Gravity Particles in a circle

Classical Mechanics Level 3

Two particles of equal mass 'm' go around a circle of radius R, under the action of their mutual gravitational attraction.

The speed of each particle is $0.5\times\sqrt{\frac{Gm}{nR}}$

Find the value of n.

6 7 5 8 1 3 4 2

1 solution

Sravanth C.
Feb 10, 2015

We know: $\begin{aligned} F&=\frac { G{ m }^{ 2 } }{ { (2R)}^{ 2 } }\\ F &= \frac { { mv }^{ 2 } }{ R } \end{aligned}$

Since the gravitational force of attraction between the two masses is the factor which is contributing to the centripetal force, we can equate them:

$\begin{aligned} \frac { { Gm }^{ 2 } }{ { 4R }^{ 2 } }&=\frac { { mv }^{ 2 } }{ R }\\ { v }^{ 2 }&=\frac { Gm }{ 4R }\\ v&=\frac { 1 }{ 2 } \sqrt { \frac { Gm }{ R } } \end{aligned}$

Therefore $n=1$ .

Please update solution. $F = \dfrac{GM^2}{(2R)^2}$

Vishwak Srinivasan - 5 years, 10 months ago

Thanks done!

Sravanth C. - 5 years, 10 months ago

Please explain why are you equating the forces. Or else its a bit difficult to understand

Md Zuhair - 4 years, 3 months ago

Edited. Thanks for writing in! :-)

Sravanth C. - 4 years, 3 months ago

Gravity Particles in a circle

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