Question

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

• A. R
• B. $R^0$ (independent of R)
• C. $1/R^2$
• D. $1/R$

Answer 1

Answer: (B)

$R^0$ (independent of R)

Answer 2

Centripetal force (F) $= \frac{m v^2}{ R}$ and the
gravitational force (F) $\frac{ GMm}{R^2} =\frac{ GMm}{R}$ (where
$R^2 \rightarrow R )$ . Since $\frac{ mv^2}{R} = \frac{ GMm}{R}$, therefore $v= \sqrt{GM }$.

Thus velocity v is independent of R.