[Geritol]

"would that reduce the damage of the impact enough to merit interception?"

That is an interresting question. The first mistake is in thinking that if we use explosives or nuclear something to convert it from a bowling ball traveling at us at 15,000 miles an hour into a pile of BBs weighing the same as the bowling ball and still traveling at us at 15,000 miles an hour, only now containing some quantity of radioactive fallout, that it will make a really huge difference in what it would do to us.

Making broad assumtions, a sphere 100 feet in diameter at 2.5 the density of water would weigh about, well, a bunch. something with 7 zeroes, 40,000 tons or so. According to this site at http://personals.galaxyinternet.net/tunga/I2.htm,
KE = 6.256 x 10-8 x (diameter)3 x (velocity)2 x (density)
Where KE is expressed in Megatons of TNT
Where diameter is expressed in meters
Where the velocity relative to Earth is expressed in kilometers per second
Where density is expressed in grams per cubic centimeter
Assuming 1 megaton of TNT is equivalent to 4.185 x 1022 ergs

6.7 kilometers per second closing speed
2.5 times density
30 meter diameter gives us
.19megaton in kinetic energy.

Added to this would be the energy released from whatever chemical reactions (fires) occur to both the object and the terrain and geology surrounding the impact site. This "secondary" reaction would be multiplied by fracturing the object. I have no idea how to calculate the energy released from vaporizing and instantly combusting several city blocks and igniting hundreds or thousands of acres. Nor do I know how to compare that energy to the energy released from instantly igniting 10 or 100 times that area with the fragmented object's spread.

[Centurion2000]

If it is mostly nickel iron like earth .. try 5.5 gr/cm3 .... comes closer to .714 MT of KE

[Lonesome in Massachussets]

I'll see your 0.19 Megaton and raise you 8 Megatons. Typical asteroid velocities are ~44 km/sec, which gives 8.19 MT. If the density is higher by half add another 8 or so MT. Either way, DUCK.

[Liberty Tree Surgeon]

An excellent analysis. When I was watching the end of one of those asteroid movies a few years ago, I almost stood up to cry "BullS$#%" in the theater. My back of the envelope calculations indicated that smashing a large asteroid would just mean the energy is released in the atmosphere, rather than a surface impact where some is absorbed by the earth itself. I figured the facing side of the earth would lose all of its atmosphere within a very short time.

[Dave Olson]

The first mistake is in thinking that if we use explosives or nuclear something to convert it from a bowling ball traveling at us at 15,000 miles an hour into a pile of BBs weighing the same as the bowling ball and still traveling at us at 15,000 miles an hour,

True, but there is an atmosphere between us and the object. It may not be enough to burn up something of any significant size that's in one chunk. However, if that object was broken up into small enough pieces that are spread out, each of those pieces would burn up.

Consider that every year, the earth is showered with thousands meteors each about the size of a grain of sand. If all those grains were combined into one large and solid hunk of rock, it could do considerable damage. But because that same amount of rock is in tiny pieces, we're treated to spectacular shows a couple of times a year instead.