Question

a small planet is revolving around a very massive star in a circular orbit of radius R with a period of revolution T.if the gravitational force between the planet and the star is proportional to the R raised to minus 5 by 2 then T would be proportional to​

Answer 1

Answer:

$T\propto  R^{\frac{7}{4}$

Explanation:

We are given that a small planet  is revolving around a very massive star in a circular orbit of radius R with a period of revolution T.

Gravitational force between the planet and the star  is proportional to $R^{\frac{5}{2}}$

We have to find T would be proportional to what.

$F\propto R^{\frac{-5}{2}}$

$F=GmMR^{-\frac{5}{2}}$

Where m=Mass of small planet

M=Mass of massive star

Centripetal force=Gravitational force

$mRw^2=GmMR^{-\frac{5}{2}}$

$w=\frac{2\pi}{T}$

$mR(\frac{2\pi}{T})^2=GmMR^{-\frac{5}{2}}$

$\frac{1}{T^2}=GMR^{-\frac{7}{2}}$

$T^2=\frac{R^{\frac{7}{2}}}{GM}$

$T\propto \sqrt{R^{\frac{7}{2}}}$

$T\propto R^{\frac{7}{4}$

Hence, T would be proportional to $R^{\frac{7}{2}}$

Answer 2

ANSWER:

The time taken to complete one rotation of massive star by the planet is proportional to $\mathbf{R}^{7 / 4}$.

EXPLANATION:

A small planet revolving around a massive star will be exhibiting centripetal force as the planet is moving in circular motion. The centripetal force exerted on the small planet will be directed towards the massive star. Thus in circular motion of planet around a massive star will have their gravitational force acting between them equal to the centripetal force exerted on the planet.

Thus,

Gravitational force between planet and star=Centripetal force exerted on planet

It is given that gravitational force is proportional to $\mathrm{R}^{-5 / 2}$, where R is the radius of circular orbit.

$\text {Gravitational force } \propto R^{-5 / 2}$

The centripetal force of the planet is  

$\text { Centripetal force }=m R \omega^{2}$

Here m is the mass of the planet, ω is the frequency of rotation and R is the radius of the planet.

We also know that

$\omega=\frac{2 \pi}{T}$

Thus,

$\text { Centripetal force }=\frac{4 \pi^{2} m R}{T^{2}}$

So,

Gravitational force between planet and star=Centripetal force exerted on planet

$\text { Centripetal force } \propto R^{-5 / 2}$

$\frac{4 \pi^{2} m R}{T^{2}} \propto R^{-5 / 2}$

The mass of the planet and 4π^2 are constant so we can ignore them.

$\frac{R}{T^{2}} \propto R^{-5 / 2}$

Thus,

$\frac{R}{R^{-5 / 2}} \propto T^{2}$

$T^{2} \propto R^{1+5 / 2}$

$T^{2} \propto R^{7 / 2}$

So taking square root on both sides will give the proportional value of T, which is  

$T \propto R^{\frac{7}{2} * \frac{1}{2}}$

$T \propto R^{7 / 4}$

So the time taken to complete one rotation of massive star by the planet is proportional to $\mathrm{R}^{7 / 4}$.