The gravitational force between two objects is 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 toA. `1/(R^(2))`B. `R^(0)`C. `R^(1)`D. `1/R`

The gravitational force between two objects is proportional to `1//R`

1.	The gravitational force between two objects is 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 toA. `1/(R^(2))`B. `R^(0)`C. `R^(1)`D. `1/R`
Answer» Correct Answer - B `F_(G)=(GMm)/R` (as `F_(G)prop1/R` given) So `(mv^(2))/R=(GMm)/Rimpliesv^(2)propR^(0)`

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The gravitational force between two objects is 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 toA. `1/(R^(2))`B. `R^(0)`C. `R^(1)`D. `1/R`

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