Question

Kepler's third law states that the square of period of revolution $\left(T\right)$ of a planet around the sun is proportional to the third power of average distance, $r$ between the sun and the planet i.e $T^{2}=Kr^{3 \, }$ .Here, $K$ is constant.
If masses of the sun and the planet are $M$ and $m$ respectively, then as per Newton's law of gravitation force of attraction between them is $F=\frac{G M m}{r^{2}}$ , where $G$ is gravitational constant. The relation between $G$ and $K$ is described as

• A. $GK=4\pi ^{2}$
• B. $GMK=4\pi ^{2}$
• C. $K=G$
• D. $K=\frac{1}{G}$

Answer 1

Answer: (B)

As, $\frac{G M m}{r^{2}}=\frac{m v^{2}}{r}=$ centripetal force
$\Rightarrow v^{2}=\frac{G M}{r}\Rightarrow T=\frac{2 \pi r}{v}$
$\Rightarrow T^{2}=\frac{4 \pi ^{2} r^{2}}{v^{2}}$
Out putting the value of $v^{2}$ , we get
$T^{2}=\frac{4 \left(\pi \right)^{2} r^{2}}{\left(\frac{G M}{r}\right)}\Rightarrow T^{2}=\frac{4 \left(\pi \right)^{2} r^{3}}{G M}$
$\Rightarrow T^{2}=Kr^{3}$
Here, $K$ is given by,
$\frac{4 \pi ^{2}}{G M}=K\Rightarrow GMK=4\pi ^{2}$