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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 $\mathrm{r}$
between sun and planet i.e. $\mathrm{T}^{2}=\mathrm{Kr}^{3}$ here $\mathrm{K}$ is constant. If the masses of sun and planet are $\mathrm{M}$ and $\mathrm{m}$ respectively then as per Newton's law of gravitation force of attraction between them is
$\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{r}^{2}}$, here $\mathrm{G}$ is gravitational constant. The relation between $\mathrm{G}$ and $\mathrm{K}$ is described as
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between sun and planet i.e. $\mathrm{T}^{2}=\mathrm{Kr}^{3}$ here $\mathrm{K}$ is constant. If the masses of sun and planet are $\mathrm{M}$ and $\mathrm{m}$ respectively then as per Newton's law of gravitation force of attraction between them is
$\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{r}^{2}}$, here $\mathrm{G}$ is gravitational constant. The relation between $\mathrm{G}$ and $\mathrm{K}$ is described as
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Verified Answer
The correct answer is:
$\mathrm{GMK}=4 \pi^{2}$
As we know, orbital speed, $V_{\text {orb }}=\sqrt{\frac{\text { GM }}{\text { r }}}$
Time period $\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{v}_{\text {orb }}}=\frac{2 \pi \mathrm{r}}{\sqrt{\mathrm{GM}}} \sqrt{\mathrm{r}}$
Squarring both sides,
$\mathrm{T}^{2}=\left(\frac{2 \pi \mathrm{r} \sqrt{\mathrm{r}}}{\sqrt{\mathrm{GM}}}\right)^{2}=\frac{4 \pi^{2}}{\mathrm{GM}} \cdot \mathrm{r}^{3}$
$\Rightarrow \quad \frac{\mathrm{T}^{2}}{\mathrm{r}^{3}}=\frac{4 \pi^{2}}{\mathrm{GM}}=\mathrm{K} \Rightarrow \mathrm{GMK}=4 \pi^{2}$
Time period $\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{v}_{\text {orb }}}=\frac{2 \pi \mathrm{r}}{\sqrt{\mathrm{GM}}} \sqrt{\mathrm{r}}$
Squarring both sides,
$\mathrm{T}^{2}=\left(\frac{2 \pi \mathrm{r} \sqrt{\mathrm{r}}}{\sqrt{\mathrm{GM}}}\right)^{2}=\frac{4 \pi^{2}}{\mathrm{GM}} \cdot \mathrm{r}^{3}$
$\Rightarrow \quad \frac{\mathrm{T}^{2}}{\mathrm{r}^{3}}=\frac{4 \pi^{2}}{\mathrm{GM}}=\mathrm{K} \Rightarrow \mathrm{GMK}=4 \pi^{2}$
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