Gravitation is inspiration

Classical Mechanics Level 3

Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

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

Rajdeep Dhingra by Rajdeep Dhingra , 20, India

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1 solution

Rajdeep Dhingra
Sep 7, 2014

N e t f o r c e o n a n y o n e o f t h e p a r t i c l e = G M 2 ( 2 R ) 2 + G M 2 ( R 2 ) 2 c o s 45 o + G M 2 ( R 2 ) 2 c o s 45 o = G M 2 R 2 [ 1 4 + 1 2 ] T h i s f o r c e w i l l b e e q u a l t o c e n t r i p e t a l f o r c e s o M u 2 R = G M 2 R 2 [ 1 4 + 1 2 ] u = 1 2 G M R ( 2 2 + 1 ) Net\quad force\quad on\quad any\quad one\quad of\quad the\quad particle\\ =\quad \frac { G{ M }^{ 2 } }{ { (2R) }^{ 2 } } \quad +\quad \frac { G{ M }^{ 2 } }{ { (R\sqrt { 2 } ) }^{ 2 } } cos\quad { 45 }^{ o }\quad +\quad \frac { G{ M }^{ 2 } }{ { (R\sqrt { 2 } ) }^{ 2 } } cos\quad { 45 }^{ o }\quad \\ =\quad \frac { G{ M }^{ 2 } }{ { R }^{ 2 } } \left[ \frac { 1 }{ 4 } +\frac { 1 }{ \sqrt { 2 } } \right] \\ This\quad force\quad will\quad be\quad equal\quad to\quad centripetal\quad force\quad so\quad \\ \quad \quad \quad \quad \frac { M{ u }^{ 2 } }{ R } =\quad \dfrac{ G{ M }^{ 2 } }{ { R }^{ 2 } }\left[ \frac { 1 }{ 4 } +\frac { 1 }{ \sqrt { 2 } } \right] \\ \quad \quad \quad \quad u\quad =\quad \frac { 1 }{ 2 } \sqrt { \frac { GM }{ R } (2\sqrt { 2 } +1) }

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@Rajdeep Dhingra nice solution, you are really good in physics, mind telling me resources you use.

Mardokay Mosazghi - 6 years, 8 months ago

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I study mostly from a site called "coursera.org". Hope you find it resourceful.

Rajdeep Dhingra - 6 years, 5 months ago

Typo :- M u 2 R = G M 2 R 2 [ 1 4 + 1 2 ] \dfrac{M u^{2}}{R} =\dfrac{G M^{2}}{R^{2}} [ \dfrac{1}{4} + \dfrac{1}{\sqrt{2}} ] .

Venkata Karthik Bandaru - 6 years, 1 month ago

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Corrected. Thanks

Rajdeep Dhingra - 5 years, 8 months ago

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