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The temperature within the disk decreased with increasing distance from the Sun, much as the planets’ temperatures vary with position today. As the disk cooled, the gases interacted chemically to produce compounds; eventually these compounds condensed into liquid droplets or solid grains. This is similar to the process by which raindrops on Earth condense from moist air as it rises over a mountain.

Let’s look in more detail at how material condensed at different places in the maturing disk ( [link] ). The first materials to form solid grains were the metals and various rock-forming silicates. As the temperature dropped, these were joined throughout much of the solar nebula by sulfur compounds and by carbon- and water-rich silicates, such as those now found abundantly among the asteroids. However, in the inner parts of the disk, the temperature never dropped low enough for such materials as ice or carbonaceous organic compounds to condense, so they were lacking on the innermost planets.

Chemical condensation sequence in the solar nebula.

A figure showing the chemical condensation sequence in the solar nebula. At the upper left of the figure is the Sun, and from left to right across the top are the planets and bodies Mercury, Venus, Earth, Mars, Asteroid, Jupiter, Saturn, Uranus, Neptune, and Pluto. At the bottom of the figure is an axis labeled “Temperature (K)” ranging from 1700 on the left to 0 on the right. A label above 1600 K reads “Metal oxides (e.g. Al 2 O 3)”. A label above 1400 K reads “Iron-nickel alloy”. A label above 1200 to 900 K reads “Silicate minerals”. A label above 1000 to 700 K reads “Iron oxide (Fe O) and Olivine (Fe 2 Si O 4 and Mg 2 Si O 4”. A label above 600 K reads “Triolite (Fe S)”. A label above 450 to 350 K reads “Hydrated minerals (containing H 2 O”. A label above 250 K reads “Water (H 2 O)”. A label above 150 K reads “Ammonia (N H 3)”. A label above 100 K reads “Methane (C H 4)”.
The scale along the bottom shows temperature; above are the materials that would condense out at each temperature under the conditions expected to prevail in the nebula.

Far from the Sun, cooler temperatures allowed the oxygen to combine with hydrogen and condense in the form of water (H 2 O) ice. Beyond the orbit of Saturn, carbon and nitrogen combined with hydrogen to make ices such as methane (CH 4 ) and ammonia (NH 3 ). This sequence of events explains the basic chemical composition differences among various regions of the solar system.

Rotation of the solar nebula

We can use the concept of angular momentum to trace the evolution of the collapsing solar nebula. The angular momentum of an object is proportional to the square of its size (diameter) times its period of rotation ( D 2 / P ). If angular momentum is conserved, then any change in the size of a nebula must be compensated for by a proportional change in period, in order to keep D 2 / P constant. Suppose the solar nebula began with a diameter of 10,000 AU and a rotation period of 1 million years. What is its rotation period when it has shrunk to the size of Pluto’s orbit, which Appendix F tells us has a radius of about 40 AU?

Solution

We are given that the final diameter of the solar nebula is about 80 AU. Noting the initial state before the collapse and the final state at Pluto’s orbit, then

P final P initial = ( D final D initial ) 2 = ( 80 10,000 ) 2 = ( 0.008 ) 2 = 0.000064

With P initial equal to 1,000,000 years, P final , the new rotation period, is 64 years. This is a lot shorter than the actual time Pluto takes to go around the Sun, but it gives you a sense of the kind of speeding up the conservation of angular momentum can produce. As we noted earlier, other mechanisms helped the material in the disk lose angular momentum before the planets fully formed.

Check your learning

What would the rotation period of the nebula in our example be when it had shrunk to the size of Jupiter’s orbit?

Answer:

The period of the rotating nebula is inversely proportional to D 2 . As we have just seen, P final P initial = ( D final D initial ) 2 . Initially, we have P initial = 10 6 yr and D initial = 10 4 AU. Then, if D final is in AU, P final (in years) is given by P final = 0.01 D final 2 . If Jupiter’s orbit has a radius of 5.2 AU, then the diameter is 10.4 AU. The period is then 1.08 years.

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Joy Reply
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Source:  OpenStax, Astronomy. OpenStax CNX. Apr 12, 2017 Download for free at http://cnx.org/content/col11992/1.13
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