a)In the early morning hours of June 14, 2002, the Earth had a remarkably close

Question

a)In the early morning hours of June 14, 2002, the Earth had a remarkably close encounter with an asteroid the size of a small city. The previously unknown asteroid, now designated 2002 MN, remained undetected until three days after it had passes the Earth. At its closest approach, the asteroid was 71500 miles from the center of the Earth. Calculate the speed of the asteroid at closest approach, assuming its speed at infinite distance to be zero and considering only its interaction with the Earth.


b)Observations indicate the asteroid to have a diameter of about 2.06 km. Estimate the kinetic energy of the asteroid at closest approach, assuming it has an average density of 3.00 g/cm3. (For comparison, a 1-megaton nuclear weapon releases about 5.60E+15 J of energy.)

Explanation / Answer

 use conservation of energy; at infinity, it had zero total energy, so must have zero total energy throughout its journey

at a distance of r from the earth, the potential energy due to the earth's gravitational field is 

PE = - GMm/r

G=newtonian grav constant, M, m are the masses of the earth and the asteroid, r is the distance from the center of the earth

the loss in PE must equal the gain in KE, or

1/2 m v^2 = GMm/r

so that v=Sqrt[2GM/r]

G=6.67x10^-11 in MKS
M=5.98x10^24kg
r=71500 miles =

= 1.15*10^8 m

assuming the asteroid is spheroidal, we find its mass from m = density x volume

volume = 4/3 pi r^3 = 4/3 pi (1030m)^3 (remember to use radius, not diameter)

volume = 4.58x10^9m^3

density = 3000 kg/m^3 so that total mass = 1.37x10^13kg

KE = 1/2 x 1.37x10^13kg x (2600.33m/s)^2 = 4.63x10^19J

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