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 $r^{5/2}$, 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)

$r^{3.5}$

Let a planet of mass $m$, revolving in a circular orbit of radius $r$ around a massive star.

For equilibrium, $F_{centrifugal}=F_{gravitation}$

According to the question, $F_{gravitation}=\frac{GMm}{r^{5/2}}$

Thus, $\frac{m{v}^2}{r}=\frac{GMm}{r^{5/2}}$

$⟹v^2=\frac{GM}{r^{3/2}}$

Time period, $T=\frac{2\pi r}{v}$

$⟹T^2=\frac{4{\pi }^2{r}^2}{v^2}$

$=\frac{4{\pi }^2{r}^2}{\frac{GM}{r^{3/2}}}$

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