USAAAO 2021, I кръг — Задача 15
USAAAO 2021 - First Round · 30 януари 2021 г.
PHYSICAL AND ASTRONOMICAL CONSTANTS
| c | Speed of light in vacuum | |
| e | Elementary charge | |
| m | Neutron rest mass | |
| m | Proton rest mass | |
| m | Electron rest mass | |
| m | Helium-4 rest mass | |
| h | Planck’s constant | |
| H | Hubble’s constant | 70 (km/s)/Mpc |
| k | Boltzmann’s constant | |
| b | Wien’s constant | |
| G | Gravitational constant | |
| Stefan-Boltzmann constant | ||
| c | First Radiation Constant | |
| c | Second Radiation Constant | |
| N | Avogadro constant | |
| R | Gas constant | |
| a | Bohr radius | |
| Bohr magneton | ||
| M | Solar mass | |
| R | Solar radius | |
| L | Solar luminosity | |
| T | Solar temperature | 5770 K |
| M | Earth mass | |
| R | Mean Earth radius | |
| I | Earth moment of Inertia | |
| R☾ | Mean Moon radius | |
| M♃ | Mean Jupiter mass | |
| R♃ | Mean Jupiter radius | |
| a♃ | Mean orbital radius of Jupiter | 5.2 AU |
| a☾ | Mean semimajor axis Moon orbit | |
| 1 light year | ||
| 1 AU | Astronomical Unit | |
| 1 pc | Parsec | |
| 1 year | ||
| 1 sidereal day | 86164 s | |
| 1 erg | ||
| 1 bar |
ENERGY CONVERSION : 1 joule (J) = electronvolts (eV)
Условие
An interesting phenomena that happens in the Solar System is the capture of comets in the interstellar medium. Assume that a comet with a mass of kg is captured by the solar system. The perihelion of this comet’s orbit after it is captured is equal to 4.64 AU, and its velocity with respect to the Sun before being captured by the Solar System was very small. Calculate the velocity of the comet at the perihelion.
(a) 87.1 km/s
(b) 45.9 km/s
(c) 5.67 km/s
(d) 105.4 km/s
(e) 19.6 km/s
Отговори
- E
Решение
First, it is important to determine the shape of the comet’s orbit.
When the comet is still far from the Sun, its velocity is very small, so its kinetic energy is negligible. At this point, the distance to the Sun is large, so the potential gravitational energy is also close to zero. Therefore, since the total mechanical energy is zero, the orbit is parabolic.
It is important to highlight that the velocity was not exactly zero and the distance to the Sun was not infinite, so the mechanical energy might not be exactly equal to zero, but it is close enough to assume that the orbit is parabolic or at least almost parabolic.
For parabolic orbits, since the total mechanical energy is equal to zero, it is possible to use the following formula to calculate the velocity of a body in a parabolic orbit:
Answer: E
Оригинал в Архива: usaaao2021_exam1.pdf · официални решения: usaaao2021_exam1_solutions.pdf