Question

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 sun and planet i.e. ${T^2} = K{r^3}$ , here K is constant. If the masses of sun and planet are M and m respectively then as per Newton's law of gravitation force of attraction between them is $$F = \dfrac{{GMm}}{2}$$ , here G is gravitational constant. The relation between G and K is described as
A. $GMK = 4{\pi ^2}$
B. K=G
C. $K = \dfrac{1}{G}$
D. $GK = 4{\pi ^2}$

Answer 1

Hint: The planets’ orbital circular motion is controlled by the centripetal force exerted by gravitation. Therefore, $\dfrac{{GMm}}{{{r^2}}} = \dfrac{{m{v^2}}}{r}$. So, the time period of the planets for one complete revolution is $T = \dfrac{{2\pi r}}{v} = \dfrac{{2\pi r}}{{\sqrt {\dfrac{{GM}}{r}} }}$. Now ${T^2} = K{r^3}$, given. Equate all these relations and simplify to get the relation between K and G.
Complete step-by-step solution:
The planets’ orbital circular motion is controlled by the centripetal force exerted by gravitation.
Therefore, $\dfrac{{GMm}}{{{r^2}}} = \dfrac{{m{v^2}}}{r}$
$ \Rightarrow \dfrac{{GM}}{r} = {v^2}$
$ \Rightarrow v = \sqrt {\dfrac{{GM}}{r}} $
So, the time period of the planets for one complete revolution is $T = \dfrac{{2\pi r}}{v} = \dfrac{{2\pi r}}{{\sqrt {\dfrac{{GM}}{r}} }}$
Squaring each side, we have
${T^2} = \dfrac{{4{\pi ^2}{r^2}}}{{\dfrac{{GM}}{r}}}$
$ \Rightarrow {T^2} = \dfrac{{4{\pi ^2}{r^3}}}{{GM}}$
$ \Rightarrow KGM = 4{\pi ^2}$

Therefore, the correct answer is option (A).

Note:
Note that orbital circular motion of the planets is controlled by the centripetal force exerted by gravitation. Again, the time period of the planets for one complete revolution is $T = \dfrac{{2\pi r}}{v} = \dfrac{{2\pi r}}{{\sqrt {\dfrac{{GM}}{r}} }}$.