Question

If the gravitational force between two objects were proportional to $\frac{1}{R}$ (and not as $\frac{1}{R^2}$) 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. $R$
• D. $\frac{1}{R}$

Answer 1

The correct option is B $R^0$

$F\propto \frac{1}{R}or,F\propto \frac{k}{R}$

Now, $F_{centrifugal}=F_c=\frac{{mv}^2}{R}\therefore \frac{{mv}^2}{R}=\frac{k}{R}orv=\sqrt{\frac{k}{m}}\therefore v\propto R^0$