*Summary*:

Yavin IV probably orbits Yavin at about 500,000 km radius, which is close enough to have
Yavin cover apparently 22° of its sky.The red gas giant planet Yavin is actually a
"brown dwarf", radiating cozy infrared radiation which is Yavin IV's main heat source.
The sun, while it illuminates Yavin IV nicely, is a minor source of heat. Yavin IV
turns about its axis once every 24 standard hours (1 ) and completes
one orbit around the gas giant Yavin in about 12 ¼ hours.

Well, you say, we already know how big Yavin looks from Yavin IV! Ralph McQuarrie's
concept painting of the X-Wing launch, and the scene in *A New Hope* which
is derived from it, show the red planet Yavin taking up a very goodly portion
of the sky. Although we see only a sliver of the actual planet, its curvature
implies that it takes up anything from 22 degrees to 50 degrees of sky arc
(for reference, our moon takes up 1/2 degree of sky arc). (We consider horizon-to-horizon
to be 180 degrees.) This apparent size
is determined by/will determine Yavin IV's orbital radius, as indicated below.
**West End Games'** *Galaxy Guide 2: Yavin and Bespin*
(1 ) states that the gas giant Yavin is 192,478 km in diameter, or 96,239
km in radius, which is what we use here.

This figure shows that the radius of Yavin IV's orbit, and the radius of the planet Yavin
perpendicular to this line, form a right triangle which sweeps out an angle, the
'apparent angular width' of half of Yavin, as seen from Yavin IV. The ratio of these
lengths forms the tangent of angle **Theta**. Just multiply theta by 2, to get Yavin's
full "angular width" in the sky.

r | Theta | Yavin covers: |

7 million km | 0.78 degrees | 1.56 degrees |

3 million km | 1.84 degrees | 3.68 degrees |

1 million km | 5.5 degrees | 11.0 degrees |

500,000 km | 10.9 degrees | 21.8 degrees |

250,000 km | 21 degrees | 42 degrees |

206,301 km | 25 degrees | 50 degrees |

96,200 (impossible) | 45 degrees | 90 degrees |

Here we immediately hit a problem with West End Games' data. The WEG GG2 states that Yavin IV
orbits Yavin roughly 3 times "in a standard year" (which we can take as being equivalent to
a Terran, or Earth year). This gives us an orbital **period** of some 4 months for the
moon. We toss **Kepler's Third Law** at the question, to determine Yavin IV's orbital
radius from this:

T² = 4 x PI² x (r³)/(G x M) | (2 ) |

T = period (time to complete one orbit)

r = radius at which our body is orbiting the big body

M = mass of the big body (in our case mass of gas giant Yavin)

G = Newtonian constant of gravitation, 6.670e-11 m²/kg-s²

The WEG GG2 states that Yavin's mass is 323.7 x the "Galactic standard" for humanoid-habitable planets-- if this is 323.7x mass of Earth, then Yavin weighs 1.93e27 kg. This is pretty close to the size of Jupiter, which weighs 1.9e27 kg. (3) Translate 4 months into seconds, and solve for r (radius of the orbit) which comes out in meters... translate it back into kilometers, and we get-- 7 million km?? This clearly cannot be correct, as at that range (see table above), Yavin would look only 3 times as big as our own moon! So, where the West End Games' data and the movie's data differ, we have to toss WEG out and go with what we saw in the movie. Leaving us with an orbital radius of 200,000-500,000 km.

As those of us who follow space probes know, the moon **Io**, Jupiter's closest Galilean (big)
satellite, is pretty tectonic. This tectonism is thought to be caused by heat induced from
tidal stresses, which in turn come from the moon being so close to Jupiter. Io orbits
Jupiter at 421,600 km (3 ). In the fictional accounts of Yavin IV
(*The Jedi Academy* trilogy by Kevin J. Anderson; and some of the
Dark Horse comics),
tectonic activity or volcanism is not really mentioned, although the WEG GG2 does state that Yavin IV has "fairly strong tectonic
action that lets heat escape from the very depths of the satellite."
(1 )(p.13). So we can probably assume some but not a tremendous amount of
volcanism... which means we prefer to put Yavin IV about 500,000 km out so as not to tear
it up too bad, but also have Yavin hanging large in the sky as portrayed.

(E = I/d²)((2 )p.383)

all other things being equal,
the amount of energy E received per area decreases with the inverse square of the distance d.
In other words, a planet twice as far away gets only ¼ as much sunlight.

We know that Earth, on the average, receives about 1370 W/m² from the sun .
Since Jupiter, whose distance we will use for this, is ~5x as far from the sun as Earth, it gets 1/d²
or 1/25 as much sunlight as we do ---> 54.8W/m². (5 ) This is not enough
to keep the flagstones warm in the Massassi courtyard.

So, being that we have our planetary system so far from the sun so the gas giant won't blow away, and we
want to keep Yavin IV from being torn apart by tectonics, how are we supposed to heat it warm enough to
be a tropical jungle Earthlike world? Here, our butts get saved by a neat little phenomenon called the
**Kelvin-Helmholtz mechanism**. (6 ) And if you can find an equation for
this puppy, let me know where, because I'd like to see it. To sum up, the Kelvin-Helmholtz mechanism
says that "if you have a big heavy gas planet, it squashes down on its own center so hard that it actually
generates heat." As a result of this, Jupiter's core temperature is estimated at 20,000K and its mean
cloud temperature is -121°C or 152K, which is warmer than we'd expect for its distance from the
sun.(6 )

So, OK, great, our Jupiter-sized gas giant Yavin might be marginally warmer than the liquid nitrogen in the
chemistry department's vacuum flask. How exactly is this supposed to keep Yavin IV warm? At this point, we
commence to explore the **black-body properties** of Yavin, to determine just how warm this gas giant
would have to be.

We'll use several equations. **Stefan-Boltzmann's Law:**

E = x T^4 |

= the Stefan-Boltzmann Constant, 5.57e-8 W/m² K^4

T = the temperature in Kelvin K.

An equation for **Calculation of Solar Constants:**

Flux = Total Power/Area At the Radius |

The "Flux" is the energy/unit area, or the W/m² seen above,

and "Total Power" comes from the

"Area at the Radius" is the surface area of a sphere calculated at the radius r where we want to know what the energy flux is. Note that "energy flux", and "solar constant," are the same thing when we are looking at how much sunlight energy passes through a certain area.

The **Body's Total Power Emitted** comes from

Power = x T^4 x (surface area of body) |

the same constant as above;

And, of course, the **Surface Area of a Sphere** is

A = 4 x PI x r² |

We have established that Yavin IV orbits at about 500,000 km from Yavin. So now, we can calculate the
total energy that Yavin has to emit, in order to give Yavin IV an energy flux of 1300 W/m² at this
distance. We use the **Solar Constant** equation above, solving for Total Power.

With the distance r converted to meters, we get 4.084e21 W emitted out of Yavin at large. (Meaning
4.084 x 10^21 watts)

We plug in Yavin's radius to the **Surface Area of a Sphere** to get Yavin's surface area.
Then we put this back into the **Solar Constant** equation, this time solving for E, the energy flux
coming off Yavin's surface. We get 35090 W/m².

And using the **Stefan-Boltzmann** equation, this gives us a surface temperature for Yavin
of 890 K. Remember that Jupiter's temperature is 152 K! Water boils at 373 K, so Yavin's surface would
be pretty toasty. Is this possible?

Also, in size (diameter), a brown dwarf can be quite close to that of Jupiter. In fact the known brown dwarf Gliese 229B is about Jupiter's size, and has an estimated surface temperature of 900 K. Perfect! All these brown dwarf facts are (9 ).

What sort of radiation is the ruddy "brown" dwarf Yavin emitting? We can consider the following diagrams:

The bottom diagram is derived from the **Wien's Displacement Formula:**

max emittance = 2897 / T |

with max emittance being the wavelength of peak energy emission (in µm) at temperature T (in Kelvin).

And when we do the math we find that indeed Yavin at 890 K would have its peak energy emission at about 3.25 µm, which is in the near infrared (see figure above).

Thus Yavin is not only a red gas giant planet, it is an infrared gas giant.

Flux = Total Power/Area At the Radius |

As "Flux", use a 100W light bulb for "Total Power" (=100W) and solve for the radius from "Area At The Radius", where "Area At the Radius" is the surface area of a sphere:

A = 4 x PI x r² |

We find that this amount of solar flux turns out to be equivalent to the light from a 100W bulb about 1 foot away. So Yavin and Yavin IV are lit a little better than your living room.

We plug these numbers into the equation for **Kepler's Third Law:**

T² = 4 x PI² x (r³)/(G x M) | (2 ) |

T = period (time to complete one orbit)

r = radius at which our body is orbiting the big body

M = mass of the big body (in our case mass of gas giant Yavin)

G = Newtonian constant of gravitation, 6.670e-11 m²/kg-s²

Solving for period T, we find that Yavin IV must orbit Yavin in 12 ¼ hours if Yavin weighs 20 times
as much as Jupiter, and the orbital period would be 7.5 hours if Yavin weighed 50x as much. Whee!

We'll choose the more conservative estimate, as we wouldn't want the Jedi students to get vertigo.

2)

3) Calvin J. Hamilton's Jupiter Info page

5) Todd Barber's Online From Jupiter

6) Bill Arnett's "Nine Planets: Jupiter"

9) All brown dwarf facts are:

Berkeley Astronomy Department and Berkeley Stars 'n' Dwarves Picture Cornell's Astronomy 201 David Griffin On Brown Dwarves |

11)Brown Dwarf Gliese 229B in Lepus, CalTech and Johns Hopkins astronomers

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