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**Hint:**Law of Gravitation – Every object in the universe attract other object along the line of center for two objects that is proportional to the product of their masses and inversely proportional to the square of the separation between the two objects.

**Complete step by step answer:**

According to the law of Gravitation,

$ F = \dfrac{{GMm}}{{{r^2}}} $ where $ F $ is the gravitational force, $ M $ is the mass of the sun, $ m $ is the mass of the planet, $ r $ is the distance between the sun and the planet.

It is stated that the planet moves round the sun in a nearly circular orbit.

Let ,

$ T = K{r^x}{M^y}{G^z} $ …(i) where $ K $ is a dimensionless constant. The dimensions of various quantities are-

$ \left[ T \right] = T $ , $ \left[ r \right] = L $ , $ \left[ M \right] = M $

$ \left[ G \right] = \dfrac{{F{r^2}}}{{Mm}} $ $ = \dfrac{{ML{T^{ - 2}}{L^2}}}{{MM}} $

$ = {M^{ - 1}}{L^3}{T^{ - 2}} $

By substituting these dimensionally in equation (i), we get

$ {M^0}{L^0}{T^1} = {M^{y - z}}{L^{x + 3z}}{T^{ - 2z}} $

Equating the dimensions,

$ y - z = 0 $ , $ x + 3z = 0 $ , $ - 2z = 1 $

On solving,

$ x = \dfrac{3}{2} $ , $ y = - \dfrac{1}{2} $ , $ z = - \dfrac{1}{2} $

$ T = K{r^{\dfrac{3}{2}}}{M^{ - \dfrac{1}{2}}}{G^{ - \dfrac{1}{2}}} $

Squaring both sides,

$ {T^2} = \dfrac{{{K^2}{R^3}}}{{MG}} $

$ \Rightarrow {T^2} \propto {r^3} $ .

**Note:**

Planets revolve in elliptical orbits around the sun. It is closest to the sun at perihelion and farthest from the sun at aphelion.

Applying Kepler’s Third Law – We assume that the planets revolve in circular orbits around the sun.

$ {F_C} = {F_G} $ where $ {F_C} $ is the centripetal force and $ {F_G} $ is the gravitational force. The centripetal force arises due to the gravitational force.

$ \dfrac{{m{v^2}}}{r} = \dfrac{{GMm}}{{{r^2}}} $

$ v = \sqrt {\dfrac{{GM}}{r}} $ where $ v $ is the velocity with which the planet moves in an orbit around the sun.

$ T = \dfrac{{2\pi r}}{v} $ where $ T $ is the time period.

$ T = \dfrac{{2\pi r}}{{\sqrt {\dfrac{{GM}}{r}} }} $

$ T = \dfrac{{2\pi {r^{\dfrac{3}{2}}}}}{{{G^{\dfrac{1}{2}}}{M^{\dfrac{1}{2}}}}} $

$ T \propto {r^{\dfrac{3}{2}}} $

Squaring both sides,

$ {T^2} \propto {r^3} $

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