If the gravitational force were proportional to $\frac{1}{r}$, then a particle in a circular orbit under such a force would have its original speed $v$
[$r$ is the radius of circular orbit]

The correct option is A Independent of $r$
To move in a circular orbit, the gravitational force must provide the centripetal force

$\therefore F_c=F_{grav}.....\left(1\right)$

Given, $F_{grav}\propto \frac{1}{r}$

$F_{grav}=\frac{K}{r}$; where $K$ is constant

and we know that,

$F_c=\frac{m{v}^2}{r}$

putting in eq. (1)

$\Rightarrow \frac{K}{r}=\frac{m{v}^2}{r}$

$\Rightarrow v=\sqrt{\frac{K}{m}}$

Thus, speed will become independent of $r$.

Hence, option $\left(a\right)$ is correct.