Test Index

Gravitation Part 1

Section: Physics

Question : 4 of 100

Marks: +1, -0

Solution:

Let us consider the gravitational force acting on each mass M by adjacent particles be F.

and the gravitational force acting on each mass M diagonally be

F_1

The net force,

F_{\text{net}}=\frac{Mv^{2}}{R}

Along the centre of circle,

\sqrt{2}F+F_{1}=\frac{Mv^{2}}{R}

\sqrt{2}\left(\frac{GMM}{(\sqrt{2}R)^{2}}\right)+\left(\frac{GMM}{(2R)^{2}}\right)

=\frac{Mv^{2}}{R}

⇒

\frac{GM}{R^{2}}\left(\frac{1}{\sqrt{2}}+\frac{1}{4}\right)=\frac{v^{2}}{R}

⇒

v=\frac{1}{2}\sqrt{\frac{GM(2\sqrt{2}+1)}{R}}

Hence, the speed of each particle is

\frac{1}{2}\sqrt{\frac{GM(2\sqrt{2}+1)}{R}}

Go to Question:

Prev Question

Next Question