IOAA 2015, Теория, кратки задачи, английски — Задача 3

SHORT PROBLEMS

Условие

On 27 May 2015 at 02:18:49, the occultation of the star HIP 89931 (δ\delta −Sgr) by the asteroid 1285 Julietta was observed from Borobudur temple, which was located at the center of the asteroid shadow path. It lasted for only 6.201 s. Assume that Earth’s orbit is circular and the orbit of Julietta is on the ecliptic plane and revolves in the same direction as Earth. At the occultation, Julietta is near its aphelion. At the time of the occultation the distances of Julietta from the Sun and the Earth are 3.076 AU and 2.156 AU respectively. Find the approximate diameter of asteroid Julietta, if the semi major axis of Julietta is a=2.9914a = 2.9914 AU.

Решение

13. Earlier marking-scheme variant
  1. 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+ΔE→D+e++νep + p + \Delta E \rightarrow D + e^{+} + \nu_e [20]

2 ΔE=(me+mD−2mp)c2=(me+(mp+mn)−2mp)c2=(me+mn−mp)c2\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 ΔE=(9.109×10−31+1.675×10−27−1.673×10−27)kg (2.99792×108ms−1)2=2.6198×10−13 J\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 ΔE∼kBT→T∼ΔEkB\Delta E \sim k_BT \rightarrow T \sim \frac{\Delta E}{k_B} [20]

5 T=2.6198×10−13 J8.315Jmol K=2.6198×10−13mol K8.315×6.02×10231mol=1.894×1010 KT = \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