Автор: Idoru
Дата: 09-10-15 17:40
По въпроса за кратера:
http://www.braeunig.us/apollo/LMcrater.htm
или накратко:
"As the exhaust gas impacts the ground, a portion of its kinetic energy is transferred to the soil. The kinetic energy transferred to the soil causes the soil particles to fly away at high velocity, leaving behind a void where the soil once was. From this news article we find that the velocity of the lunar dust has been estimated to be between 0.6 and 1.5 miles per second (about 1,000 to 2,400 m/s). It is logical to assume the gas velocity (after losing energy to the soil) is the same as the dust. That is, the dust particles are propelled by the gas and are carried along with the gas flow at the same speed. We'll assume the low end of the velocity range, i.e. 1,000 m/s, because this is ultra-conservative. The lower the final gas velocity, the more energy it gives up to the soil. Furthermore, the lower the soil velocity, the more mass is required to carry the energy. All this adds up to a bigger crater.
To be further conservative, we'll assume no loss of kinetic energy. In reality, the kinetic energy at the end of the process will be less than that before because some of the energy will be used to perform other actions, such as dislodging or detaching soil particles. There will also be some dissipation of energy due to friction, turbulence, etc.
From the preceding section, we know the mass and kinetic energy of the exhaust gas as it contacts the ground. After colliding with the ground and losing energy to the soil, the gas will slow to 1,000 m/s. Knowing the final velocity of the gas we can calculate its final kinetic energy. Using the maximum mass value from Figure 5, we have
KEgas = 6.332 × 10002 / 2 = 3.17×106 J/m2
The amount of energy transferred to the soil is equal to the amount given up by the gas (assuming no loses). Using the maximum kinetic energy from Figure 6, we have the following. Note that in this scenario 89% of the gas kinetic energy is transferred to the soil.
KEsoil = 29.08×106 – 3.17×106 = 25.91×106 J/m2
Now that we know the kinetic energy and velocity of the soil, we can calculate the mass of soil needed to carry this energy,
m = 2 × 25.91×106 / 10002 = 51.82 kg/m2
To convert the above to a depth, we simply divide by the soil density. From Geotechnical Properties of Lunar Soil (Page 6), we read "On average, the bulk density, r, is approximately 1.30 g/cm3 at the surface, increases rapidly to 1.52 g/cm3 at a depth of 10 cm, then more gradually to 1.83 g/cm3 at a depth of 100 cm." Since we're dealing essentially with surface material, we'll use the low number of 1.30 g/cm3 (1,300 kg/m3), which is also conservative because a lower density means a larger volume of material is excavated. We have,
Depth = 51.82 / 1300 = 0.0399 m, or 39.9 mm (1.57 inches)"
Казано с думи прости спецификата на дюзата, изходящата струя газ и лунната почва са такива, че кратера който се е издълбал при приземяването е с голям диаметър и дълбочина от ... 4см.
Ако ви е прекалено сложно, ще ви го опростя малко - задайте си въпроса защо котката като ходи по паркета той скърца като да ходи човек по него? Отговорът се крие в "специфичния натиск" - теглото се разпределя върху определена площ.
Публикацията е редактирана (09-10-15 17:42)
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