Question

Hello friends!!

Derive the inverse-square rules for gravitation Force .
That is ,

$= > f \infty \frac{1}{ {r}^{2} }$
Prove this !!

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Answer 1

$\textbf{Newton's Law of Gravitation :}$

In this the force between two bodies is directly proportional to their masses and inversally proportional to the square of distance between them..

$\textbf{Newton's Inverse Square Law :}$

According to it the force between two bodies is inversally proportional to the square of distance between them..

i.e.

$F \: \infty \: 1 \: \div \: d {}^{2}$

Or

$F \: \infty \: 1 \: \div \: r {}^{2}$

$\textbf{Derivation is in Attachment!!}$

$\textbf{What is Kepler's Third Law...?}$

The square of time period of revolution of planet around sun is directly proportional to the cube of length of semi major axis...

i.e.

$= > {T}^{2} \infty {R}^{3}$

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Answer 2

Newton’s Law of Gravitation : 

In this the force between two bodies is directly proportional to their masses and inversally proportional to the square of distance between them..

\textbf{Newton's Inverse Square Law :}Newton’s Inverse Square Law : 

According to it the force between two bodies is inversally proportional to the square of distance between them..

i.e.

F \: \infty \: 1 \: \div \: d {}^{2}F∞1÷d2 

Or

F \: \infty \: 1 \: \div \: r {}^{2}F∞1÷r2 

\textbf{Derivation is in Attachment!!}Derivation is in Attachment!! 

\textbf{What is Kepler's Third Law...?}What is Kepler’s Third Law...? 

The square of time period of revolution of planet around sun is directly proportional to the cube of length of semi major axis...

i.e. 

= > {T}^{2} \infty {R}^{3}=>T2∞R3 

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