| ▲ | sebastiennight 8 months ago | |||||||
OK I'm curious about this bit. How do you deduce the range of height of the ejecta from a calculation of the kinetic energy of the impactor? | ||||||||
| ▲ | belter 8 months ago | parent [-] | |||||||
You would start with kinetic energy of the impactor using the Kinetic Energy Formula. https://www.calculatorsoup.com/calculators/physics/kinetic.p... Let's go with 1.0E+16 Kg for mass and 25000 m/s for impact speed what gives about KE = 3.125E+24 J Assume 10% of that goes into ejecta.Then use the formula for Velocity from Kinetic Energy. https://calculator.academy/kinetic-energy-to-velocity-calcul... What gives about 8000 m/s of initial speed. Follow that using Maximum Height Formula from kinematics based on a Gravitational acceleration at g: 9.8 m/s² https://www.omnicalculator.com/physics/maximum-height-projec... We are looking at eject reaching orbits of 2500 Km to 3000 Km above Earth. Heat and corresponding vaporization effects excluded for simplification purposes. Now we wait for an event to validate my calculations.... | ||||||||
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