IOAA 2015, Теория, кратки задачи, английски — Задача 5

SHORT PROBLEMS

Условие

An observer is trying to determine an approximate value of the orbital eccentricity of a man-made satellite. When the satellite was at apogee, it was observed to have moved by Δθ1=2′44′′\Delta\theta_1 = 2'44'' in a short time. When the radius vector connecting Earth and the satellite is perpendicular to the major axis (true anomaly is equal to 90∘90^{\circ}), within the same duration of time, it was observed to have moved by Δθ2=21′17′′\Delta\theta_2 = 21'17''. Assume that the observer is located at the center of the Earth. Find an approximate value of the eccentricity of the satellite’s orbit.

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