Question

Imagine a light planet revolving around a very massive star in a circular orbit of radius r with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to $r^{5/2}$, then the square of the time period will be proportional to.

• A. $r^3$
• B. $r^2$
• C. $r^{2.5}$
• D. $r^{3.5}$

Answer 1

The correct option is D $r^{3.5}$

Gravitational force,$F_G\Rightarrow \frac{1}{r^{5/2}}$

revolvong in circular orbit, so, $F_G=$centripetal force

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

$mr{\omega }^2\propto r^{-5/2}$

Mass is constant, so,

${\omega }^2\propto \frac{r^{-5/2}}{r}$

${\omega }^2\propto r^{-7/2}\omega =\frac{2\pi }{T}$

${\left(\frac{2\pi }{T}\right)}^2\propto r^{-7/2}$

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