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Project Cyclops, A Design... - Department of Earth and Planetary ...

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Black at NASA's Ames Research Center reanalyzed Van<br />

de Kamp's data <strong>and</strong> found that an even better fit<br />

between the theoretical <strong>and</strong> observed wobbles could be<br />

obtained by assuming three bodies in circular orbits. Van<br />

de Kamp's two planets <strong>and</strong> Suffolk <strong>and</strong> Black's three<br />

planet solutions are given in Table 2-3.<br />

Barnard's star is an M5 star with only 15 percent <strong>of</strong><br />

the Sun's mass. It is tempting to regard Suffolk <strong>and</strong><br />

Black's solution as evidence for a scaled-down solar<br />

system with three gas-giant planets at 1.8, 2.9, <strong>and</strong> 4.5<br />

AU, rather than four at 5.2, 9.5, 19, <strong>and</strong> 30 AU, as in<br />

our Sun's family. However, the data do not appear to<br />

support this simple interpretation.<br />

A single planet in orbit about a star causes the star to<br />

execute an orbit <strong>of</strong> the same shape (same eccentricity)<br />

about the common center <strong>of</strong> gravity. If the planetary<br />

orbit is circular, the observed stellar motion should be<br />

simple harmonic in both right ascension <strong>and</strong> declination,<br />

<strong>and</strong> the Fourier spectrum <strong>of</strong> both motions should<br />

therefore contain a single frequency. If the orbit is<br />

highly elliptical, the motion <strong>of</strong> the primary will occur<br />

rapidly when the two bodies are near periapsis, with<br />

relatively slow motion for the long intervening period.<br />

Thus, the spectrum <strong>of</strong> the motion will contain harmonics<br />

<strong>of</strong> the fundamental frequency <strong>of</strong> revolution.<br />

TABLE 2-3<br />

CHARACTERISTICS OF POSSIBLE PLANETS<br />

ORBITING BARNARD'S STAR<br />

Mass Orbit<br />

Planet Distance (AU) (Jupiter = 1) Period (Trs)<br />

Source<br />

BI 4.7 1.1 26 van de Kamp<br />

4.5 1.26 24.8 Suffolk & Black<br />

B2 2.8 0.8 12 van de Kamp<br />

2.9 0.63 12.5 Suffolk & Black<br />

B3 1.8 0.89 6.1 Suffolk & Black<br />

Examination <strong>of</strong> the periods <strong>of</strong> the "planets" in Table<br />

2-3 shows them to be very nearly in the ratio <strong>of</strong> 2:4 for<br />

Van de Kamp's solution <strong>and</strong> l:2:q '_,r Suffolk <strong>and</strong><br />

Black's solution. This raises the ql, estion as to whether<br />

the perturbations ascribed to planets/32 <strong>and</strong> B3 may not<br />

in reality be harmonics produced by a highly elliptical<br />

orbit for B1 as originally proposed by Van de Kamp.<br />

It is significant that the "harmonic content" <strong>of</strong> the<br />

wobble <strong>of</strong> Barnard's star is different in right ascension<br />

<strong>and</strong> declination. This could not be the case for multiple<br />

planets in coplanar circular orbits. The reduction in<br />

residual errors in Suffolk <strong>and</strong> Black's solution as<br />

compared with Van de Kamp's solutions is obtained<br />

only if the orbit <strong>of</strong> BI is steeply inclined (_> 40 °) to the<br />

orbits <strong>of</strong> B2 <strong>and</strong> B3.<br />

Thus, it appears likely that Barnard's star has more<br />

than one orbiting companion <strong>and</strong> that either the orbits<br />

are not coplanar or at least one is highly elliptical. The<br />

evidence for three bodies will not be conclusive until a<br />

longer observation time shows that the periods <strong>of</strong> B1,<br />

B2, <strong>and</strong> B3 are indeed incommensurate. A solution that<br />

allowed ellipticity but required coplanarity would be<br />

interesting to pursue.<br />

Table 2-4 lists several other stars known to have dark<br />

companions. For the first six the companion has from<br />

1/10 to 1/35 the mass <strong>of</strong> the visible star. These we might<br />

classify as binary stars in which the smaller companion<br />

has too little mass to initiate thermonuclear reactions in<br />

its core. (It is generally agreed that this requires about<br />

0.05 solar masses.) The last examples, 61 Cyg A <strong>and</strong><br />

Proxima Centauri, are borderline cases where the star is<br />

55 or more times as massive as the companion. (The Sun<br />

is 1000 times as massive as Jupiter.) Like Barnard's star,<br />

70 Ophiuchi appears to have more than one unseen<br />

companion. We infer from these examples that a more or<br />

less continuous spectrum <strong>of</strong> systems exists between<br />

symmetrical binaries at one extreme <strong>and</strong> single stars with<br />

a giant planet, or planets, at the other.<br />

It will be noticed that with one exception all the stars<br />

listed in Table 2-4 are K5 or smaller. The size <strong>of</strong> the<br />

wobbling to be expected from planets orbiting stars <strong>of</strong><br />

one solar mass or larger is too small to be detected with<br />

present instruments. That we find an unseen companion<br />

<strong>of</strong> giant planetary size about nearly every star for which<br />

we could hope to detect the perturbations argues that<br />

most single stars have their retinue <strong>of</strong> planets.<br />

ATMOSPHERIC<br />

EVOLUTION<br />

If all planets had simply condensed from the early<br />

disk one would expect them to have essentially the same<br />

composition as the Sun: less than 2 percent heavy<br />

elements <strong>and</strong> the remainder hydrogen <strong>and</strong> helium. The<br />

giant planets appear to have approximately this composi.<br />

tion. The inner planets, on the other h<strong>and</strong>, are composed<br />

almost entirely <strong>of</strong> heavy elements. (It is perhaps worth<br />

noting that, without their hydrogen <strong>and</strong> helium atmospheres,<br />

the outer planets would be comparable in size to<br />

the inner planets. Jupiter would be about 4 times as<br />

massive as <strong>Earth</strong>, Saturn about 2 times, Uranus 0.3<br />

times, <strong>and</strong> Neptune 0.35 times.) <strong>Earth</strong> has almost no<br />

helium <strong>and</strong> relatively little hydrogen, most <strong>of</strong> it in the<br />

form <strong>of</strong> water. It is believed that the inner planets either<br />

formed without hydrogen or helium atmospheres or else<br />

lost these gases soon after formation.<br />

15

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