Question

The gravitational force between two objects is proportional to $1/R$(and not as $1/R^2$) where R is separation between them, then a particle in circular orbit under such a force would have its orbital speed v proportional to.

• A. $\frac{1}{R^2}$
• B. $R^0$
• C. $R^1$
• D. $\frac{1}{R}$

Answer 1

The correct option is B $R^0$

Given that $\frac{F}{G}\propto \frac{1}{R}$

So, $\frac{F}{G}=\frac{k}{R}$

Rotating in circular orbital, so centripetal force is

$F_{centripetal}=\frac{m{v}^2}{R}$

Gravitational force is balanced by centripetal force

So,

$=\frac{m{v}^2}{R}$

$\frac{k}{R}=\frac{m{v}^2}{R}$

$v^2=\frac{k}{m}\Rightarrow v=\sqrt{\frac{k}{m}}\Rightarrow v\propto \frac{1}{\sqrt{m}}$

So, v does not depend on $R_0$. Hence, $v\propto R^0$