A simple temperature model for the solar nebula disk is
Tdisk(a) = 20 × SQRT 100 AU/ a K, (1) where a should be expressed in AU in this equation.
(a) Find the temperature in the solar nebula disk at the orbit of Mer- cury, Earth and Jupiter.
(b) Where is the water ice line located in this disk? [Use the conden- sation sequence values provided in the textbook]
(c) Mercury is mostly made of metal, the Earth mostly made of rock and Jupiter mostly made of gas. Are these three planetary bulk compositions consistent with the condensation sequence, if one follows the above temperature model for the Solar nebula?
To answer these questions, we need to substitute the values of 'a' into the temperature equation and solve for each case. Let's go through each part step by step.
(a) Find the temperature in the solar nebula disk at the orbit of Mercury, Earth, and Jupiter:
1. Mercury: The average distance of Mercury from the Sun is approximately 0.39 AU. Substitute this value into the temperature equation:
Tdisk(0.39) = 20 × sqrt(100/0.39)
Using a calculator, calculate the square root and divide 100 by 0.39, then multiply it by 20:
Tdisk(0.39) ≈ 20 × sqrt(256.41) ≈ 20 × 16.02 ≈ 320.34 K
So, the temperature at the orbit of Mercury in the solar nebula disk is approximately 320.34 Kelvin.
2. Earth: The average distance of Earth from the Sun is about 1 AU. Substitute this value into the temperature equation:
Tdisk(1) = 20 × sqrt(100/1)
Again, using a calculator:
Tdisk(1) ≈ 20 × sqrt(100) ≈ 20 × 10 ≈ 200 K
Therefore, the temperature at the orbit of Earth in the solar nebula disk is approximately 200 Kelvin.
3. Jupiter: The average distance of Jupiter from the Sun is roughly 5.2 AU. Substitute this value into the temperature equation:
Tdisk(5.2) = 20 × sqrt(100/5.2)
Applying the calculations in the same manner as before:
Tdisk(5.2) ≈ 20 × sqrt(19.23) ≈ 20 × 4.38 ≈ 87.62 K
So, the temperature at the orbit of Jupiter in the solar nebula disk is approximately 87.62 Kelvin.
(b) To determine the location of the water ice line in the disk, we need to refer to the condensation sequence values provided in the textbook. The water ice line is the distance from the Sun beyond which water can exist as a solid (ice).
The condensation sequence values describe the temperatures at which different compounds condense or transition from a gas to a solid form. Water condenses out of the gas phase at a temperature close to 170 Kelvin.
Substituting this value into the temperature equation, we can solve for 'a':
170 = 20 × sqrt(100/a)
Divide both sides by 20 and square both sides to isolate 'a':
(170/20)^2 = 100/a
(8.5)^2 = 100/a
Take the square root of both sides:
8.5 = 10 / sqrt(a)
Multiply both sides by sqrt(a):
8.5 √a = 10
Divide both sides by 8.5:
√a = 10/8.5
Square both sides:
a = (10/8.5)^2
Calculating this value will give you the location of the water ice line in the disk.
(c) To determine if the bulk compositions of Mercury, Earth, and Jupiter are consistent with the condensation sequence, we need to compare the temperatures in the solar nebula to the condensation temperatures of various materials.
Mercury, being mostly made of metal, would have condensed and formed closer to the Sun where temperatures were higher. Therefore, its composition is consistent with the higher temperatures near the inner regions of the solar nebula.
Earth, being mostly made of rock, would have formed farther from the Sun where temperatures were lower. This is consistent with its location in the habitable zone, where temperatures are suitable for liquid water and the formation of rocky planets.
Jupiter, being mostly made of gas, would have formed in the outer regions of the solar nebula where temperatures were much colder. This is consistent with its location as a gas giant planet.
In summary, the compositions of Mercury, Earth, and Jupiter align with the condensation sequence and are consistent with the temperature model for the solar nebula disk.