Gravitation - Result Question 2
2. 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., $T^{2}=Kr^{3}$ 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 $F=\frac{GMm}{r^{2}}$, here $G$ is gravitational constant.
The relation between $G$ and $K$ is described as
(a) $GMK=4 \pi^{2}$
(b) $K=G$
(c) $K=\frac{1}{G}$
(d) $GK=4 \pi^{2}$
[2015]
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Answer:
Correct Answer: 2. (a)
Solution:
- (a) As we know, orbital speed, $v _{\text{orb }}=\sqrt{\frac{G M}{r}}$
Time period $T=\frac{2 \pi r}{v _{\text{orb }}}=\frac{2 \pi r}{\sqrt{G M}} \sqrt{r}$ Squarring both sides,
$T^{2}=(\frac{2 \pi r \sqrt{r}}{\sqrt{G M}})^{2}=\frac{4 \pi^{2}}{G M} \cdot r^{3}$
$\Rightarrow \frac{T^{2}}{r^{3}}=\frac{4 \pi^{2}}{GM}=K$
$\Rightarrow GMK=4 \pi^{2}$
