An asteroid remained undetected until three days after it hadpassed the Earth. A

Question

An asteroid remained undetected until three days after it hadpassed the Earth. At its closet approach, the asteroid was 73,600miles from the center of the Earth -- About a third of the distanceto the moon. (a) Find the speed of the asteroid at closestapproach, assuming its speed at infinite distance to be zero andconsidering only its interaction with the Earth. (B) Observationsindicate the asteroid to have a diameter of about 2.0 km. Estimatethe kinetic energy of the asteroid at closest approach, assuming ithas an average density of 3.33g/cm3. (For comparison, a1-meaton nuclear weapon releases about 5.6 *1015 J of energy.)

Explanation / Answer

if v is the speed at closest approach and r is the distance atclosest approach, energy conservation gives total E at closest approach = KE + PE = (1/2)mv^2 - GmM/r =total E at = 0 + 0 = 0 v = (2GM/r) plug in the values of G, M (mass of earth) and r (convert itto meters) to get v. To get KE you need to estimate m. m = (4/3)R^3 * where R = radius of asteroid (=1000 m), = density =3330 kg/m^3 Now you can get KE as KE = (1/2)mv^2

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