I think the answer is Dust. Moons and stars definitely don't seem likely and dark particles, I am not even sure what those are. But I have seen rings on other planets before. Hope this helps. :)
The answer is A ..........
Answer:
D = 2.38 m
Explanation:
This exercise is a diffraction problem where we must be able to separate the license plate numbers, so we must use a criterion to know when two light sources are separated, let's use the Rayleigh criterion, according to this criterion two light sources are separated if The maximum diffraction of a point coincides with the first minimum of the second point, so we can use the diffraction equation for a slit
a sin θ = m λ
Where the first minimum occurs for m = 1, as in these experiments the angle is very small, we can approximate the sine to the angle
θ = λ / a
Also when we use a circular aperture instead of slits, we must use polar coordinates, which introduce a numerical constant
θ = 1.22 λ / D
Where D is the circular tightness
Let's apply this equation to our case
D = 1.22 λ / θ
To calculate the angles let's use trigonometry
tan θ = y / x
θ = tan⁻¹ y / x
θ = tan⁻¹ (4.30 10⁻² / 140 10³)
θ = tan⁻¹ (3.07 10⁻⁷)
θ = 3.07 10⁻⁷ rad
Let's calculate
D = 1.22 600 10⁻⁹ / 3.07 10⁻⁷
D = 2.38 m
According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>
In other words, this law states a relation between the orbital period 
of a body (moon, planet, satellite) orbiting a greater body in space with the size 
of its orbit.
This Law is originally expressed as follows:
<h2>
(1)
</h2>
Where;
is the Gravitational Constant and its value is 
is the mass of Jupiter
is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)
If we want to find the period, we have to express equation (1) as written below and substitute all the values:
<h2>
(2)
</h2>
Then:
<h2>
(3)
</h2>
Which is the same as:
<h2>
</h2>
Therefore, the answer is:
The orbital period of Io is 42.482 h
