MCQ
If the gravitational force between two objects were proportional to $\frac{1}{R}$ (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
    $\frac{1}{{R^2}}$
  • ${R^0}$
  • C
    ${R^1}$
  • D
    $\frac{1}{R}$

Answer

Correct option: B.
${R^0}$
b
(b)Gravitational force provides the required centripetal force for orbiting the satellite
$\frac{{m{v^2}}}{R} = \frac{K}{R}$    because $\left( {F \propto \frac{1}{R}} \right)$
 $v \propto R^\circ $

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