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**Hint:**Analyze the question by trying to find a relation between ${T^2}$ and ${r^3}$ following the motion of planets around the sun in their orbits. One of Keppler’s laws justifies such a relation. Here, $T$ is the time period of revolution of the planet around the Sun and $r$ is the radius of its orbit, i.e., distance from the Sun.

**Complete step by step solution:**

The three laws of Kepler about planetary motion are:

The Law of Ellipses: The path of the planets about the sun is elliptical, with the centre of the sun being located at one focus.

The Law of Equal Areas: An imaginary line drawn from the centre of the sun to the centre of the planet will sweep out equal areas in equal intervals of time.

The Law of Harmonies: The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun.

Since,

$T = \dfrac{{2\pi }}{\omega }$ where, $\omega$ is the angular velocity of the body in circular motion

$v = \omega r$

$\Rightarrow \omega = \dfrac{v}{r}$

where, $v$ is the linear velocity of the body and $r$ is the radius of revolution

The orbital velocity of a planet revolving around the sun is given as

$\Rightarrow v = \sqrt {\dfrac{{G{M_S}}}{r}}$

Now,

$\Rightarrow T = \dfrac{{2\pi r}}{v}$

We put $v = \sqrt {\dfrac{{G{M_S}}}{r}}$ and get-

$\Rightarrow T = \dfrac{{2\pi r}}{{\sqrt {\dfrac{{G{M_S}}}{r}} }}$

where,

${M_s}$ is mass of Sun as given

$G$ is the gravitational constant i.e., $6.673 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}$

Squaring both sides,

$\Rightarrow {T^2} = \dfrac{{4{\pi ^2}{r^3}}}{{G{M_S}}}$

Therefore, the slope of the graph would be $\dfrac{{4{\pi ^2}}}{{{M_S}}}$ and it would be a straight-line graph as ${T^2} \propto {r^3}$

**The correct answer is [A], $\dfrac{{4{\pi ^2}}}{{{M_S}}}$.**

**Note:**Orbital velocity is the velocity at which a body revolves around another body in space. The orbital velocity is directly proportional to the mass of the body for which it is being calculated and inversely proportional to the radius.

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