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Kepler's third law states that square of period of revolution (T) of a planet around the sun is proportional to third power of average distance r between the sun and planet, i.e. T² = Kr³, here K is constant. If the masses of the sun and planet are M and m respectively, then as per Newton's law of gravitational force of attraction between them is $F = \frac{G M m}{r^{2}}$ , here G is gravitational constant. The relation between G and K is described as [CBSE AIPMT 2015]

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$G K = 4 π^{2}$

$G M K = 4 π^{2}$

$K = G$

$K = I / G$

answer is B.

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Detailed Solution

The gravitational force of attraction between the planet and sun provides the centripetal force,

i.e., $\frac{G M m}{r^{2}} = \frac{m v^{2}}{r} \Rightarrow v = \sqrt{\frac{G M}{r}}$

The time period of planet will be

$T = \frac{2 π r}{v}$

$\Rightarrow T^{2} = \frac{4 π^{2} r^{2}}{\frac{G M}{r}} = \frac{4 π^{2} r^{3}}{G M}$ …(i)

Also, from the Kepler's third law, $T^{2} = K r^{3}$ …(ii)

From Eqs. (i) and (ii), we get

$\frac{4 π^{2} r^{3}}{G M} = K r^{3} \Rightarrow G M K = 4 π^{2}$

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