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

Answer: (D)

$\frac{m v^{2}}{r}=\frac{G M m}{r^{5 / 2}}$
$\left[\because F \propto \frac{1}{r^{5 / 2}}\right]$
or $r \omega^{2}=\frac{G M}{r^{5 / 2}}$
or $r \frac{4 \pi^{2}}{T^{2}}=\frac{G M}{r^{5 / 2}}$
$T^{2} \propto r^{5 / 2}$ or $T^{2} \propto r^{3.5}$