Describe the different processes that lead to substantial internal heat sources for Jupiter and Saturn. Since these two objects
generate much of their energy internally, should they be called stars instead of planets?
1 answer:
Explanation:
The internal heat sources for Jupiter and Saturn derive from primordial heat resulting from the initial gravitational contraction of each planet. Jupiter also generates heat by slow contraction, which liberates substantial gravitational energy. A significant part of Saturn’s heat comes from the release of gravitational energy from helium separating from the lighter hydrogen and sinking to its core. What one considers to be a star is a matter of definition, as we discuss in more detail in the chapter on The Birth of Stars and the Discovery of Planets outside the Solar System. While both Jupiter and Saturn generate much of their energy internally, they are not large enough (by a significant factor) to support nuclear reactions in their interiors, and so are not considered to be stars.
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a) Total power output:
b) The relative percentage change of power output is 1.67%
c) The intensity of the radiation on Mars is
Explanation:
a)
The intensity of electromagnetic radiation is given by
where
P is the power output
A is the surface area considered
In this problem, we have
is the intensity of the solar radiation at the Earth
The area to be considered is area of a sphere of radius
(distance Earth-Sun)
Therefore
And now, using the first equation, we can find the total power output of the Sun:
b)
The energy of the solar radiation is directly proportional to its frequency, given the relationship
where E is the energy, h is the Planck's constant, f is the frequency.
Also, the power output of the Sun is directly proportional to the energy,
where t is the time.
This means that the power output is proportional to the frequency:
Here the frequency increases by 1 MHz: the original frequency was
so the relative percentage change in frequency is
And therefore, the power also increases by 1.67 %.
c)
In this second case, we have to calculate the new power output of the Sun:
Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is
Now we can find the radiation intensity with the equation
Where the area is
And substituting,
Learn more about electromagnetic radiation:
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