Kepler’s second law is based on:
A) Newton’s first law
B) Newton’s second law
C) Special theory of relativity
D) Conservation of angular momentum

Hint: Kepler gave three laws. His laws of planetary motion tell us how a planet moves in an elliptical orbit, equal areas of space are swept in equal intervals of time and the time period of a planet is proportional to the size of the orbit i.e. the size of semi-major axis.

Complete solution:
Kepler’s second law of planetary motion states that if an imaginary line is drawn from the centre of any planet to the centre of the sun then the equal amount of areas will be swept in equal time intervals.
It means the area with respect to time is always constant.
That is, $\dfrac{{dA}}{{dt}} = $ constant ……..(i)
Now, let us assume a wedge swept by the line in a small time interval $dt$. Let $r$ be the distance between the centre of the plant to the centre of the sun. Also, assume that the angle made at the centre of the sun by this wedge in a small time interval be $d\theta $ and the angular velocity of the planet is assumed to be $\omega $.
As, we now that,
Length of the arc $ = r\theta $
Now, the length of wedge arc $ = rd\theta $
As the angle is very small, it can be consider as a triangle in shape
The area of wedge triangle is90 given by,
$dA = \dfrac{1}{2}$(radius of arc)(length of wedge arc)
$ \Rightarrow dA = \dfrac{1}{2}{r^2}d\theta $ ……….(ii)
Now divide the above equation by $dt$ on both sides, we get
$\dfrac{{dA}}{{dt}} = \dfrac{1}{2}\dfrac{{{r^2}d\theta }}{{dt}}$
As we also know that the angular velocity is given by
$\omega = \dfrac{{d\theta }}{{dt}}$
Now put that in the equation (ii), we get
$ \Rightarrow \dfrac{{dA}}{{dt}} = \dfrac{{\omega {r^2}}}{2}$ ……..(iii)
Also, the angular momentum of the body of mass $m$ is given by,
$L = m\omega {r^2}$
Here, rewrite this equation, we get
$ \Rightarrow \omega {r^2} = \dfrac{L}{{2m}}$
Now put the above term in the equation (iii), we get
$ \Rightarrow \dfrac{{dA}}{{dt}} = \dfrac{L}{{2m}}$
Now compare this with the equation (i), we get
$\dfrac{L}{{2m}} = $ constant
As here the mass of the planet is the constant term
So, $ \therefore L = $ constant
Hence, the angular momentum remains constant which means that the angular momentum is conserved.
Therefore, the correct option is (D).

Note: Remember that this Kepler’s second law of planetary motion is also known as the law of equal areas. Also, Newton’s first law of motion states that the body which is at rest or in motion will always remain in that state until and unless any external force is applied to it. The planets sweep equal areas in equal intervals of time on both the sides of the focus.