physics
posted by janie on .
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 passed the Earth. Suppose that at its closest approach, the asteroid was 84600 miles from the center of the Earth  about a third of the distance to the Moon.
(a) Find 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 2.0 km. Estimate the kinetic energy of the asteroid at closest approach, assuming it has an average density of 3.33 g/cm3. (For comparison, a 1megaton nuclear weapon releases about 5.6 1015 J of energy.)

F = G m Me/r^2
potential energy relative to zero at infinity = G m Me/r
here r = 84600 miles
convert that to meters
then
potential energy lost = Kinetic energy gained
G m Me/r = (1/2)m v^2
so
v^2 = 2 G Me/r
for part b you need m
diameter = 2*10^3
so
diameter/2 = radius = 1,000 meters
convert density to kg/m^3
3.33 g/cm^3 * (1kg/10^3g) * (10^6 cm^3/m^3) = 3330 kg/m^3
m = density * (4/3)pi (diameter/2)^3
then do Ke = (1/2) m v^2 
This is where I posted the reply to the multiple postings.