If the gravitational force between two objects were 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

Options

A. `1/(R^(2))`
B. `R^(0)`
C. `R^(1)`
D. `1/R`

Answer

B

Explanation

`F_(G)=(GMm)/R` (as `F_(G)prop1/R` given)
So `(mv^(2))/R=(GMm)/Rimpliesv^(2)propR^(0)`