M=27\ days,7\ hours,43\ minutes=0.0749\ years$ ,
Mean orbital radius of the Moon: $a_M=60.3R_E$ ; $R_E$ : radius of the Earth.
Mass of the Sun ( $M$ )
Kepler’s Third Law: $G(M+m)=4\pi^2\frac{a^3}{T^2}\rightarrow$
Mass of the Planet: $m_P$
Radius of the Planet: $R_P$
Semi major axis of the Planet = Mean orbital radius of the Planet: $a_P$
Semi major axis of the Satellite = Mean orbital radius of the Satellite: $a_S$
Semi major axis of the Moon = Mean orbital radius of the Moon: $a_M$
Mass of the Earth: $m_E$
Radius of the Earth: $R_E$
Semi major axis of the Earth = Mean orbital radius of the Earth: $a_E$

Planet – Satellite:

G(m_P+m_S)=4\pi^2\frac{a_S^3}{P_S^2},m_P\gg m_S\rightarrow G(m_P)=4\pi^2\frac{a_S^3}{P_S^2}\qquad(1)

(5)

The Earth–the Moon:

G(m_E+m_M)=4\pi^2\frac{a_M^3}{P_M^2},m_E\gg m_M\rightarrow G(m_E)=4\pi^2\frac{a_M^3}{P_M^2}\qquad(2)

(5)

Planet–Sun:

G(M+m_P)=4\pi^2\frac{a_P^3}{P_P^2},M\gg m_P\rightarrow G(M)=4\pi^2\frac{a_P^3}{P_P^2}\qquad(3)

(5)

Earth–Sun:

G(M+m_E)=4\pi^2\frac{a_E^3}{P_E^2},M\gg m_E\rightarrow G(M)=4\pi^2\frac{a_E^3}{P_E^2}\qquad(4)

(5)

From equation (1) and (3):

\frac{m_P}{M}=\frac{4\pi^2\frac{a_S^3}{P_S^2}}{4\pi^2\frac{a_P^3}{P_P^2}}=\frac{(a_S)^3}{(a_P)^3}\frac{(P_P)^2}{(P_S)^2}=\frac{(15.3R_P)^3}{(a_P)^3}\frac{(P_P)^2}{(P_S)^2}=\frac{(15.3R_P)^3}{(a_P)^3}\frac{(P_P)^2}{(0.0196)^2}

(20)

From equation (2) and (4):

\frac{m_E}{M}=\frac{4\pi^2\frac{a_M^3}{P_M^2}}{4\pi^2\frac{a_E^3}{P_E^2}}=\frac{(a_M)^3}{(a_E)^3}\frac{(P_E)^2}{(P_M)^2}=\frac{(60.3R_E)^3}{(a_E)^3}\frac{(P_E)^2}{(P_M)^2}=\frac{(60.3R_E)^3}{(a_E)^3}\frac{(P_E)^2}{(0.0749)^2}=\frac{(60.3R_E)^3}{(1)^3}\frac{(1)^2}{(0.0749)^2}

(20)

\rightarrow\quad\frac{m_E}{m_P}=\frac{\frac{(15.3R_P)^3}{(a_P)^3}\frac{(P_P)^2}{(0.0196)^2}}{\frac{(60.3R_E)^3}{(1)^3}\frac{(1)^2}{(0.0749)^2}}=\frac{(15.3R_P)^3/(0.0196)^2}{(60.3R_E)^3/(0.0749)^2}

(20)

\rightarrow\quad\frac{\rho_P}{\rho_E}=\frac{m_P/Vol_P}{m_E/Vol_E}=\frac{m_P/\frac43\pi(R_P)^3}{m_E/\frac43\pi(R_E)^3}=\frac{(15.3R_P)^3/(0.0196)^2}{(60.3R_E)^3/(0.0749)^2}\cdot\frac{\frac43\pi(R_E)^3}{\frac43\pi(R_P)^3}=\frac{\frac{(15.3)^3}{(0.0196)^2}}{\frac{(60.3)^3}{(0.0749)^2}}=0.24

(20)

13. Earlier marking-scheme variant

Two protons collide to produce deuteron, positron, and neutrino in the center of a star. Write down the reaction equation and calculate the minimum temperature needed for this reaction. Assume that the neutrino mass is negligible.

Answer

1 $p + p + \Delta E \rightarrow D + e^{+} + \nu_e$ [20]

2 $\Delta E = (m_e + m_D - 2m_p)c^2 = (m_e + (m_p + m_n) - 2m_p)c^2 = (m_e + m_n - m_p)c^2$ [20]

3 $\Delta E = (9.109\times10^{-31} + 1.675\times10^{-27} - 1.673\times10^{-27})\mathrm{kg}\,(2.99792\times10^{8}\mathrm{ms^{-1}})^2 = 2.6198\times10^{-13}\ J$ [20]

4 $\Delta E \sim k_BT \rightarrow T \sim \frac{\Delta E}{k_B}$ [20]

5 $T = \frac{2.6198\times10^{-13}\ J}{8.315\frac{J}{mol\ K}} = \frac{2.6198\times10^{-13}mol\ K}{8.315}\times6.02\times10^{23}\frac{1}{mol} = 1.894\times10^{10}\ K$ [20]

Оригинал в Архива: Short_problem_Final_EN.pdf · официални решения: Short Problem Solution-Marking Scheme 30.07.pdf · допълнителни данни: Short_Problem_Solution.docx