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 inversely proportional to $r^{\frac{5}{2}}$, 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 $r^{3.5}$

$\frac{m{v}^2}{r}=\frac{GMm}{r^{\frac{5}{2}}}[\because F\propto \frac{1}{r^{\frac{5}{2}}}]$
$r{\omega }^2=\frac{GM}{r^{\frac{5}{2}}}$
$r\frac{4{\pi }^2}{T^2}=\frac{GM}{r^{\frac{5}{2}}}$
$T^2\propto r^{\frac{7}{2}}$

$T^2\propto r^{3.5}$