Four particles of equal masses M move along a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle.

Question

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Answer 1

Force on M at C due to gravitational attraction.

$F_{C B} = \frac{G m ^{2}}{2 R ^{2}} \hat{i}$

$F_{C D} = \frac{G M ^{2}}{4 R ^{2}} c o s 45 \hat{j} + \frac{G M ^{2}}{4 R ^{2}} s i n 45 \hat{j}$

So resultant force on C,
$∴ F_{c} = F_{C A} + F_{C B} + F_{C D}$

$= - \frac{G M ^{2}}{4 R ^{2}} (2 + \frac{1}{2}) \hat{i} + \frac{G M ^{2}}{4 R ^{2}} (2 + \frac{1}{2})$

$∴ F_{c} \frac{G M ^{2}}{4 R ^{2}} (22 + 1)$

For moving along the circle , $F = \frac{m v ^{2}}{R}$

or $\frac{G M ^{2}}{4 R ^{2}} (22 + 1) = \frac{M V ^{2}}{R} o r V = \frac{G M}{R} (\frac{2 2 + 1}{4})$

Answer 2

Force on M at C due to gravitational attraction.

F_{C B} = \frac{G m ^{2}}{2 R ^{2}} \hat{i}

F_{C D} = \frac{G M ^{2}}{4 R ^{2}} c o s 45 \hat{j} + \frac{G M ^{2}}{4 R ^{2}} s i n 45 \hat{j}

So resultant force on C,

∴ F_{c} = F_{C A} + F_{C B} + F_{C D}

= - \frac{G M ^{2}}{4 R ^{2}} (2 + \frac{1}{2}) \hat{i} + \frac{G M ^{2}}{4 R ^{2}} (2 + \frac{1}{2})

∴ F_{c} \frac{G M ^{2}}{4 R ^{2}} (22 + 1)

For moving along the circle ,

F = \frac{m v ^{2}}{R}

or

\frac{G M ^{2}}{4 R ^{2}} (22 + 1) = \frac{M V ^{2}}{R} o r V = \frac{G M}{R} (\frac{2 2 + 1}{4})