1.

If the gravitational force between two objects were proportional to 1//R (and not as 1//R^(2)), where R is the distance between them, then a particle in a circular path (under such a force) would have its orbital speed v, proportional to

Answer»

REMAINS unchanged
`R_(0)` (INDEPENDENT of R)
`1//R^(2)`
`1//R`

Solution :Centripetal force(F)`=(mv^(2))/(R )` and the gravitational force (F) `=(GMm)/(R^(2))=(GMm)/(R )`(where `R^(2) to R)`.
Since `(mv^(2))/(R )=(GMm)/(R )`, therefore `V=sqrt(GM)`.
Thus velocity v is independent of R.


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