Question

If g is inversely proportional to R cube, instead of R square ,then the relation between time period of the satellite near earth's surface and radius R will be

Answer 1

Answer:

Given:

g is inversely proportional to R³

To find:

Relationship between time period of satellite and Radius of Earth.

Concept:

When a satellite is revolving around the Earth , the gravitational force is responsible for providing Centripetal Force.

Calculation:

$\therefore \: g \: \propto \dfrac{1}{ {R}^{3} }$

Introducing any constant k

$\therefore \: g \: = \dfrac{k}{ {R}^{3} }$

As per our "Concept" :

$mg = \dfrac{m {v}^{2} }{R}$

$= > m\dfrac{k}{ {R}^{3} } = \dfrac{m {v}^{2} }{R}$

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

Now , Time period is as :

$time = \dfrac{distance}{speed}$

$= > time = \dfrac{2\pi R}{v}$

$= > time = \dfrac{2\pi R}{ \bigg( \dfrac{ \sqrt{k} }{R} \bigg)}$

$= > time = \dfrac{2\pi}{ \sqrt{k} } \times {R}^{2}$

$= > time \: \propto \: {R}^{2}$

So final answer :

$\boxed{\boxed{ \sf{ \red{ \bold{ \huge{ time \: \propto \: {R}^{2}}}}}}}$

Answer 2

$\large{\underline{\underline{\mathbf\green{Question\::}}}}$

If g is inversely proportional to R cube, instead of R square ,then the relation between time period of the satellite near earth's surface and radius R will be.

$\large{\underline{\underline{\mathbf\red{SOLUTION\::}}}}$

$\implies$ g $\propto$ 1/r³

$\implies$ g = k/r³ $\pink{( K\: is\: a\: constant)}$

$\implies$ mg = mv²/r $\orange{( f = mg, p = mv)}$

$\implies$ m. k/r³= mv /r

$\implies$ v = √k/r

$\implies$ we know that,

$\implies$ time = distance / speed

$\implies$ time = 2πr /(√k/r)

$\implies$ time = 2π/√K × r²

$\implies$ $\huge\red{\boxed{ time \propto r^2}}$