In the early morning hours of June 14, 2002, the Earth had a remarkably close en
ID: 2024313 • Letter: I
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 unkown asteroid, now designated 2002 MN, remained undetected until three days after it had passed the Earth. At its closest approach, the asteroid was 73,600 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 interatction 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 desntiy of 3.33 g/cm3. (For comparison, a 1-megaton nuclear weapon releases about 5.6 x 1015 J of energy.)
Explanation / Answer
the loss of gravitational potential energy is equal to gain in kinetic energyU = -mM2/R = 1/2 * m1 v^2 v= 2GM2/R plugging in the values as G = 6.67*10-11 (SI units)
M2 = mass of earth = 5.98*1024 kg
R = distance of closed approach of asteroid = 73600 miles =118422400 meters
we get v = 2590 m/s b) KE = (1/2)mv2
where m = *V = 3330kg/m3 *(4/3)**10003 m3 =3.927*1019 J
= 7013 megatons