VocExile Main Menu /Comet

By Ed Perley

In recent years there has been considerable discussion about the possibility of a comet hitting our world and causing widespread destruction. Some have discussed ways in which a comet on a collision course with the earth could be destroyed or diverted. The most obvious suggestion has been to use missiles with nuclear warheads to shatter the comet or change it’s course. Another suggestion was to deposit giant rockets on the comet’s surface and fire them to divert it. Both of these ideas have problems. First, just breaking up the comet would not be much benefit, unless the resulting fragments were very small. The Earth would get hit by a shotgun instead of a bullet. Using giant rockets might work, but the logistics of placing rockets with sufficient thrust to divert a comet would be very difficult and expensive.

Actually there is a serious problem with trying to blow up a comet with nuclear explosives. Thermonuclear explosions in space do not work the same as explosions in the Earth’s atmosphere. A nuclear explosion results from the sudden disintegration of uranium 235 to form cesium 140 and rubidium 33. The nuclear reaction produces an extremely dense flux of gamma and Xrays. Because the atmosphere is opaque to this radiation, a relatively small volume of atmosphere absorbs all of it. This in turn causes an almost instantaneous heating of this air to hotter than the surface of the sun. The sudden expansion of this extremely hot gas is what causes the fire and blast we witness when a nuclear device is detonated.

But in the vacuum of space, the blast effects from a nuclear explosion would be much weaker. There would be no gasses to absorb the radiation and expand rapidly, except for those from the vaporized bomb. Some blast and heat energy would be absorbed by the comet, but the bulk of it would shoot off into space, maybe frying the electronics of dozens of earth satellites. So, for a nuclear explosive to be effective, it would have to be buried deep in the interior of the comet. Again, the logistics of drilling a deep shaft in a comet millions of miles from the earth would be a big problem. Also, the calculations below suggest that a single warhead, which only contains a few Kg of nuclear explosive, does not produce sufficient explosive energy to have a significant effect on a large comet.

There is another possible way to use nuclear power to divert or even destroy a comet. Remember the Chernobyl disaster in Russia, when a nuclear power plant melted down. The movie, “The China Syndrome” introduced the idea of what would happen if a meltdown could not be stopped. According to this theory, the molten core would eventually melt it’s way through the earth’s crust and sink to the center of the earth, spewing out superheated gasses as it descended.

What if we could put a very large mass (much more than found in a bomb) of uranium 235 or plutonium on the surface of a comet? Then deliberately let it go critical and melt down. If the meltdown was initiated slowly, the heat generated by the core would vaporize the ice under it, and one would hope that it would sink into the comet far enough so that escaping gas from a complete meltdown would not overcome the weak gravity of the comet head and blow it into space.

When the nuclear material would have sunk sufficiently deep, a complete meltdown would be initiated. Gravity would draw the hot molten core toward the center of the comet. And at the same time, hot gasses would be blasted out the hole into space. You would, in effect, have a giant nuclear rocket capable of firing for a very long time. The energy output of this super rocket would be much greater than that of a nuclear explosive because of the enormous mass of nuclear fuel and an unlimited supply of propellent (steam). If a large enough quantity of nuclear material was used, it might be capable of destroying the entire comet from the inside out.

There would, of course, be some danger in this type of operation. We would have to be absolutely certain that the comet would be diverted far from the Earth’s path. Otherwise, we could be hit by highly radioactive debris. It would be necessary to do a great deal of experimentation with “safe” comets before such a system could be used with any confidence. Even so, a danger of radiation would probably be preferable to certain annihilation from a big comet.

This type of process could be used for transporting payloads weighing thousands of tons inexpensively through space. Just intercept a comet or ice asteroid, load it with ores mined from metallic asteroids, induce a controlled meltdown, and send it on it’s way. Of course, you would have to find a way to steer it.

First we will calculate the mass of a comet with a volume of one cubic kilometer. We definitely would not want something that big to hit the earth. Scientists believe that a comet is composed mostly of ice, and we know that one cubic centimeter of ice weighs aproximately one gram. So, 1 Cubic Km = 10 E15 Cubic cm, which is equivalent to 10 E12 Kg of ice.

In conclusion, our comet weighs 10 E12 Kg, or 1000 billion Kg

*(Note that '10 E12' stands for ten to the twelfth power, or ten multiplied by itself
twelve times) *

The nuclear reaction associated with fission of Uranium 235 is:

U (235) + n --> Cs (140) + Rb (37) + 3 n + 200 Mev

One Uranium 235 atom is hit by a neutron. It disintigrates, producing Cesium 140, Rubidium 37, three neutrons and 200 million electron volts. The neutrons produced hit other Uranium 235 atoms, making them undergo the same reaction. The chain reaction continues until the remaining uranium atoms are spread so far apart that neutrons can no longer hit them.

First, we will calculate the amount of energy produced by one Kg of Uranium 235.

1 Mev = 1.6027 x 10 E-13 joules200 Mev = 3.204 x 10 E-11 joules (for one atom)

1 mole of Uranium 235 (weighing 235 gm) contains 6.0228 x 10 E23 atoms (Avogadro's Number)

Energy produced by 235 gm = (6.02228 x 10 E23) X (3.204 x 10 E-11) = 4.53 x 10 E9 joules

Joules per Kg of Uranium 235 = (4.53 x 10 E9) x (1000/235) = 1.928 x 10 E10 joules

Let's estimate how much useful energy we could generate if we planted one metric ton of
Uranium 235 on a comet. Also let's assume that only 10 per cent of the energy produced is
actually converted into motion. Therefore, the energy of motion generated by one ton of Uranium 235 will be:

(1.928 x 10 E10) x 1000 x .10 = 1.928 x 10 E12, or close to 2 x 10 E12 joules.

Now that we know the weight of the ice and energy produced, we can calculate how much the velocity of the comet can be changed. We will want to move it 90 degrees from it's original trajectory so that it will miss the earth by 100,000 km. We will want to move this object this far because the trajectories of comets are rather unpredictable. Therefore, we will want to have a large margin of error.

We will use the Kinetic Energy Equation, solving for Velocity:

velocity = SQRT( ( 2 x energy)/mass))

(SQRT means Square Root)

velocity = SQRT( ( 2 x 2 x 10 E12) / (10 E12)) = 2 meters per second.

Converting this velocity to km per day, we get a velocity of 173 km per day. But we want to move the comet 100,000 km. So, the number of days needed before impact is:100,000 km / 173 km per day = 578 days.

So, our calculations show that the comet's velocity must be changed about one and half years before the comet is scheduled to impact. I am not able to calculate how long it would take to change the velocity of the comet. That would depend on the speed of the nuclear reaction. What if you use more Uranium 235? If you inspect the equation, you can see that to double the velocity change, you will need four times as much Uranium 235.

Consider a smaller object, for instance one 100 meters across. Smaller, but still big enough to take out a city. The mass will be one one thousandth as much as that of the one cubic km object, one billion Kg. Using the above equation:

velocity = SQRT( 2 x (2 x 10 E12) / (10 E9)) = 44.7 m per sec, or 3862 km per day.

So, the number of days needed before impact = 100,000 km / 173 km per day = 25 days. Since this object will be close enough to accurately calculate it's motion, you would probably be able to get away with moving it only a fraction of that distance.

The above calculations provide only rough approximations. A number of factors are not taken into account. As I noted above, I do not have the information to calculate how much time is needed for ten percent of the nuclear energy to be converted into actual motion. Also, a good deal of energy is needed to raise the temperature of the ice to it's boiling temperature. That energy is lost. Secondly, the boiling off of the ice will reduce the mass of the ice chunk, allowing a greater change in velocity than the calculations indicate. And the arbitrary efficiency of ten per cent is, of course, only a guess.

More calculations concerning this problem may be added in the future.

Encarta 98 Encyclopedia, a PC application, provided the nuclear reaction data.

Any good physics textbook will explain the physics equations.

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Date last updated: November 24, 1999