- Home
- Standard 11
- Physics
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.
$\therefore \;{T^2} = k{r^3}$
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 $F = \frac{{GMm}}{{{r^2}}}$ , here $G$ gravitational constant . The relation between $G$ and $K$ is described as
$GK=4$${\pi ^2}$
$GMK=4$${\pi ^2}$
$K=G$
$K=$$\frac{1}{G}$
Solution
Gravitational force of attraction between sun and planet provides centripetal force for the orbit of planet.
$\therefore \frac{{GMm}}{{{r^2}}} = \frac{{m{v^2}}}{r}\,;\,{v^2} = \frac{{GM}}{r}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( i \right)$
Time period of the planet is given by
$T = \frac{{2\pi r}}{v},{T^2} = \frac{{4{\pi ^2}{r^2}}}{{{v^2}}} = \frac{{4{\pi ^2}{r^2}}}{{\left( {\frac{{GM}}{r}} \right)}}$ $(Using\,\,(i))$
${T^2} = \frac{{4{\pi ^2}{r^3}}}{{GM}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( {ii} \right)$
According to question,
${T^2} = K{r^3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( {iii} \right)$
Comparing equation $(ii)$ and $(iii)$, we get
$K = \frac{{4{\pi ^2}}}{{GM}}\,\,\,\therefore GMK = 4{\pi ^2}$
