If the gravitational force between two objects were proportional to 1/R; where R is separation between them, then a particle in circular orbit under such a force would have its orbital speed v proportional to

The correct option is R⁰.

Let us analyze the problem step-by-step to understand how the orbital speed $v$ of a particle in a circular orbit depends on the orbital radius $R$ when the gravitational force $F$ is proportional to $\frac{1}{R}$.

First, we are given that the gravitational force $F$ between two objects is proportional to $\frac{1}{R}$. We can write this relation as:
$F=\frac{k}{R}$
where $k$ is a constant of proportionality.

For a particle of mass $m$ executing a circular orbit of radius $R$ with a constant speed $v$, the necessary centripetal force is provided by this gravitational force. The formula for the centripetal force is:
$F_c=\frac{m{v}^2}{R}$

Equating the centripetal force to the gravitational force, we get:
$\frac{m{v}^2}{R}=\frac{k}{R}$

We can simplify this equation by multiplying both sides by $R$:
$m{v}^2=k$

Solving for $v$:
$v^2=\frac{k}{m}$
$v=\sqrt{\frac{k}{m}}$

Since $k$ and $m$ are constants, the orbital speed $v$ is a constant value and does not depend on the radius $R$ at all. In mathematical terms, a quantity that is independent of $R$ is proportional to $R^0$ (since $R^0=1$). Thus:
$v\propto R^0$

Therefore, the orbital speed $v$ is proportional to R⁰.