Question

Q]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. T²= Kr³
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 =(GMm
)/r² , here
G is gravitational constant. The relation between G
and K is described as?​

Answer 1

$\begin{gathered}\large {\boxed{\sf{\mid{\overline {\underline {\star ANSWER ::}}}\mid}}}\end{gathered}$

CORRECT ANSWER :-

$\underline{\large \boxed{ \bold{ \star \: GMK = 4 \pi {}^{2} }}}$

EXPLAINATION :-

⟼ The gravitational force of attraction provides the centripetal force to the planet's orbital circular motion. Which can be noted as :

$\dag \: \: \bold{ \frac{GMm}{ {r}^{2} } = \frac{m {v}^{2} }{r} } \\ \\ \bold{ \frac{GM}{ {r} } = \frac{ {v}^{2} }{ 1} } \\ \\ {\bold{v = \sqrt{ \frac{GM}{r} } }}$

⟼ The time period (T) for the planet for one revolution can be given as :

$\dag \: \: \bold{ T = \frac{2 \pi r}{v} } \\ \\ \bold{T = \frac{2 \pi r}{ \sqrt{ \frac{GM}{r} } } } \\ \bold{(by \: substituting \: the \: value \: of \: v)} \\ \\ \bold{T {}^{2} = \frac{(2 \pi r) {}^{2} }{ ({\sqrt{ \frac{GM}{r}})^{2} } } } \\ \bold{(by \: squaring \: on \: both \: sides)} \\ \\ \bold{T {}^{2} = \frac{(2 \pi r) {}^{2} }{ {{\frac{GM}{r}} } } } \\ \\ \dag \: \:\boxed{\bold{T {}^{2} = \frac{4 \pi {}^{2} r {}^{3} }{ {{GM} } } } \: \: \: \: \: \: \:... \bold{equation \: 1}}$

⟼ Given Equation :

$\dag \: \: \boxed{\bold{T {}^{2} = K {r}^{3} } \: \: \: \: ... \bold{equation \: 2}}$

⟼ Comparing equation 1 and equation 2 :

$\bold{\frac{4 \pi {}^{2} {r}^{3} }{GM} = K {r}^{3} } \\ \\ \bold{\frac{4 \pi {}^{2} }{GM} = K } \\ \\ \star \: \underline{\bold{GMK = 4 { \pi}^{2} }}$

@SweetestBitter

Answer 2

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