Question

Kepler's law of plantery motion ​

Answer 1

$\large{\underline{\boxed{\purple{\tt{\dag Kepler's\:Law\:of\:Planetary \:Motion:-: }}}}}$

Kepler gave three laws , which are :

• ❒ The orbit of a planet is an ellipse with the Sun at one of the two foci.
• ❒ A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
• ❒ The square of a planet's orbital period is proportional to the cube of the semi-major of its orbit.

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$\Large\purple{\underline{\orange{\mathbb{\leadsto MorE \:\:To\:\:KnoW:-}}}}$

Using third Kepler's law of Planetary motion , we can derive the inverse square relaⁿ :-

✤ The law states that gravitational attraction force between two point masses is inversely proportional to the square of the distance between them.

✤ The Newton's inverse square law is deduced from Kepler's 'Third law' of planetary motion.

We know that centripetal force acting on a body can be calculated as ,

$\large\blue{\underline{\boxed{\pink{\tt{\dag Force_{centripetal}\:\:=\:\:\dfrac{mv^2}{r} }}}}}$

Where ,

• ✒️ m is mass.
• ✒️ r is radius.
• ✒️ v is Velocity.

$\tt:\implies Force =\dfrac{mv^2}{r}$

$\tt:\implies Force =\dfrac{m}{r}\times v^2$

$\tt:\implies Force =\dfrac{m}{r}\times\bigg(\dfrac{2\pi r}{T}\bigg)^2$

$\tt:\implies Force =\dfrac{m}{r}\times \dfrac{4\pi^2r^2}{T^2}$

$\tt:\implies Force=\dfrac{4\pi^2mr^2}{T^2r}$

$\tt:\implies Force=\dfrac{4\pi^2m}{r^2}\times\dfrac{r^3}{T^2}$

$\bf Here\:4\pi^2r^2m\:and\:\dfrac{r^3}{T^2}\:is\: constant.$

$\purple{\underset{\green{\bf Inverse\:square\:relation}}{\underbrace{\underline\blue{\boxed{\red{\tt{\longmapsto Force\:\:\:\:\propto\:\:\:\:\dfrac{1}{r^2}}}}}}}}$

Answer 2

Answer:

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