Question

if the gravitational force between the two masses varies as F is inversely proportional to R^3 then
,the orbital velocity of this satellite varies with the radius of orbit as
ITS ANS IS V INVERSELY PROPORTIONAL TO R.
how? can anyone help plz.​

Answer 1

Answer:

Given:

Gravitational force is inversely proportional to R³.

To find:

Relationship between Orbital velocity and radius

Concept:

Whenever a satellite revolves around a planet , the gravitational force is responsible for providing a Centripetal component for revolution.

Calculation:

$gravitation \: f = centripetal \: f$

$= > G \dfrac{m1 \times m2}{ {R }^{3} } = \dfrac{(m1) {v}^{2} }{R }$

$= > {v}^{2} = G \dfrac{m2}{ {R}^{2} }$

$= > v = \sqrt{ G \dfrac{m2}{ {R}^{2} } }$

$= > v =\dfrac{ \sqrt{Gm2}}{ R}$

$= > v \: \propto \: \dfrac{1}{R }$

So final answer :

$\boxed{ \red{ \sf{ \bold{ \huge{ v \: \propto \: \dfrac{1}{R } }}}}}$

Answer 2

$\huge \underline {\underline{ \mathfrak{ \green{Ans}wer \colon}}}$

If R³ is inversely proportional to the gravitational force then,

$\large{\boxed{\sf{g \: = \: G \dfrac{m_1 m_2}{R^3}}}}$

And formula for centripetal acceleration is :

$\large{\boxed{\sf{g \: = \: \dfrac{m_1 v^2}{r}}}}$

Now, equate both the equations :

$\implies {\sf{G \dfrac{m_1 m_2}{R^3} \: = \: \dfrac{m_1 v^2}{R}}} \\ \\ \implies {\sf{\dfrac{G m_2}{R^2} \: = \: \dfrac{v^2}{1}}} \\ \\ \implies {\sf{v \: = \: \sqrt{\dfrac{G m_2}{R^2}}}} \\ \\ \implies {\sf{v \: = \: \dfrac{\sqrt{G m_2}}{R}}} \\ \\ \implies {\sf{v \: \propto \: \dfrac{1}{R}}}$