Asteroid 2005YU55

This page is about a near earth crossing asteroid.  It is about 400 m in diameter and will pass about 324,605 km (201,700 miles) from Earth's surface (this is between 50 and 51 Earth radi).  Although it is very unlikely that this will hit the Earth this time (November 11th, 2011) as predicting orbital trajectories is just Kepler's law, it may hit on a later pass.  It's semi-major axis is about 1.06 A.U. and its period is about 1.10 years.  This orbit takes it inside Venus' at perihelion and out to Mars' aphelion.  Due to perturbations of the planets, the orbit will deviate slightly from Kepler's law over time.  That is becasue this is not really a two body problem anymore.  If this asteroid were to hit Earth, the size of the impact crater would be about 4.4 km.  See below for work.

D = 2*ρ_m^.11*ρ_p^-1/3*g_p^-.22*R^.13*E_k^.22*sin^1/3θ

This formula is derived from empirical results [Planetary Sciences Imke de Pater and Jack J. Lissauer second edition page 180].  It gives the diameter of the impact crater.

ρ_m is the planet density = 3500 kg/m³

ρ_p is the asteroid density = 2080 kg/m³ (assumed to be this-stone with some gaps)

g_p is the surface gravity = 9.81 m/s²

R is the radius of the asteroid (assumed to be spherical) = 200 m

E_k is the kinetic energy of the impactor (2*π*R³*ρ_m*v²/3) = 3.485*10¹⁸ J

v is the impactor velocity = 10000 m/s (typical impactor speed is assumed)

θ is the impact angle = 90^o (assumed to give maximum crater diameter)

The rule of ten (the diameter impact crater is about 10 times greater than the diameter of the impactor indicates the result should be about 4 km). 

Answer using formula: 4.4 km