Kepler's third law states that the square of | vedclass

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(B) The gravitational force of attraction between the sun and the planet provides the necessary centripetal force for the planet's orbit.$	herefore \

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$GMK = 4\pi^2$

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(B) The gravitational force of attraction between the sun and the planet provides the necessary centripetal force for the planet's orbit.$	herefore \frac{GMm}{r^2} = \frac{mv^2}{r} \implies v^2 = \frac{GM}{r} \quad \dots(i)$The time period $(T)$ of the planet is given by $T = \frac{2\pi r}{v}$.Squaring both sides,we get $T^2 = \frac{4\pi^2 r^2}{v^2}$.Substituting the value of $v^2$ from equation $(i)$:$T^2 = \frac{4\pi^2 r^2}{(\frac{GM}{r})} = \frac{4\pi^2 r^3}{GM} \quad \dots(ii)$According to the problem,Kepler's third law is given as:$T^2 = Kr^3 \quad \dots(iii)$Comparing equations $(ii)$ and $(iii)$,we get:$K = \frac{4\pi^2}{GM} \implies GMK = 4\pi^2$.

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