Съдържание

IOAA 2015, Теория, кратки задачи, английски — Задача 13

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Всички материали от IOAA 2015

SHORT PROBLEMS

Условие

Volcanic activity on Io, whose rotation period synchronizes with its orbital period, was proposed to be the result of tidal heating mainly from Jupiter. The resultant tidal force on a body is the difference in gravitational force experienced by the near and far sides of that body due to another body. Measurements of the surface distortion of Io via satellite radar altimeter mapping indicate that the surface rises and falls by up to 100m during one-half orbit. Only the surface layers will move by this amount. Interior layers within Io will move by a smaller amount, and thus we assume that on average the entire mass of Io is moved through 50m. Assume that Io is considered as two hemispheres each treated as a point mass. Calculate the average power of the tidal heating on Io.

Hint: you can use the following approximation $(1+x)^n \approx 1+nx$ for small x.

The mass of Io is $m_{Io} = 8.931938\times10^{22}\ kg$

The perijove distance is $r_{peri}$ = 420000 km

The apojove distance is $r_{apo}$ = 423400 km

The orbital period of Io is 152853 s

The radius of Io is $R_\mathrm{Io}$ = 1821.6 km.

Решение

Answer and Marking Scheme

1 Calculate the tidal force

The tidal force (assuming the point mass of hemisphere is located at $R_{Io}/2$ ):

F_{tidal}=F(r-R_{Io})-F(r+R_{Io})

=-\frac{GM_J\frac{m_{Io}}2}{(r-\frac{R_{Io}}2)^2}+\frac{GM_J\frac{m_{Io}}2}{(r+\frac{R_{Io}}2)^2}=-GM_J\frac{m_{Io}}2\left[\frac1{(r-\frac{R_{Io}}2)^2}-\frac1{(r+\frac{R_{Io}}2)^2}\right]

=-\frac{GM_J\frac{m_{Io}}2}{r^2}\left[\frac{\frac{2R_{Io}}r}{1-(\frac{R_{Io}}r)^2}\right]

[40]

The term $\left(\frac{R_{Io}}r\right)^2$ is too small and it can be ignored , so that

F_{tidal}=-\frac{GM_Jm_{Io}R_{Io}}{r^3}

[10]

2 Calculate the difference of the tidal force

The difference in the tidal force experienced at perijove (closest to Jupiter) and apojove (furthest from Jupiter)

\Delta F=\left|F_{tidal}(r_{peri})-F_{tidal}(r_{apo})\right|

[15]

= GM_Jm_{Io}R_{Io}\left[\frac1{r_{peri}^3}-\frac1{r_{apo}^3}\right]=6.65\times10^{18}\ N

[15]

3 Calculate the average power of work done

Hence, the average power of the work done on the rock during one-half orbit is

\overline P=\frac{\Delta F\ d}{t_{Io}/2}

[10]

\overline P=4.35\times10^{15}W

[10]

Alternate solution

If assume the mass of the hemisphere is located at the center of mass of a solid hemisphere, $3R_{Io}/8$ , then

1 Calculate the tidal force

The tidal force (assuming the point mass of hemisphere is located at $3R_{Io}/8$ ):

F_{tidal}=F(r-R_{Io})-F(r+R_{Io})

=-\frac{GM_J\frac{m_{Io}}2}{(r-\frac{3R_{Io}}8)^2}+\frac{GM_J\frac{m_{Io}}2}{(r+\frac{3R_{Io}}8)^2}=-GM_J\frac{m_{Io}}2\left[\frac1{(r-\frac{3R_{Io}}8)^2}-\frac1{(r+\frac{3R_{Io}}8)^2}\right]

=-\frac{GM_J\frac{m_{Io}}2}{r^2}\left[\frac{\frac{3R_{Io}}{2r}}{1-(\frac{6R_{Io}}{8r})^2}\right]

[40]

The term $\left(\frac{R_{Io}}r\right)^2$ is too small and it can be ignored , so that

F_{tidal}=-\frac{3GM_Jm_{Io}R_{Io}}{4r^3}

[10]

2 Calculate the difference of the tidal force

The difference in the tidal force experienced at perijove (closest to Jupiter) and apojove (furthest from Jupiter)

\Delta F=\left|F_{tidal}(r_{peri})-F_{tidal}(r_{apo})\right|

[15]

=\frac34 GM_Jm_{Io}R_{Io}\left[\frac1{r_{peri}^3}-\frac1{r_{apo}^3}\right]=4.97\times10^{18}\ N

[15]

3 Calculate the average power of work done

Hence, the average power of the work done on the rock during one-half orbit is

\overline P=\frac{\Delta F\ d}{t_{Io}/2}

[10]

\overline P=3.25\times10^{15}W

[10]

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