In [ ]:
import numpy as np
from astropy.table import QTable
from astropy import units as u
from astropy import constants as const
In [ ]:
In [ ]:
The asteroid [Name] would hit the Earth with the energy of [X] million tons of TNT
In [ ]:
It would take [X] 1 ton yield nuclear weapons to destory the asteroid [Name]
In [ ]:
Kepler's first law says: The orbit of every planet is an ellipse with the sun at one focus. The Semimajor axis a and the eccentricity ecc parametrize the size and shape of the ellipse. The units of a in our dataset are Astronomical Units (AU), the average distance between the Sun and the Earth.
For a closed elliptical orbit (orbits gravitationally bound to the Sun), $ecc = \sqrt{1 - {b^2}/{a^2}}$, where a and b are the semimajor and semiminor axes. As you can see from the equation, when a = b, ecc = 0 (a circle), and when a $>>$ b, ecc approaches 1. When ecc = 1, the orbit is a parabolic orbit (just bound). When ecc $>$ 1 the orbit is a hyperbolic orbit (unbound).
The speed of an object on an elliptical orbit around the Sun at a distance r from the Sun is:
$$ \large v\ =\ \sqrt{GM_{\odot}\ \left(\frac{2}{r} - \frac{1}{a}\right)} $$The encounter speed of an asteroid meeting the Earth at 1 AU is (assuimg aligned prograde orbits):
$$ \large V_{\textrm{encounter}}\ =\ V_{\textrm{asteroid at 1AU}}\ -\ V_{\textrm{Earth}} $$Where $V_{\textrm{Earth}}\ =\ 30\ \textrm{km/s}$
The impact speed of an asteroid hitting the Earth is:
$$ \large V_{\textrm{impact}}\ =\ \sqrt{V_{\textrm{encounter}}^2 + V_{\textrm{escape}}^2} $$Where $V_{\textrm{escape}}\ = 11.2\ km/s$
The self gravitational potential energy of a uniform sphere of mass (m) and diameter (d) is:
$$ \large PE \ = \ \frac{6}{5} \cdot \frac{Gm^2}{d} $$This is the amount of energy you need to give the sphere to move all of its components pieces infinitely far away (i.e. to destroy it!).
Remember that the mass and diameter of the asteroid is derived from its absolute magnitude H and albedo A.