In the early morning hours of June 14, 2002, the Earth had a remarkably close encounter with an asteroid?

Question

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 78300 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 1-megaton nuclear weapon releases about 5.6

Answer 1

a. v = (2GM/r)^.5 = (6.67E-11*6E24/1.24E8)^0.5 = 1800 m/sec


b. KE = 0.5mv^2
...m = 3330*Pi*1000^2 = 10,500,000,000 kg
...KE = 0.5*10.500E9*1800^2= 17E6 Gj

Each megaton is 4184000 Gj so the impact as suggested would have a KE in megatons of about 4 megatons.

Seems low.

Answer 2

DO u really need help on this one???