Four identical particles of mass M are located at the corners of a square of side 'a'. What should be their speed if each of them revolves under the influence of others' gravitational field in circular orbit circumscribing the square ?


 

  • Option 1)

    1.41\sqrt{\frac{GM}{a}}

  • Option 2)

    1.16\sqrt{\frac{GM}{a}}

  • Option 3)

    1.21\sqrt{\frac{GM}{a}}

     

  • Option 4)

    1.35\sqrt{\frac{GM}{a}}

 

Answers (1)

   \\2r=\sqrt{2}a\\\; \; \; \; \\r=\frac{a}{\sqrt{2}}\; \; \; \; \; \; \; (1)

\\F_{1}=\frac{Gmm}{a^{2}}(due \ \ to \ two \ adjacent \ \ masses)\\\; \; \; \\F_{2}=\frac{Gmm}{(\sqrt{2}a)^{2}} (Due \ \ to \ two \ masses \ which \ are \ dimetrically \ opposite)

\\F_{net}=\sqrt{2}\; F_{1}+F_{2}=\frac{mV^{2}}{r}

\Rightarrow \frac{\sqrt{2}\times Gm^{2}}{a^{2}}+\frac{Gm^{2}}{(\sqrt{2}\; a)^{2}}=\frac{mV^{2}}{(\frac{a}{\sqrt{2}})}

\Rightarrow \frac{Gm}{\sqrt{2}\; a}(\sqrt{2}+\frac{1}{2})=V^{2}

V^{2}=\frac{Gm(2\sqrt{2}+1)}{2\sqrt{2}\; a}

V=1.16\sqrt{\frac{Gm}{a}}


Option 1)

1.41\sqrt{\frac{GM}{a}}

Option 2)

1.16\sqrt{\frac{GM}{a}}

Option 3)

1.21\sqrt{\frac{GM}{a}}

 

Option 4)

1.35\sqrt{\frac{GM}{a}}

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