Gravitation - Result Question 58
61. 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
(a) $1 / R^{2}$ (b) $R^{0}$ (c) $R^{1}$ (d) $1 / R$
[1989]
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Solution:
(b) $F=\frac{k}{R}=\frac{M v^{2}}{R}$. Hence $v \propto R^{0}$
