2. MORE MECHANICS OF THE SOLAR SYSTEM


  2.1 A walk on asteroid Icarus
      surface gravity, escape velocity

The problem: The asteroid Icarus has a diameter of 1.4 km, a density typical of rock, 3 x the density of liquid water on Earth, and rotates once in 2.3 hours. Determine the weight on Icarus of an astronaut of 200 kg, including space equipment. Determine the escape speed from Icarus and the speed of an object in circular orbit around Icarus, just above the surface of Icarus. How fast must an astronaut progress around Icarus such as to remain on the night-side of Icarus? Discuss whether an astronaut can walk around Icarus at the appropriate speed.

The astronomical setting: The famous asteroids have diameters of several hundred km. The largest, Ceres, has a diameter of about 1000 km. Much smaller asteroids are known only if they happen to approach Earth closely, i.e. if they cross r = 1 A.U. during their orbits. These are known as "Earth-crossing" asteroids and are of interest because one or another may collide with the Earth in the future.

   Icarus is one member of this class. It came within 107km = 0.07 A.U. of the Earth on June 14, 1968. Icarus is one of only two known asteroids that approach the Sun closer than the planet Mercury. It has a very eccentric orbit, between 0.2 A.U. and 2.2 A.U. Clearly, the surface on Icarus becomes very hot when it is only 0.2 A.U. from the Sun. Arthur C. Clarke wrote a short science-fiction story "Summertime on Icarus" in which the asteroid is used as a platform for machinery used to survey the inner solar system. At a time when Icarus is near the Sun, an astronaut is dropped by a space ship on the nightside of Icarus, in order to repair some equipment. The rotation of Icarus carries him toward the hot dayside. His heat shield fails. He experiences nearly lethal heat when the Sun rises, but fortunately the space ship rescues him just in time. We shall compute the temperature on Icarus in problem 4.1. Here the question is: since Icarus is so small, can the astronaut walk fast enough to remain on the night side permanently?

The solution: The surface gravity on an object of mass M and radius R is GM/R2. With MI = (4pi/3)rhoRI3 = 4x1012kg, the surface gravity is 5x10-4m/s2, compared to 10m/s2on Earth. An astronaut with space suit totaling 200 kg would weigh on Icarus what merely 10g weigh on Earth! [The ratio of gravities on Icarus and Earth can also be obtained from (MI/ME)(RE/RI)2 with Earth mass and radius given in the introduction or, more elegantly and with less chance of making an error, from (rhoI RI)/(rho E R E) where Earth's density is rho E = 5.5 times liquid water.]

   The escape speed is (2G MI/ RI)1/2 = 0.9 m/s = 3.2 km/hour, the speed of an object in circular orbit just above the surface is (G MI/ RI)1/2 = 0.63 m/s = 2.3 km/hour. The astronaut needs a speed 2piRI/P = 1.9 km/hour to keep up with the asteroid's rotation. This is almost the orbital speed!

Discussion: The danger of walking on Icarus is that the astronaut accidentally will push off from the ground enough so as to escape the asteroid, thus getting into the full sunshine that is so dangerous. Since the speed with which the astronaut must walk is almost the speed of the circular orbit, each step is really a sub-orbital flight. Each step must be calculated carefully. It seems like a risky maneuver, but perhaps OK when the alternative, staying in place, is lethal.

Regarding the Earth's seasons: If you are at moderate latitudes on Earth and experience "summer" and "winter", ask your students to explain these seasonal temperature differences. If they argue that summer occurs because Earth in summer is closer to the Sun, ask them to explain why summer occurs in July in the northern hemisphere but January in the southern hemisphere. This problem is a good opportunity to make sure that students know the correct explanation of Earth's summers and winters, in terms of the inclination of the Earth's axis of rotation. Icarus, of course, is just the opposite, in that the hot season is caused by proximity to the Sun (see problem 4.1).


2.2 The planetary slingshot effect
      frames of reference - the question of accuracy

Physics goals: change of reference frames; discussion of appropriate measurement accuracies.

The problem: Consider a space craft approaching Jupiter. Jupiter's orbit is circular, at r = 5.2 A.U. The space craft is in a minimum-energy orbit from Earth to Jupiter. See the left diagram below. Using the energy equation, evaluate Jupiter's velocity V and the space craft velocity v(arr) when it reaches r = 5.2 A.U. and arrives near Jupiter. Which moves faster? Transform to the frame of reference of Jupiter. In this frame, what is the initial velocity of the space craft, that is, the velocity of the space craft relative to Jupiter before gravitational attraction by Jupiter? Now assume that, after acceleration and deceleration by Jupiter, the final velocity is reversed relative to the initial velocity. See the right diagram below. Transform back into the solar system. What is the new velocity, v(dep) of the space craft, in the frame of the solar system, when it departs from Jupiter? Using the energy equation for the new orbit, estimate a and describe the new orbit.

   Write down a formula for the change in v2as evaluated in the frame of the solar system, in terms of Jupiter's velocity V and the velocity of the space craft just before arrival near Jupiter, v(arr).

 


 
 


 

The setting: Once a space craft is free of Earth's gravity, it is in an orbit around the Sun similar to the orbit of Earth. Energy is needed so that the space craft moves on an orbit toward its destination in the solar system (see problem 1.2). Given limited rocket power, the further we want to send a space craft, the less equipment it can carry. However, it is possible to boost the orbital energy by having the spacecraft fly past a planet. For instance, the space craft Cassini, launched on October 15, 1997, will arrive at Saturn on July 1, 2004, after flying twice past Venus, in April 1998 and June 1999, once past Earth in August 1999, and once past Jupiter in December 2000, with each encounter changing the orbital energy.

   The physics of the problem is quite simple: Kinetic energy depends on the frame of reference. It is often worthwhile making a simple computation in a frame in which the energy is constant, and then transforming to the frame of reference of the observer. The brief orbit of the space probe around a planet can be described ignoring the gravity of the Sun. In the frame of reference of the planet, the orbit is a hyperbola. The total (kinetic plus potential) energy is constant. Also, the kinetic energies before and after the encounter are the same, but the direction of the velocity has changed. Upon converting into the inertial frame of the solar system, the change in velocity means that the orbital energy is changed, and thus the orbit about the Sun is changed. This process is called gravity-assist and, more colloquially, the slingshot effect.

The solution: The spacecraft orbit in the solar system is given by the energy equation v2 = GMo(2/r-1/a). For Jupiter we have r = a = 5.2 A.U. and V = 13 km/s; for the arriving space craft we have r = 5.2 A.U., a = 3.1 A.U., and v(arr) = 7.5 km/s. The relative velocity is 13 - 7.2 = 5.5 km/s. Jupiter is faster. It catches up to the space craft. After the fly-by, v(dep) = 13 + 5.5 = 18.5 km/s. For the new orbit, again r = 5.2 A.U.. Then v2 (dep) is very nearly 2GMo/r, and a must be large. To within the accuracy of these computations, the probe achieves the escape speed from the solar system! Mathematically, v(dep) = v(arr) + 2 [V-v(arr)] = 2V - v(arr), v2 (dep) - v2 (arr) = 4V[V-v(arr)].

Interpretation: This example shows the effectiveness of the planetary slingshot. The main assumption is the deflection by 180o in the frame of Jupiter. Mathematically, it applies only in the limit of an infinitely long hyperbola approaching the focus arbitrarily closely. Physically, it is a good approximation if the velocity of approach to the planet is much less than the escape speed from that planet. For our example, the velocity of approach is 5.5 km/s, Jupiter's escape speed is 60 km/s, and 5.5 << 60 is not a bad approximation. Indeed, when the Pioneer and Voyager space craft encountered Jupiter, they experienced significant energy changes and deflections (though with encounter geometries more complicated than the example here). In the opposite limit, a very high velocity of approach means little time is available for gravity to act and a small deflection results. This problem is quite analogous to the electrostatic deflection of electrons by protons.

   The terrestrial planets generally produce rather small deflections. But these deflections can still be very useful, especially if one spacecraft experiences several planetary deflections. The spacecraft NEAR (= Near Earth Asteroid Rendezvous), launched February 16, 1996, came past Earth on January 23, 1998 and will arrive at the asteroid 433 Eros in October 1999. A direct orbit to Eros would have required a larger launch rocket, adding $5x107to the actual cost of about $1.3x108 (which is considered relatively cheap).

Didactics: The criterion for a large planetary deflection, (velocity of approach)/(escape speed) <<1, is "obvious" to most experienced scientists because there are no other dimensionless ratios in the problem. But such judgments are not obvious to students and must be substantiated a few times before they get in the habit of looking for such dimensionless ratios. In this case, the proof starts with the hyperbolic orbit, r = a(e2-1) /(1+ecostheta), r(min) = a(e-1) with e close to 1, and L = r 2dtheta/dt = r(min)v(max). Then dr/dt = r2 e sintheta dtheta/dt [a(e2-1] -1 = 1/2 sintheta v(max). Far away, a highly elongated orbit requires sintheta << 1. Therefore, the velocity of approach must satisfy dr/dt << v(max) proportional escape speed.

Astrophysical versus astronomical accuracy: The previous exercises have stressed "astrophysical accuracy", in which 10% accuracy is considered quite good and a factor of two is often no cause for worry. In contrast, the sending of space craft through the solar system is a process of "astronomical accuracy". Enormous accuracy was needed for the successful orbits of the two Voyager space craft launched in 1977. Both reached Jupiter in 1979 (not on a minimum-energy orbit), both reached Saturn, in 1980 and 1981 respectively, and one continued to Uranus in 1986 and Neptune in 1989. The tracking of the space craft is accomplished by carefully measuring the Doppler shift of the radio signal from the space craft. Even for the older Pioneer 10, the Doppler frequency shifts of its 2.1x109Hz signal can be measured accurate to 10-3Hz.

   The space craft NEAR, on the way to its Earth fly-by and later to asteroid 433 Eros, flew past the small asteroid 253 Mathilde, approaching within 1200 km on June 27, 1997. The measured Doppler frequency shift showed that the velocity of the space craft changed by only 0.23mm/s! This was enough to measure the mass of the asteroid, 1017kg. Given its size of about 46x48x66 km, the density turns out to be only 1.3 times that of liquid water. This is surprising since rocks (such as on our Moon, on two of Jupiter's moons, and in the outer part of the Earth) generally have a density of about 3; even mixtures of rock and ices (such as the two other Galilean moons of Jupiter and the largest asteroid Ceres) have a density of about 2. The asteroid's lower density suggests that it is composed of rather loose rubble, perhaps created or accumulated from past collisions. Indeed, the asteroid shows one truly enormous (relative to the asteroid) crater 30 km wide, 6 km deep, and four more craters at least 20 km wide. The five craters cover nearly half the surface. The craters suggest collisions in the past that nearly shattered the asteroid. Yet the craters are sharply edged and do not seem to consist of rubble. Thus we have both a surprise and a puzzle. Both are made possible by measurement of a change in velocity of only 0.23mm/s.
   Photo Credit: NASA.


2.3 Can a dust grain destroy a space probe?
      collisions, kinetic energy

The problem: Estimate the kinetic energy of a dust grain that might hit a camera on the space probe Giotto flying past the comet Halley in 1986. Compare your result to the energy of a stone thrown by a human. Is the camera likely to survive such a collision? (Assume the comet is in a highly elongated orbit, releases dust grains that nearly continue to move with the comet, and this dust is like the dust that pollutes Earth's cities when it is not raining. For all needed parameters, use your general knowledge of the solar system.)

The setting: From Earth, we cannot photograph the actual body of a comet because it is shrouded in gases and dust that reflect sunlight to us. In 1986, the European space craft Giotto flew past comet Halley as near as was technically possible, in order to photograph the actual cometary body and transmit the picture back to Earth by radio. The precise choice of Giotto's orbit was aided by several other space craft, also watching Halley.

   The comet is made of a mixture of solid particles and ices; on approaching the Sun, the evaporating ices escape from the comet in the form of gaseous jets, and solid particles are also blown off the comet by these jets. This explains the two kinds of tails, made of gas and "dust", respectively, both of which reflect sunlight to us. The released dust grains have orbits through the solar system similar to that of the parent comet. Comet Halley is on a highly elliptical orbit, and so are its dust grains. The orbit of Giotto and its camera is quite different. Is an impact of one of these dust grains likely to destroy the camera on Giotto before it can take the desired photograph? Before Halley and Giotto, we knew very little about the range of sizes and masses of the dust grains, and it was difficult to answer this question.

A reasonable solution: Regarding the dust grain that might hit Giotto : The dust in our cities is made of small grains that are barely visible. Imagine a spherical grain of radius 0.1 mm. Use the density of rocks, say 4 times the density of liquid water, to get a grain mass of 1.6x10-8kg.

   What is the likely relative velocity of the grain and the camera? The grain, pushed off Halley by thermal forces, has a small velocity relative to Halley, so the grain's orbit through the solar system is almost that of Halley. Halley is in a highly elliptical orbit. Giotto meets it not very far from 1 A.U. Let us use for Halley the speed it would have if it fell from infinity to Earth's orbit. That speed is 21/2 times the Earth's orbital speed, about 42 km/s. (Halley actually "falls" from a maximum orbital distance of 35 A.U.) If the space craft is relatively slow, say 30 km/s or less, the major component of the relative velocity is the velocity of Halley. So let us use as relative speed 50 km/s. Then the kinetic energy of the grain hitting the camera is 20j.

   Is 20j important? What does it take to damage a camera on Earth? Perhaps a thrown stone. How fast can a man throw a stone? If you visualize the thrown stone, it may have the speed of a car moving at, say, 50 km/hour (note the unit: hour, not seconds.). With the stone's mass about 0.1 kg, you get an energy of 10j. The energy of the stone is similar to the kinetic energy of the grain hitting the camera. The stone would damage the camera. Conclusion: The impact of the comet grain on the Giotto camera is likely to damage the camera.

The actual situation: Giotto met Halley on March 13-14, 1986 at a distance of merely 0.9 A. U. from the Sun, when the comet was moving away from the Sun. Giotto came to within 600 km from the comet. The camera took its last picture at the distance of about 1700 km. Then the space craft was violently tilted, presumably from the impact of large dust particles. Many instruments became inoperable after the sand-blasting by the dust during the encounter. Equipment on Giotto measured about 12x103impacts of dust grains with masses between 10-20and 10-7kg. Satisfactorily, this range includes our estimate of 10-8kg. We do not know the actual mass of the grain that hit the camera. Giotto's velocity relative to Halley actually was 68.4km/s, and that was also very nearly the velocity of Giotto's camera relative to the dust grain.

 


 

Credit: ESA

 


 

Interpretation: The last picture of Halley taken by the Giotto camera showed the comet to have the shape of a peanut, with a length of about 16 km, width about 8 km, covered with a very dark (presumably carbon-rich) surface. Jets of gases reflecting bright sunlight were escaping from sunlit parts of the surface. The jets imply the vigorous evaporation of ices within the comet. In the future, once Halley has approached the Sun several hundred times more, these gases will be gone and Halley's spectacle every 76 years will cease. The grains that now are still encased in ices then will be loose and acquire their own orbits. The orbit of each grain will be similar to the orbit of the comet. But the grains will spread out along this orbit until they constitute an elongated cloud of particles with the shape of Halley's orbit. Whenever the Earth passes through this cloud, it will intercept some of the particles. At those times, we observe a meteor shower. This process is already beginning: Meteor showers around Oct. 21 and May 4 are associated with Halley.

   Recent radar observations of the 1996 comet Hyakutake imply that as much as one sixth of all the material ejected from that comet (comet diameter no more than 3 km) is in dust grains larger than 1 mm. Grains with cm sizes are probably loose assemblies of smaller grains, not solid rocks.

Didactics: This problem is supposed to be an enjoyable exercise, giving some opportunity to the students' imaginations. Definitely, the students should estimate the needed parameters on their own, preferably by discussion within groups of 2 to 4 students. Of course, if the students are not used to making estimates based on their general knowledge, then this exercise takes time. When they seek the authoritative knowledge of the professor, they should receive only suggestive questions. This is the same kind of exercise that the experts went through only a few years ago. The time used by this discussion imparts more science to the students than would some additional part of a lecture imposing a few more facts .

   The most uncertain parameter in the problem is the size of the dust grain. Class members might well choose a much smaller grain, for instance the size of the wavelength of visible light. In that case, the grain does no damage whatsoever. But then class members should be reminded that we do not know the numbers of dust grains at larger sizes, and danger should be evaluated taking into account "worst cases". Thus they might consider larger grains. At the other extreme, our physical intuition (albeit based on city winds rather than gaseous comet jets) suggests that grains larger than some value (1 mm? 1 cm?) are unlikely to be blown out of the comet. Thus it is rather surprising to detect cm-sized grains from comet Hyakutake.


2.4 An asteroid impact on Earth
      collisions - a new scientific judgment

General science goal: A realistic example of the uneven progress of frontier science.

The problem: Estimate the energy of an asteroid of radius 1 km hitting the Earth, and compare with the energy released by the largest bomb exploded on Earth. (Assume the asteroid is just a large rock. Estimate whatever parameters you need. A "100 Megaton" bomb yields about 4x1017joules.)

A reasonable solution: With a density of rock four times that of liquid water, the mass of the asteroid is 1.6x1013kg. Since most asteroids orbit the Sun in the same sense as the Earth, the relative velocity is probably less than the Earth's orbital velocity. If we take 20 km/s, the energy of the impact is 3x1021j, comparable to the energy from several thousand of the largest bombs.

Didactics: As in the previous problem, students benefit from discussing the problem and making their own estimates of the density and impact velocity of the object. What matters is the physics, and that the students figure out what parameters need to be estimated. Numerical factors of two are not important.

The setting: Scientists often participate in fads, which are more politely called paradigms. In the last decade, collisions in the solar system have caught the attention of many scientists and of the public.

   For most of the past century, astronomers knew that objects from space occasionally hit the Earth, but these impacts were not considered important. Then, in 1978, came the claim that the dinosaurs became extinct because of the impact of an asteroid about 65 million years ago. The evidence for the asteroid was a layer of iridium-rich material in rock strata, about 65 million years old, found in several places all over the Earth. Iridium is a very rare element, and especially so on Earth because the Iridium originally distributed throughout the Earth has largely sunk down to the Earth's core. Asteroids contain relatively more iridium than the Earth's surface. If the impacting asteroid had a diameter about 8 km, it probably contained 2x108kg of Iridium, enough to explain the observations. This argument created much debate.

   Now there is compelling evidence for the asteroid impact 65 million years ago, namely an impact crater of the correct age, found on the edge of Mexico's Yucatan peninsula with a diameter of 200 km. There is general agreement that this impact caused the extinction of roughly 70% of all then living species. (There is not yet complete agreement about the dinosaurs, because they might have died out slightly earlier.). At least one additional major extinction event is probably caused by an impact (probably a comet rather than an asteroid, 2.5x108years ago), but some extinctions may be caused by episodes of intense volcanism. The evidence is still debated.

   Major international attention focused on comet Shoemaker-Levy 9. This comet was broken into roughly twenty pieces by Jupiter's gravity in 1992. The pieces fell onto Jupiter in July 1994. During the year before these impacts, theoreticians tried to predict the effects of the collisions on Jupiter. Their predictions depended strongly on the unknown mass and solidity of each of the comet pieces. The actual events, especially plumes of gases rising far above Jupiter's surface, were more energetic and violent than had been predicted. Their effects can be seen on Jupiter even five years later.

A new scientific judgment. All the old evidence of impact craters on Earth and Moon is now seen with a different scientific attitude. Possible collisions of celestial objects with the Earth have now become a major scientific topic. An object about 1 km in diameter passed Earth at the distance of only 725x103km, less than twice the Moon's distance. Yet we detected it only after it passed. We are really much more ignorant of the problem than we realized merely ten years ago.

   An estimated two thousand near-Earth asteroids exist larger than 1 km. Efforts are beginning to detect systematically all such asteroids and to find which travel on orbits that might some day intersect with the Earth. Unfortunately, such a prediction is not possible with newly arriving comets.

   The emphasis will be on searching for asteroids of size 10 km or larger. When they impact on Earth, they tunnel into the Earth and then explode so powerfully that much material is blown into the high atmosphere, above all rain clouds. This material remains up there long enough to be dispersed over the entire Earth. It returns to Earth surface only after months or even years later. The effects of a global cloud of dust, acid rains, and secondary effects on the oceans can cause an extinction episode such as occurred 65 million years ago.

   What happens if we actually determine that a large asteroid or comet will hit Earth? The most common suggestion is that we send rockets to the asteroid or comet and use the rocket or bombs to deviate the comet's path. Since Earth is so tiny, even a small deviation will be sufficient if we do it early enough. But this technology is still many years away.

   Discussion of large Earth impacts raises strong emotions. To many people, such impacts seem dreadful, indicating a possible future disappearance of humans from the Earth. We are learning, however, that past extinctions spurred the subsequent rapid evolution of new species. The impact 65 million years ago spurred the development of mammals and thus, indirectly, of humans.

 


  2.5 The clean-up of the solar system
      collision cross sections - mean free path - kinetic theory

The problem: Estimate the number of years needed for an asteroid flying freely and randomly throughout the inner solar system (r < 2 A.U.) before it is likely to crash into a terrestrial planet. Assume there are three planets in this volume, each with radius 2RE (to include captures by the planet's gravity). You may use one (or both) of two analogies: 1) A person is given a ball, blind-folded, turned around many times, and then is asked to throw the ball at a target 1m2 in area, at a distance of 10m. How often does this experiment have to be repeated before the target is likely to have been hit once? How frequently might asteroids orbiting through the solar system "try" this experiment? 2) Use the concept of mean-free-path from kinetic theory.

The astronomical setting: Our Moon shows enormous numbers of craters from past impacts. The lunar dark areas (the "mare") now consist solidified lavas. The lavas long ago rose out of the lunar interior and flowed into giant craters (several hundred km diameter) caused by impacts even longer ago. The radioactive dating of rocks brought back from the Moon (by the U.S. Project Apollo, 1969-1972) shows that most of the impacts, especially the very large ones, occurred in the first 0.5x109 years of the Moon's life, roughly in the period from 4.5 to 4.0 x 109 years ago. Why did the collisions stop? Apparently, because most of the rocks roaming the solar system at that time either had collided with some planet or had been deflected into orbits such that they left the inner solar system.

   A few impacts still happen in the inner solar system even today, that is, in the last 108 years. The impact 65 million years ago (problem 2.4) is a very energetic example. Some recently fallen meteorites can be analyzed for the amount of cosmic rays that have hit these meteorites while they were in space. (One measures the isotopes 3He, 21Ne, and 38Ar which result from the collision of cosmic rays with rock.) The result is a "space-exposure" age. Typically this age is of the order of 108 years. Before the meteorites were exposed to space and the cosmic rays, these objects must have been part of a larger asteroid body. Presumably a collision with the parent asteroid released the objects into space, onto an orbit that ultimately reached the Earth. The exposure ages of 108years inform us that the parent asteroids suffered a collision at that time in the past.

   Indeed there is evidence for such asteroidal collisions. The meteorite from Mars that may carry evidence for long-ago primitive life on Mars (highly debatable) was knocked off Mars 16 million years ago. The composition of the asteroid Vesta (diameter 530 km) led to the prediction that Vesta is the source of a distinctive class of tiny asteroids and some meteorites. Now the Hubble Space Telescope has observed a suitably large crater on Vesta with a diameter of 459 km.

   Here we estimate whether a collision time with the terrestrial planets is reasonably on the order of 108 years.

One reasonable solution:

  1. The target analogy: The area of a sphere 10 m in radius is 400pi m2. If the ball is thrown truly randomly in all directions, a target of 1 m2at 10 m is likely to be hit once in 400pi attempts. In the solar-system, the target is Earth at a distance of, say, 1 A.U. from the randomly moving asteroid. The Earth at that distance occupies a fraction of the sky (piRE2)/{4pi(1A.U.) 2} = 4x10-10. Given three planets and gravitational radius 2RE, the probability becomes about 5x10-9. Hence about 2x108 attempts are needed before there is a reasonable chance of success. If there is one attempt per year, it takes roughly 2x108years.

  2. Kinetic theory: The mean free path is 1/nsigma (ignoring factors of order unity), where n = number of objects per unit volume = 3 / [4pi/3 x 23] A.U. -3 = 0.1 A.U. -3 and sigma is the cross section of an object in the same units, pi(2RE)2 = 2x10-8 A.U. 2 resulting in a mean free path 1/nsigma = 5x108A.U. The mean collision time is 1/nsigmav, where v is the relative velocity, say 3 A.U./year (half the Earth's speed), which yields a collision time of about 1.5x108 years.

Interpretation: Obviously the individual numbers are very approximate. One of the uncertainties for the very earliest period of the solar system is the number of terrestrial planets. At least one additional Mars-sized object probably existed for a while, until it hit the Earth and created our Moon. Another uncertainty is the effective cross section of the planet, here taken four times the geometrical cross section. The effective cross section depends on the speed of the asteroid relative to the planet: the lower the relative speed, the more the planet's gravity can deflect the asteroid such as to hit the planet, and the larger is the cross section. There is an additional cross section, of similar magnitude, for the gravitational deflection of an asteroid by a planet that is sufficient for the asteroid to leave the inner solar system.

   The main physical uncertainty is the "random" path of the asteroid. During any few thousand years, the orbit of an asteroid is constant and may cross the plane of the Earth's orbit only well inside or well outside 1 A.U., so there is no chance of collision. However, we are learning from "chaos theory" that the orbits of asteroids are changing more often than we had expected, due to "resonances" with the planets, mainly Jupiter. Resonances with Jupiter occur when the ratio of the object's and Jupiter's orbital periods is a ratio of two small integers, so that Jupiter affects the object's orbit repeatedly in the same way. When an orbit is resonant, it may be nearly constant for millions of years, then change "suddenly", perhaps in a mere few thousand years, and then become nearly constant again. Such sudden, large orbital changes may occur every few millions years. They are called "chaotic" changes in orbit because even slightly different initial conditions will lead to different sudden orbital changes at different times. Since we do not know the initial conditions, we must deal with probabilities of sudden orbital changes. After each such change, there is a chance that the new orbit sweeps through Earth's orbital plane at 1 A.U. for many thousands of years, permitting a collision when Earth is in the way. It is this chance that is simulated by the calculation.

   The terrestrial and Jovian planets will maintain their orbits indefinitely. The orbits of the Trojan asteroids, situated at Jupiter's distance from the Sun and making an equilateral triangle with Jupiter and the Sun, appear to be stable. Pluto is in a stable resonant orbit with Neptune, that is, Pluto orbits the Sun twice while Neptune orbits it three time. When Pluto crosses Neptune's orbit, Neptune is always a quarter orbit away from Pluto.

Didactics: This should be an enjoyable exercise for the students, allowing some freedom to the imagination. If students come up with parameters that yield 109 years, it is more important for them to have chosen some reasonable parameters than being numerically "correct". Certainly, this kind of calculation cannot be expected to agree with reality to better than a factor of, say, three. The two methods used here involve somewhat different assumptions about the "random" velocities and object separations and could easily differ by a factor of two. The answer achieved here, 2x108 years, is to be considered attractively similar to the observational data regarding the first 5x108 years and the most recent 108 years of the inner solar system.