Answer:
Orbital period, T = 1.00074 years
Explanation:
It is given that,
Orbital radius of a solar system planet,
The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :
M is the mass of the sun
T = 31559467.6761 s
T = 1.00074 years
So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.
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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